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*Appendix: For Online Publication

<Ref> presents proofs omitted from the main text. <Ref>
presents extensions to our main model, as discussed in <Ref>. <Ref>  presents a non-cooperative formulation of the cooperative game studied in the main text. <Ref> presents additional empirical results and descriptions.

§ PROOFS

§.§ Proof of <Ref>

Consider any outcome $O=\{(f_g^i(v),w_i^g(v))\}_{v \in [0,1], i =1,2, g=A,B}$ in which the following hold for almost all $v\in[0,1]$ and all $g\in\{A,B\}$:

* $f_g^1(v)+f_g^2(v) = f_g(v)$, and

* for all $i\in \{1,2\}$, $w_i^g(v)=v$ if $f_g^i(v)>0$.

We establish the desired result through three lemmas regarding these two enumerated conditions.

Any outcome $O$ satisfying the two enumerated conditions in the Proof of <Ref> is a core outcome.

Suppose not for the sake of contradiction. Then there are a firm $j$ and a distinct outcome (for firm $j$)  $\tilde O_j:=\{(\tilde f_g^j(v),\tilde w_j^g(v))\}_{v \in [0,1], g=A,B}$ that blocks $O.$ In order for $\tilde O_j$ to block $O$ it must be that $\pi^{\tilde O_j}_j > \pi_j^{O_j}$. However, it is also the case that

\begin{align*}
\pi^{O_j}_j&= \beta \int_0^1 [v-w_j^A(v)]f^j_A(v)dv +  \int_0^1 [v-w_i^B(v)]f^j_B(v)dv \\
& = 0\\
& \geq \beta \int_0^1 [v-\tilde w_j^A(v)]\tilde f^j_A(v)dv +  \int_0^1 [v-\tilde w_i^B(v)]\tilde f^j_B(v)dv \\
& =  \pi^{\tilde {O}_j}_j
\end{align*}

The second equality follows because, by the construction of $O$, either $f^j_g(v)=0$ or $w_j^g(v)=v$ for almost all $v$ and all $g$, therefore, the integrand is almost always equal to zero. The inequality follows because of the following exhaustive cases for almost all $v$ and all $g$, corresponding, respectively, to Conditions 1-4 of the definition of block:

* Suppose $\tilde w_j^g(v) \ge  w_j^g(v)$ and $\tilde w_j^g(v) >  w_{-j}^g(v)$, then it must be that $\tilde w_j^g(v) \ge v$ since $\max\{w_j^g(v),w_{-j}^g(v)\}=v$, which makes the integrand weakly negative,

* Suppose $\tilde w_j^g(v) \ge  w_j^g(v)$ and $\tilde f_g^j(v)+f_g^{-j}(v) \le f_g(v)$. If $\tilde f_g^j(v)=0$ then the integrand is weakly negative. If $\tilde f_g^j(v)>0$ then it must be that $f_g^{-j}(v) < f_g(v)$, and by the construction of $O$ that  $f_g^j(v)+f_g^{-j}(v) = f_g(v)$, it must be that $f_g^j(v)>0.$ Therefore, it must be that $ w_j^g(v)=v$, and the requirement that $\tilde w_j^g(v) \ge  w_j^g(v)$ makes the integrand weakly negative.

* Suppose $\tilde w_j^g(v) >  w_{-j}^g(v)$ and $\tilde f_g^j(v)+f_g^{j}(v) \le f_g(v)$. If $\tilde f_g^j(v)=0$ then the integrand is weakly negative. If $\tilde f_g^j(v)>0$ then it must be that $f_g^{j}(v) < f_g(v)$, and by the construction of $O$ that  $f_g^j(v)+f_g^{-j}(v) = f_g(v)$, it must be that $f_g^{-j}(v)>0.$ Therefore, it must be that $w_{-j}^g(v)=v$, and the requirement that $\tilde w_j^g(v) >  w_{-j}^g(v)$ makes the integrand strictly negative.

* Suppose $\tilde f_g^j(v)+f_g^j(v)+f_g^{-j}(v) \le f_g(v)$ then it must be that $\tilde f_g^j(v)=0$ since by the construction of $O$ it is the case that $f_g^j(v)+f_g^{-j}(v) = f_g(v)$. Therefore, the integrand is weakly negative.

These cases reveal a contradiction with the premise that $\pi^{\tilde O_j}_j > \pi_j^{O_j}$. Therefore, $O$ is a core outcome.

There exist a continuum of (non-equivalent) core outcomes satisfying the two enumerated conditions in the Proof of <Ref>.

For any $v^*\in[0,1]$ there is an outcome $O$ such that $f_g^1(v)=f_g(v)$ and $w_1^g(v)=v$ for all $g$ and $v\leq v^*$, and $f_g^2(v)=f_g(v)$ and $w_2^g(v)=v$ for all $g$ and $v> v^*$.

It remains to show that any core outcome must be such that all workers are hired and all workers receive a wage equal to their productivity.

There exist no core outcomes which do not satisfy the two enumerated conditions in the Proof of <Ref>.

Suppose for contradiction that there is a core outcome $O=\{(f_g^i(v),w_i^g(v))\}_{v \in [0,1], i =1,2, g=A,B}$ such that there exists a set with positive measure $V$ where the following fails for some $g\in\{A,B\}$ and all $v\in V$:

* $f_g^1(v)+f_g^2(v) = f_g(v)$, and

* for all $i\in \{1,2\}$, $w_i^g(v)=v$ if $f_g^i(v)>0$.

Throughout, we assume that the above conditions are violated for $g=B$, and we construct a blocking outcome $\tilde O_j=\{(\tilde f_g^j(v),\tilde w_j^g(v))\}_{v \in [0,1],g=A,B}$, where $(\tilde f_A^j(v),\tilde w_j^A(v))=( f_A^j(v), w_j^A(v))$ (or $\tilde O_{-j}=\{(\tilde f_g^{-j}(v),\tilde w_{-j}^g(v))\}_{v \in [0,1],g=A,B}$ where $(\tilde f_A^{-j}(v),\tilde w_{-j}^A(v))=( f_A^{-j}(v), w_{-j}^A(v))$), i.e. we do not change the outcome for $A-$group workers for either firm. The argument in which the violations occur for $g=A$ is analogous, where terms related to the firm's profit must be multiplied by $\beta$.
We show a contradiction by considering six exhaustive cases: By countable additivity of measure, the set of productivities that fails one of the above two enumerated points  has positive measure if and only if at least one of the sets in the following six cases has a positive measure.

First, suppose there exist a firm $j$ and a subset of productivities $V \subset [0,1]$ with positive measure such that $w_j^g(v)>v$ for all $v \in V$. Then $\tilde O_j$ where for all $v$:

\begin{align*}
\tilde f_g^j(v) & := \begin{cases}
f_g^j(v) & \text{if }  v\notin V,  \\
0 & \text{if }  v\in V.
\end{cases}
~~~~~~ \tilde w_j^g(v) := \begin{cases}
w_j^g(v) & \text{if }  v\notin V,  \\
0 & \text{if }  v\in V.
\end{cases}
\end{align*}

blocks $O$ as $j$'s profit increases and Condition 4 of the definition of block is satisfied for all $v\in V$ (i.e. the workers in $V$  are fired) and for all $v\in [0,1]\setminus V$ Condition 2 of the definition of block is satisfied (i.e. there is no change in the hiring or wages of workers in $[0,1]\setminus V$). By construction, $\tilde w_j^g(v)$ satisfies our monotonicity condition.

Therefore, we proceed with the assumption that for each firm $j$, $w_j^g(v)\leq v$ for almost all $v$.
Second, suppose there exist a firm $j$ and a subset of productivities $V$ with positive measure such that  $f_g^1(v)+f_g^2(v) < f_g(v)$ and $w_j^g(v)< v$ for all $v\in V$. Then $\tilde O_j$ where for all $v$:

\begin{align*}
\tilde f_g^j(v) :=
~~~~~~~~~~~\tilde w_j^g(v) :=  \begin{cases}
0 & \text{if } \tilde f_g^j(v)=0, \\
\underset{v' \leq  v}{\sup~} w_{j}^g(v') &  \text{otherwise.}
\end{cases}
\end{align*}

blocks $O$ as firm $j$'s profit increases  as some previously unemployed workers are hired at a wage strictly less than their productivity while all existing workers at $j$ continue to be employed at the same wage as before, and Condition 2 of the definition of a block is satisfied for all $v\in[0,1]$ (i.e. no worker receives a wage cut and no workers are poached from firm $-j$). By construction, $\tilde w_j^g(v)$ satisfies our monotonicity condition.

Third,  there exists a firm $j$ and a subset of productivities $V$ with positive measure such that
$f_g^1(v)+f_g^2(v) < f_g(v)$ and $w_j^g(v)= v$ for all $v\in V$. Then, there exists $\varepsilon>0$ and $V'$ with positive measure such that $f_g^1(v)+f_g^2(v) < f_g(v)-\varepsilon$ and $w_j^g(v)= v$ for all $v \in V'$.[The proof is as follows: Suppose for contradiction that for each $\varepsilon$, any set of productivities such that $f_g^1(v)+f_g^2(v) < f_g(v)-\varepsilon$ and $w_j^g(v)= v$ has zero measure. Then, for each $n =1,2,\dots$ define the set $V_n:=\{v \in V|   f_g(v)-f_g^1(v)-f_g^2(v)>\frac{1}{n} \text{ and } w_j^g(v)= v$}. Then, by assumption,   $V_1,V_2,...$ is an increasing sequence of sets and $\cup_{n} V_n=V^*:=\{v \in V|   f_g(v)-f_g^1(v)+f_g^2(v)>0 \text{ and } w_j^g(v)= v\}$. Therefore, by countable additivity of the Lebesgue measure, we have $\mu_g(V^*)=\lim_n \mu_g(V_n)=0$, which contradicts the assumption that $V$ has positive measure and the fact that $V \subseteq V^*$.]
Now, arbitrarily fix $p<1$ such that $\bar f_g (1-p) \le p \varepsilon$ and $\frac{1}{2} p^2 \varepsilon> \bar f_g (1-p)$ (note that those inequalities are satisfied by any sufficiently large  $p<1$). By <cit.>, there exists an interval $I:=[\underline v,\bar v] \subseteq [0,1]$ such that $\mu(V' \cap I)>p \mu(I)$, where $\mu(\cdot)$ is the Lebesgue measure. Then $\tilde O_j$ where for all $v$:

\begin{align*}
\tilde f_g^j(v) & := \begin{cases}
f_g(v)-f_g^j(v)-f_g^{-j}(v) & \text{ if } v \in I,  \\
f_g^j(v) & \text{ otherwise.}
\end{cases}
~~~~~~ \tilde w_j^g(v) := \begin{cases}
0 & \text{if } \tilde f_g^j(v)=0, \\
\underset{v' < \underline v}{\sup~} w_{j}^g(v') & \text{if } v \in I \text{ and } \tilde f_g^j(v)>0,  \\
w_{j}^g(v) & \text{otherwise.}
\end{cases}
\end{align*}

blocks $O$. To see this, note Condition 4 of the definition of block is satisfied for all $v\in I$ (firm $j$ fires all existing workers in set $I$ and hires unemployed workers of the same productivity), and Condition 2 of the definition of block is satisfied for all $v\notin I$ (no worker receives a wage cut and are poached from firm $-j$). By construction, $\tilde w_j^g(v)$ satisfies our monotonicity condition.

To see that firm $j$'s profit increases, let $\delta:=\underline v-\sup_{v \le \underline v} w^g_j(v)$. Note that the firm makes an additional profit of at least
$$\delta p \mu(I) \varepsilon+ \frac{1}{2} (p\mu(I))^2 \varepsilon,$$
from hiring workers from set $V' \cap I$ while firing existing workers from $V' \cap I$ causes no loss (because those workers were hired at wages equal to their productivities), and the loss from losing workers from $I \setminus V'$ is bounded from above by $\bar f_g [(1-p) \mu(I) \times (\delta + \mu(I))]=\bar f_g (1-p) \delta \mu(I)+\bar f_g (1-p) \mu(I)^2$.  Because $p$ satisfies $\bar f_g (1-p) \le p \varepsilon$ and $\frac{1}{2} p^2 \varepsilon> \bar f_g (1-p)$ by assumption, the total change of firm $j$'s payoff is strictly positive, as desired.

The previous two cases exhaust the possibility of a core outcome in which  $f_g^1(v)+f_g^2(v) < f_g(v)$ for any $g$ and a subset of productivities with positive measure. Therefore, we proceed with the assumption that $f_g^1(v)+f_g^2(v) = f_g(v)$ for almost all $v$.

Fourth, suppose that there exist $j$ and a set $V$ of productivities with positive measure such that $w_j^g(v)< v$ and $f_g^j(v)= f_g(v)$ for all $v\in V$. Then, there exists $\varepsilon>0$ and $V'$ with positive measure such that $w_j^g(v)< v-\varepsilon$ and $f_g^j(v)= f_g(v)$ for all $v\in V'$.[The proof is analogous to the one given in Footnote <ref>.] For any $p<1$, by <cit.>, there exists an interval $I^p:=[\underline v^p,\bar v^p] \subseteq [0,1]$ such that $\mu(V' \cap I^p)>p \mu(I^p)$, where $\mu(\cdot)$ is the Lebesgue measure.
Consider the following cases.

* Suppose that there is no $V^p \subseteq [\bar v^p, 1]$ with positive measure such that $w^g_{-j}(v) < w^g_j(\bar v^p)$ and $f^{-j}_g(v)>0$ for all $v\in V^p$ and for all $p$ sufficiently close to 1.
Let $w^p:=\sup_{v < \underline v^p} w_{-j}^g(v).$
Then for a constant $\varepsilon'>0$, $\tilde O_{-j}$ where for all $v$:

\begin{align*}
\tilde f_g^{-j}(v) & := \begin{cases}
f_g(v)-f_g^j(v)-f_g^{-j}(v) & \text{ if } v \in I^p,  \\
f_g^{-j}(v) & \text{ otherwise.}
\end{cases}
\\ \tilde w_{-j}^g(v) & := \begin{cases}
0 & \text{if } \tilde f_g^{-j}(v)=0,\\
\max\{w^p,w_{j}^g(v)+\varepsilon'\} & \text{if } v \in I^p \text { and } \tilde f_g^{-j}(v)>0,  \\
\max\{\sup_{v \le \bar v^p} w_{j}^g(v)+\varepsilon',w_{-j}^g(v)\} & \text{if } v \in [\bar v^p, 1] \text { and } \tilde f_g^{-j}(v)>0,  \\
w_{-j}^g(v) & \text{otherwise.}
\end{cases}
\end{align*}

blocks $O$ for the following reasons: First, for all $v\in I^p$, Condition 4 of the definition of block is satisfied, and second, for all $v\notin I^p$, Condition 2 of the definition of block is satisfied. Note also that $\tilde w_j^g(v)$ satisfies our monotonicity condition by construction. It therefore remains only to show that $\tilde O_{-j}$ increases the profit of firm $-j$.

Firm $-j$ benefits from poaching workers in $I^p \cap V'$, which results in gains of at least

\[
\underline f_g\stackrel[\underline{v}^p]{\bar v^p-(1-p)(\bar v^p-\underline v^p)}{\int}\min\{v-w^p,\varepsilon-\varepsilon'\}dv.\]

This follows because, in the worst case, there are at most $p$ fraction of workers in $I^p\cap V'$ who are lost by firm $-j$, and these $p$ fraction of workers loaded into the right most part of $I^p$. The loss from losing existing (at most $p$ fraction of) workers in $I^p$ is upper bounded by

\begin{equation*}
\bar f_g (1-p)(\bar v^p-\underline v^p)(\bar v^p-w^p).
\end{equation*}

Let $v^p:=\max\{\min\{ w^p+\varepsilon-\varepsilon', \bar v^p-(1-p)(\bar v^p-\underline v^p)\},\underline v^p\}$. Then we can rewrite the lower bound on the gain as

\begin{equation}
\underline f_g\stackrel[\underline{v}^p]{v^p}{\int}(v-w^p)dv+ \underline f_g\stackrel[v^p]{\bar v^p-(1-p)(\bar v^p-\underline v^p)}{\int}(\varepsilon-\varepsilon')dv. \label{eq:gain}
\end{equation}

We proceed by showing that the gain exceeds the loss over the interval $I^p$ for sufficiently large $p$.  We can rewrite (<ref>) as:

\begin{equation*}
(v^p-\underline v^p)(\underline v^p-w^p)\underline f_g+\frac{1}{2}(v^p-\underline v^p)^2\underline f_g + (\varepsilon-\varepsilon')(\bar v^p-(1-p)(\bar v^p-\underline v^p)-v^p)\underline f_g,
\end{equation*}

which is no smaller than

\begin{align*}
& (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\left[(v^p-\underline v^p)(\underline v^p-w^p)+(v^p-\underline v^p)^2+(\bar v^p-(1-p)(\bar v^p-\underline v^p)-v^p)\right] \\
= &
(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\left[(v^p-\underline v^p)(v^p-w^p)+(\bar v^p-(1-p)(\bar v^p-\underline v^p)-v^p)\right]&\\
\end{align*}

and therefore, a lower bound on the gain minus the loss is proportional to

\begin{align}
\frac{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g}{(1-p)(\bar v^p-\underline v^p)}\left[(v^p-\underline v^p)(v^p-w^p)+(\bar v^p-(1-p)(\bar v^p-\underline v^p)-v^p)\right]-\bar f_g (\bar v^p-w^p)\label{eq:combined_4.1}
\end{align}

We proceed to provide a lower bound of (<ref>). We do so by considering three exhaustive alternatives.

First, suppose that $v^p=w^p+\varepsilon-\varepsilon'.$
(<ref>) is proportional to

\begin{align*}
\frac{(v^p-\underline v^p)(v^p-w^p)+\bar v^p-v^p}{(1-p)(\bar v^p-\underline v^p)}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g},
\end{align*}

\begin{align}
\frac{(v^p-\underline v^p)(v^p-w^p)+\bar v^p-v^p}{(1-p)(\bar v^p-\underline v^p)}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} &= \frac{(v^p-\underline v^p)(v^p-w^p)+\underline v ^p -v^p}{(1-p)(\bar v^p-\underline v^p)}+\frac{1}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} \notag\\
&= \frac{(v^p-\underline v^p)\left[(v^p-w^p)-1\right]}{(1-p)(\bar v^p-\underline v^p)}+\frac{1}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} \notag\\
& \geq \frac{(v^p-w^p)-1}{(1-p)}+\frac{1}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} \notag \\
&= \frac{v^p-w^p}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} \notag\\
&=\frac{\varepsilon-\varepsilon'}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g}, \notag
\end{align}

where the first inequality follows because  $1\geq \bar v^p >\underline v^p$  and $\bar v^p\geq v^p\geq \underline v^p \geq w^p\geq 0$ for all $p$ which implies that $\frac{v^p-\underline v^p}{\bar v^p-\underline v^p}\in[0,1]$ and $v^p-w^p\leq 1$, the final equality follows because $v^p=w^p+\varepsilon-\varepsilon'$.  Renormalizing this term, a lower bound on the net profit over region $I^p\cap V'$ in this case is

(1-p)(\bar v^p-\underline v^p) (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g \left [\frac{\varepsilon-\varepsilon'}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} \right ].$$

It is clear that this expression is positive for any sufficiently large $p<1$ since $\varepsilon-\varepsilon'>0$ by assumption.

Second, suppose that $v^p=\bar v^p-(1-p)(\bar v^p-\underline v^p)$. Then (<ref>) can be rewritten as

\begin{align*}
\frac{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g}{(1-p)}p(v^p-w^p)-\bar f_g (\bar v^p-w^p).
\end{align*}

Noting that each of the terms in parentheses is non-negative by construction, then we have shown the desired result if we show that, for sufficiently large $p$,

\frac{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g}{(1-p)}\frac{p(v^p-w^p)}{\bar v^p-w^p}-\bar f_g

is bounded away from zero. We can see that for all $p$

\begin{align}
\frac{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g}{(1-p)}\frac{p(v^p-w^p)}{\bar v^p-w^p}-\bar f_g&= \frac{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g}{(1-p)}\left[p-p(1-p)\frac{\bar v^p-\underline v^p}{\bar v^p-w^p}\right]-\bar f_g \notag \\
&\geq(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\left[\frac{p}{1-p}-p\right]-\bar f_g \notag \\
&= (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p^2}{1-p}-\bar f_g \notag
\end{align}

where the first equality comes from substituting in $v^p=\bar v^p-(1-p)(\bar v^p-\underline v^p),$ and the inequality follows because $\frac{\bar v^p-\underline v^p}{\bar v^p-w^p}\leq 1$ for all $p$ because $\bar v^p >\underline v^p$  and $\bar v^p\geq v^p\geq \underline v^p \geq w^p$.  Renormalizing this term, a lower bound on the net profit over region $I^p \cap V'$ in this case is

(1-p)(\bar v^p-\underline v^p) (\bar v^p-w^p) \left [ (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p^2}{1-p}-\bar f_g \right].$$

It is clear that the final line is positive for any sufficiently large $p<1$ since $\varepsilon-\varepsilon'>0$ by assumption.

Third, suppose that $v^p=\underline v^p.$ Then (<ref>) is equal to

\begin{align*}
(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p}{1-p}-\bar f_g (\bar v^p-w^p)
\end{align*}

Noting that each of the terms in parentheses is non-negative by construction and $\bar v^p>\underline v^p\geq w^p$, then
the above equation is bounded below by

\begin{align*}
(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p}{1-p}-\bar f_g.
\end{align*}

Renormalizing this term, a lower bound on the net profit over region $I^p\cap V'$ in this case is

(1-p)(\bar v^p-\underline v^p)    \left [ (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p}{1-p}-\bar f_g \right ].$$

This expression is positive for any sufficiently  large $p<1$ since $\varepsilon-\varepsilon'>0$ by assumption.

Therefore, we have shown that firm $-j$'s change in profit from workers with $v\in I^p\cap V'$ is at least

\begin{align}
\hspace{-15mm}  (1-p)(\bar v^p-\underline v^p) \times %\notag \\
\min\left \{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g \left [\frac{\varepsilon-\varepsilon'}{1-p}-1-\frac{\bar f_g (\bar v^p-w^p)}{(\varepsilon-\varepsilon')\frac{1}{2}\underline f_g} \right ], (\bar v^p-w^p) \left [ (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p^2}{1-p}-\bar f_g \right], (\varepsilon-\varepsilon')\frac{1}{2}\underline f_g\frac{p}{1-p}-\bar f_g\right \},\notag
\end{align}

and this expression is positive for every $\epsilon' \in (0,\epsilon)$ and sufficiently large $p<1$. Moreover, it can be observed by inspection that this expression is decreasing in $\epsilon'.$ Furthermore, the wage paid for workers in $[\bar v^p, 1]$ may increase at most by $\varepsilon',$ resulting in a loss of profit from the increased wage being bounded from above by $\varepsilon' \beta.$ From these observations, for any sufficiently large $p <1$ and sufficiently small $\varepsilon'>0,$ firm $-j$ strictly profits with the block, i.e. $\pi_{-j}^{\tilde O_{-j}}>\pi_{-j}^{O_{-j}}$ as desired.

* Suppose that there exists a subset of $[0,1)$ whose supremum is $1$ such that, for each $p$ in that subset, there is a set $V^p \subseteq [\bar v^p, 1]$ with positive measure such that $w^g_{-j}(v) < w^g_j(\bar v^p)$ and $f^{-j}_g(v)>0$ for all $v\in V^p$. Fix any such $p$ and $V^p$.
Consider $\tilde O_j$ where for all $v$:

\begin{align*}
\tilde f_g^j(v) & := \begin{cases}
f_g(v)  & \text{if }  v \in V^p,\\
f_g^j(v) & \text{otherwise.}
\end{cases}
~~~~~~ \tilde w_j^g(v) := \begin{cases}
0 & \text{if } \tilde f_g^j(v)=0,  \\
w_j^g(v) & \text{otherwise.}
\end{cases}
\end{align*}

$\tilde O_j$ blocks $O_j$ for the following reasons: Condition 1 of the definition of block is satisfied for all $v\in V^p$ since $w_{-j}^g(v)<w_j^g(\bar v^p)\leq \tilde w_j^g(v)$ for all $v\in V^p$ by construction, and Condition 2 of the definition of block is satisfied for all $v\notin V^p$. Moreover, firm $j$ obtains a strictly higher profit under this outcome.[In the proposed block $\tilde O_j$, one could alternatively set $\tilde w_j^g(v) := w_j^g(v) \text{ for every } v \in [0,1],
$ and the proof works without change.]

Fifth, suppose that there exist $j$ and $V \subset [0,1]$ with positive measure such that $0\le w_{-j}^g(v)\leq w_j^g(v)<v$ and $f_g^j(v)\in(0, f_g(v))$ for all $v \in V$. Then, there exists $\varepsilon>0$ and $V' \subset V$ with positive measure such that  $0\le w_{-j}^g(v)\leq w_j^g(v)<v-\varepsilon$ and $f_g^j(v)\in(0, f_g(v)-\varepsilon)$ for all $v \in V'.$  Then for a constant $\varepsilon'>0$, consider $\tilde O_{j}$ where for all $v$:

\begin{align*}
\tilde f_g^{j}(v) & := \begin{cases}
f_g(v) & \text{ if }  v \in V', \\
f_g^j(v) & \text{otherwise.}
\end{cases}
~~~~ \tilde w_{j}^g(v)  := \begin{cases}
0 & \text{if } \tilde f_g^j(v)=0,  \\
w_j^g(v)+\varepsilon' & \text{otherwise.}
\end{cases}
\end{align*}

$\tilde O_{j}$ blocks $O_{j}$ for the following reasons: Condition 1 of the definition of block is satisfied for all $v\in V'$ since $w_{-j}^g(v)\leq w_j^g(\bar v)< \tilde w_{j}^g(v)$ for all $v\in V'$ by construction, and Condition 2 of the definition of the block is satisfied for all $v\notin V'$. To see that firm $j$'s profit increases, first note that firm $j$ benefits from hiring workers from $V'$, which results in an additional profit of at least $(\varepsilon-\varepsilon')\varepsilon \mu(V').$ Meanwhile, the firm may lose from paying more for existing workers, but the associated loss is bounded from above by $\varepsilon' \beta.$ Therefore, for any sufficiently small $\varepsilon',$ firm $j$'s profit increases, as desired. Note also that  monotonicity is satisfied by $\tilde w^g_j(\cdot)$ because $w^g_j(\cdot)$ is monotone and $\varepsilon'$ is a constant.

Cases 4 and 5 exhaust the possibility of a core outcome in which there exists a set $V'$ of positive Lebesgue measure such that $\max\{w^g_1(v),w^g_2(v)\}<v$ for almost all $v\in V'$. Therefore, we proceed with the assumption that for almost any $v\in[0,1]$ there exists a firm $j$ such that $w^g_j(v)=v$.

Sixth, suppose there exist a set $V''$ of positive Lebesgue measure and a firm $j$ such that $0\leq w^g_{-j}(v)<w^g_j(v)=v$ and $f_g^{-j}(v)\in(0, f_g(v))$ for all $v\in V''.$ Intuitively, we proceed by showing that firm $j$ can fire some subset of its workers who receive wages equal to productivity, and poach workers of the same productivity from firm $-j$. We proceed by constructing a set of workers with positive measure where such a maneuver is feasible.

Following earlier arguments, there exist $\delta>0$ and a set $V'$ with $\mu(V')>0$ such that $0\leq w^g_{-j}(v)+\delta<w^g_j(v)=v$ and $f_g^{-j}(v)\in(0, f_g(v))$ for all $v\in V'.$

Let $\mathrm{cl}(V')$ be the closure of $V'$. $\mathrm{cl}(V')$ is compact because it is a closed and bounded subset of $[0,1]$. For any $v \in [0,1]$ and $\varepsilon>0$, define $B_\varepsilon(v):=(v-\varepsilon,v+\varepsilon) \cap [0,1]$ to be the $\varepsilon$-ball around $v.$
Consider a collection of sets $\{B_{\varepsilon}(v')\}_{v' \in \mathrm{cl}(V')}$ where $\varepsilon < \frac{\delta}{2}$. It is obvious that $\{B_{\varepsilon}(v')\}_{v' \in \mathrm{cl}(V')}$ covers $\mathrm{cl}(V')$ and, because $\mathrm{cl}(V')$ is compact,  there exist $v_1,v_2,\dots,v_n \in \mathrm{cl}(V')$ such that
$\{B_{\varepsilon}(v_i)\}_{i=1}^n$ covers $\mathrm{cl}(V')$, that is,

\bigcup_{i=1}^n B_{\varepsilon}(v_i) \supseteq \mathrm{cl}(V').

Therefore, it follows that

\bigcup_{i=1}^n \left [ B_{\varepsilon}(v_i) \cap V' \right ]= V'.

Because $\mu(V')>0$, this implies that

\mu\left (\bigcup_{i=1}^n \left [ B_{\varepsilon}(v_i) \cap V' \right ]\right ) >0,$$

so there exists $i \in \{1,\dots,n\}$ such that
\mu\left (B_{\varepsilon}(v_i) \cap V'\right ) >0.$

Given the conclusion of the preceding paragraph, fix $i\in\{1,\dots n\}$ such that $
\mu\left (B_{\varepsilon}(v_i) \cap V'\right ) >0.$
We will show that there exists $v'_i\in B_{\varepsilon}(v_i)\cap V'$ such that $\mu\left ([v_i-\varepsilon,v_i'] \cap V'\right ) >0.$ To see this, suppose not for contradiction. Let $\bar v:=\sup B_{\varepsilon}(v_i) \cap V'$. Take a sequence  $(v^k)_{k=1}^\infty$ such that $v^k \in B_{\varepsilon}(v_i) \cap V'$ for each $k$ and $\lim_{k \to \infty} v^k=\bar v$ (such a sequence exists by definition of $\bar v$.) By the assumption made for the purpose of contradiction, we have that $\mu\left ([v_i-\varepsilon,v^k] \cap V'\right )=0$ for each $k=1,2,\dots$. Since the sets $\left ([v_i-\varepsilon,v^k] \cap V'\right )_{k=1}^\infty$ form an increasing sequence of measurable sets,
we have $0=\mu\left ([v_i-\varepsilon,\bar v] \cap V'\right )=\mu\left ([v_i-\varepsilon,v_i+\varepsilon] \cap V'\right )=\mu\left (B_{\varepsilon}(v_i) \cap V'\right )>0$, where the inequality is assumed at the beginning of the current paragraph. This is a contradiction.

Therefore, following the preceding paragraph,  fix $v'_i\in B_{\varepsilon}(v_i)\cap V'$ with the property that $\mu\left ([v_i-\varepsilon,v_i'] \cap V'\right ) >0.$
Because $v_i' < v_i+\varepsilon$  and $\varepsilon<\frac{\delta}{2}$, we have $[v_i'-\delta, v_i'] \supseteq [v_i-\varepsilon, v_i'].$
Hence, noting that $[v_i'-\delta, v_i'] \cap V'$ and $[v_i-\varepsilon, v_i'] \cap V'$ are measurable,
$\mu([v_i'-\delta, v_i'] \cap V') \geq \mu([v_i-\varepsilon, v_i'] \cap V')>0.$

We now show that firm $j$ can block outcome $O$ via workers whose productivities fall in $[v_i'-\delta, v_i']$. To do so, we observe that $w^g_{-j}(v_i')<v_i'-\delta$ because $v_i' \in V'$. Thus, by the monotonicity of $w^g_{-j}$, $w^g_{-j}(v)<v'_i-\delta$ for all $v \in [v_i'-\delta,v_i'].$ This implies that $w^g_{-j}(v)<v$ for all $v \in [v'_i-\delta, v'_i]$. Therefore, by the ongoing assumption (following the conclusion of Cases 4 and 5) that  $\max\{w^g_1(v),w^g_2(v)\}=v$ for almost every $v \in [0,1]$, it follows that $w^g_j(v)=v$ for almost all $v \in [v_i'-\delta,v_i'].$

Consider $\tilde O_j$ where

\begin{align*}
\tilde f_g^{j}(v) & := \begin{cases}
f_g^{j}(v) & \text{ if }  v \notin [v_i'-\delta,v_i'], \\
f_g^{-j}(v) & \text{ if }  v \in [v_i'-\delta,v_i'].
\end{cases}
~~~~~~~~~~~\tilde w_j^g(v) :=  \begin{cases}
w_{j}^g(v) &  \text{ if } v \notin [v_i'-\delta,v_i'],\\
v_i'-\delta & \text{ if }  v \in [v_i'-\delta,v_i'] \text{ and } \tilde f_g^{j}(v)>0,\\
0 & \text{ otherwise.}
\end{cases}
\end{align*}

$\tilde O_j$ blocks $O_j$ for the following reasons: First, it is obvious from construction that $\tilde w_j^g$ satisfies monotonicity.  Condition 3 of the definition of block is satisfied for all $v\in [v_i'-\delta,v_i']$ (i.e. the workers previously employed by firm $-j$ are successfully poached and some workers are fired),
and Condition 2 of the definition of block is satisfied for all $v\notin [v_i'-\delta,v_i']$ (i.e. workers in this set do not experience changes to hiring or wages). It is also the case that $\tilde O_j$ provides firm $j$ with higher profit than $O_j$: newly poached workers from $[v_i'-\delta,v_i']\cap V'$
(of whom there are a positive measure) are paid lower wages than their productivity in outcome $\tilde O_j$ while all newly-fired workers are from $[v_i'-\delta,v_i']$ and received wages equal to productivity from $j$ in outcome $O_j$. Therefore, we have shown that $O$ is not a core outcome as desired.

As these six cases are exhaustive and none of them admits a core outcome, we have completed the argument that any core outcome must be such that all workers are hired and all workers receive a wage equal to their productivity.

§.§ Proof of <Ref>

First, we show that if firm 1 does not hire any $B-$group workers, then it hires all $A-$group workers and firm 2 hires all $B-$group workers (the usual disclaimer about zero-measure sets applies).  That is, in any core outcome $O$, if $f_B^1(v)=0$ for almost all $v$, then it must be that $f_A^1(v)=f_A(v)$ for almost all $v$ and $f_B^2(v)=f_B(v)$ for almost all $v$. We show this in several exhaustive cases. Throughout, it suffices to assume $w_i^g(v)\leq v$ for all $v\in[0,1]$, all $i\in\{1,2\}$, and all $g\in\{A,B\}$ by the Individual Rationality Condition.

* We will show that if $f_B^1(v)=0$ for almost all $v$, then it must be that there exists a set $V\subset[0,1]$ of positive Lebesgue measure such that  $f_A^1(v)>0$ for all $v\in V$. Suppose not toward a contradiction so that  $f_B^1(v)=0$ for almost all $v$ and $f_A^1(v)=0$ for almost all $v$. Then,  $\pi_1^O=0$. By the Equal Profit Condition it must be that $\pi^O_2=0$.
There are two possibilities to consider. First, it may be that there exist a group $g\in\{A,B\}$ and a set $V$ with positive Lebesgue measure such that $f_g^2(v)<f_g(v)$ for all $v\in V$. Without loss of generality let $g=A$. Then firm 1 can block outcome $O$ via $\tilde O_1$ where

\begin{align*}
\tilde f_A^1(v) & :=
f_A(v)-f_A^2(v)-f_A^{1}(v) \text{ for all } v ~~~~~~~  \tilde w_1^A(v)  := 0 \text{ for all } v,
\\ \tilde f_B^1(v) & := 0 \text{ for all } v ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\tilde w_1^B(v)  := 0 \text{ for all } v.
\end{align*}

which yields a positive profit, contradiction. Second, it may be that $f_A^2(v)=f_A(v)$ and $f_B^2(v)=f_B(v)$ for almost all $v$. By group-based EPSW, it must be that firm 2 pays a common wage to all workers: there exists $w_2\geq 0$ such that $w_2=w_2^A(v)=w_2^B(v)$ for almost all $v$. Moreover, by the ongoing assumption that $w_i^g(v)\leq v$ for all $v\in[0,1]$, all $i\in\{1,2\}$, and all $g\in\{A,B\}$, it must be the case that $w_2=0$. But then for any $w^*\in(0,1)$ firm 1 can block $O$ via $\tilde O_1$ where for each $g\in\{A,B\}$:

\begin{align*}
\tilde f_g^1(v) & := \begin{cases}
f_g(v)-f_g^1(v) & \text{ if } v\geq w^*,  \\
0 & \text{ otherwise.}
\end{cases}~~~~ \tilde w_1^g(v)  := \begin{cases}
0 & \text{if } \tilde f_g^1(v)=0, \\
w^* &  \text{otherwise.}
\end{cases}
\end{align*}

which yields positive profit, contradiction.

* We will prove that if $f_B^1(v)=0$ for almost all $v$ and there exists a positive Lebesgue measure set $V$ such that  $f_A^1(v)>0$ for all $v\in V$, then there does not exist any pair of sets $V_A$ and $V_B$ with positive Lebesgue measure such that $f_A^2(v)>0$ for all $v\in V_A$ and $f_B^2(v)>0$ for all $v\in V_B$. To show this, assume for contradiction that there exist  sets $V_A$ and $V_B$ with positive Lebesgue measure such that $f_A^2(v)>0$ for all $v\in V_A$ and $f_B^2(v)>0$ for all $v\in V_B$.  Then due to the group-based EPSW, it must be that firm 2 pays a common wage $w_2$ to almost all workers it hires, i.e. $w^A_2(v)=w^B_2(v)=w_2$ for all $v$ such that $f^2_A(v)>0$ and $f^2_B(v)>0,$ respectively. By our ongoing assumption that  $w_2^g(v)\leq v$ for all $v\in[0,1]$ and all $g\in\{A,B\}$, it must be the case that $f^2_A(v)=0$ for all $v\leq w_2$. This implies that firm 2 is earning a positive profit from the $A-$group workers it hires as $w_2<v$ for almost all $v\in V_A$. By the proof of Case 4.2 of <Ref>, firm 1 can block $O$ by “poaching” some subset of these workers and increase its profits.

* Next, we will establish that if $f_B^1(v)=0$ for almost all $v$ and there exists a  set $V$ with positive Lebesgue measure such that $f_A^1(v)>0$ for all $v\in V$ then $f_A^2(v)=0$ for almost all $v$. To show this, by the previous case, it suffices to consider for contradiction that $f_B^2(v)=0$ for almost all $v$ and that there exists a positive Lebesgue measure set $V_A$ such that $f_A^2(v)>0$ for all $v\in V_A$. The conclusion of <Ref> applies, and the two firms “Bertrand” compete away profits from $A-$group workers such that they both earn zero profits, i.e. $w_1^A(v)=v$ for all $v$ such that $f^1_A(v)>0$ and $w_2^A(v)=v$ for all $v$ such that $f^2_A(v)>0$. But then an arbitrary firm $j$ can block via outcome $\tilde O_j$:

\begin{align*}\tilde f_A^j(v) & := 0 \text{ for all } v ~~~~~~~~~~~~~~~~~~~~~~~~~~~\tilde w_j^A(v)  := 0  \text{ for all } v\\
\tilde f_B^j(v) & := f_B(v)-f_B^{-j}(v) \text{ for all } v ~~~~~~~~~\tilde w_j^B(v)  := 0  \text{ for all } v
\end{align*}

The ongoing assumption that $f_B^{1}(v)=0$ for almost all $v$ and $f_B^{2}(v)=0$ for almost all $v$ demonstrates that $\tilde O_j$ indeed blocks $O$.

* We will show that if $f_B^1(v)=0$ for almost all $v$, there exists a  set $V$ with positive Lebesgue measure such that $f_A^1(v)>0$ for all $v\in V$, and $f_A^2(v)=0$ for almost all $v$ then it must be that $f_A^1(v)=f_A(v)$ for almost all $v$. This claim is shown via the second and third cases in Claim 3 of the proof of <Ref>.

Therefore, the only remaining possible core outcomes in which firms do not completely segregate involves each firm hiring both groups of workers, i.e. there exist  sets $V^1_A, V^2_A, V^1_B, V^2_B$ with positive Lebesgue measure such that $f^1_A(v)>0$ for all $v\in V^1_A,$ $f^2_A(v)>0$ for all $v\in V^2_A,$ $f^1_B(v)>0$ for all $v\in V^1_B,$ and $f^2_B(v)>0$ for all $v\in V^2_B$.

Suppose for contradiction that there exists such a non-segregation core outcome $O$. We have argued that in any such core outcome, each firm $i\in\{1,2\}$ must pay a common wage $w_i$ to almost every worker it hires. Without loss of generality, let $w_1\leq w_2$. We first claim that, if $w_1< w_2$:

* For almost all $v<w_1$, all workers of  productivity $v$ are unemployed: for almost all $v<w_1$, $f^1_A(v)=f^1_B(v)=f^2_A(v)=f^2_B(v)=0$,

* for almost all $v\in (w_2,1]$, all workers of productivity $v$ are hired by firm 2, i.e., $f^2_A(v)=f_A(v)$ and
$f^2_B(v)=f_B(v)$, and

* for almost all $v\in [w_1,w_2]$, all workers of productivity $v$ are hired by firm 1, i.e., $f^1_A(v)=f_A(v)$ and $f^1_B(v)=f_B(v)$.

Point 1 follows from our previous argument that no firm hires a positive measure of workers at wage higher than productivity in any core outcome. Point 2 is demonstrated with the following argument. As argued in the previous paragraph, firm 2 will hire almost no workers with productivity $v\in[w_1,w_2]$. If there exists a  set $V\subset[w_2,1]$ with  positive measure such that  $f^2_A(v)<f_A(v)$ or $f^2_B(v)<f_B(v)$ for all $v\in V$, then firm 2 can block outcome $O$ via $\tilde O_2$ such that for all $g\in\{A,B\}$:

\begin{align*}
\tilde f_g^2(v) & := \begin{cases}
f_g(v) & \text{ if } v\in (w_2,1],  \\
0 & \text{ otherwise.}
\end{cases}~~~~~ \tilde w_2^g(v) := \begin{cases}
0 & \text{if } \tilde f_g^2(v)=0, \\
w_2 &  \text{otherwise.}
\end{cases}
\end{align*}

which clearly blocks $O$ as firm 2's profit increases (from hiring additional workers at a positive marginal profit) and Condition 1 of the definition of block is satisfied for all $v \in (w_2,1]$. Point 3 follows from a similar argument as the proof of point 2.

We now claim that in any core outcome $O$  it must be that $w_1<w_2$ (recall that we have  assumed $w_1\leq w_2$). To see this, suppose for contradiction that $w_1=w_2$. If $w_1=w_2<1$,  then at least one firm $i\in\{1,2\}$ receives total profit $\pi^O_i<\int_{w_1}^{1}(v-w_1)[\beta f_A(v)+f_B(v)]dv$. But because $F$ is atomless, for sufficiently small $\epsilon>0$, firm $i$ would receive $\int_{w_1+\epsilon}^{1}(v-w_1-\epsilon)[\beta f_A(v)+f_B(v)]dv>\pi^O_i$ by instead setting wage $w'=w_1+\epsilon$ and hiring all workers with productivity strictly greater than $w_1+\epsilon$. This is a contradiction to the assumption that $O$ is a core outcome. If $w_1=w_2=1$, then $\pi^O_1=\pi^O_2=0$ and so either firm could block $O$ by  setting any wage $w'\in(0,1)$ and hiring all workers with productivities $v\in[w',1)$, which yields profit $\int_{w'}^{1}(v-w')[\beta f_A(v)+f_B(v)]dv>0$. This contradicts the assumption that that $O$ is a core outcome. As these two cases are exhaustive, it must be the case that $w_1<w_2$ in any core outcome in which $w_1\leq w_2$.

We now show that generically, any outcome $O$ in which  $f_A^1(v)=f_B^1(v)=f_A^2(v)=f_B^2(v)=0$ for almost all $v<w_1$, $f_A^1(v)=f_A(v)$ and $f_B^1(v)=f_B(v)$ for almost all $v\in[w_1,w_2]$, and $f_A^2(v)=f_A(v)$ and $f_B^2(v)=f_B(v)$ for almost all $v>w_2$ is not a core outcome.  We consider two exhaustive cases.

* $0<w_1<w_2$. Let $\pi_{i,g}$
represent the profit  firm $i\in\{1,2\}$ receives from all workers in group $g\in\{A,B\}$ in outcome $O$, or more formally,

\pi_{i,A}:=\beta \int_{0}^{1}(v-w_i(v))f_A^i(v)dv, ~~~~~~~
\pi_{i,B}:= \int_{0}^{1}(v-w_i(v))f_B^i(v)dv.

It must be the case that $\pi_{1,A}+\pi_{2,A}\geq\pi_{1,B}+\pi_{2,B}$, or vice versa. Without loss of generality, assume $\pi_{1,A}+\pi_{2,A}\geq\pi_{1,B}+\pi_{2,B}$. We claim that firm 1 can block outcome $O$ via outcome $\tilde O_1$ where for some small $\epsilon'>0$:

\begin{align*}
\tilde f_A^1(v) & :=
f_A(v) ~~~~~ \text{ for all } v\in [0,1]
~~~~~~\tilde w_1^A(v)  := \begin{cases}
0 & \text{if } v<w_1, \\
w_1 &  \text{if } v\in[w_1,w_2], \\
w_2+\epsilon' &  \text{if } v>w_2.
\end{cases}\\
\tilde f_B^1(v) & :=
0 ~~~~~~~~~~~ \text{ for all } v\in [0,1]~~~
~~~ \tilde w^B_1(v)  :=
0 ~~~~~~~~~~~ \text{ for all } v\in [0,1]
\end{align*}

For sufficiently small $\epsilon'$,  $\pi^{\tilde O_1}_{1}>\pi_{1,A}+\pi_{2,A}\geq \pi_1$, where the last inequality comes from the assumption that $\pi_{1,A}+\pi_{2,A}\geq\pi_{1,B}+\pi_{2,B}$ and the Equal Profit Condition $\pi_1=\pi_2=\frac{1}{2}(\pi_{1,A}+\pi_{2,A}+\pi_{1,B}+\pi_{2,B})$. For almost all $A-$group workers, Condition 1 of the definition of block is satisfied, and for almost all $B-$group workers, Condition 4 of the definition of block is satisfied. Therefore, $\tilde O_1$ blocks $O$.

* $0=w_1<w_2$. The argument is analogous to case 1 if $\pi_{1,A}+\pi_{2,A}>\pi_{1,B}+\pi_{2,B}$ or $\pi_{1,A}+\pi_{2,A}<\pi_{1,B}+\pi_{2,B}$.
Below, we argue that the condition $\pi_{1,A}+\pi_{2,A}=\pi_{1,B}+\pi_{2,B}$ is non-generic on the space of distributions:

First, we will show the openness of the set of distributions $F_A,F_B$ such that $\pi_{1,A}+\pi_{2,A}>\pi_{1,B}+\pi_{2,B}$ or $\pi_{1,A}+\pi_{2,A}<\pi_{1,B}+\pi_{2,B}$. To do so, assume that the former inequality holds at $F_A,F_B$ (the other case is analogous), where the core outcome wages of firms 1 and 2 are $0$ and $w_2$, respectively. Let $\epsilon>0$ be small enough that
\begin{align}
\pi_{1,A}+\pi_{2,A}>\pi_{1,B}+\pi_{2,B}+\epsilon.\label{equation:epsilon-inequality}
\end{align}

By the Portmanteu Theorem,[See Theorem 2.8.1 (c) of [Ash and Doléans-Dade, 2000].
] for any $\tilde w_2 \in [0,1]$ it follows that

\begin{align}
\beta \int_0^{\tilde w_2} v f'_A(v)dv \to \beta \int_0^{\tilde w_2} v f_A(v)dv, \notag
\end{align}

as $(F'_A,F'_B)$ with densities $f'_A,f_B'$ converges in weak$^*$ topology to $(F_A,F_B).$
Thus, there is a neighborhood of $(F_A,F_B)$ such that, for any $(F'_A,F'_B)$ in that neighborhood, we have

\begin{align}
\beta \int^{\tilde w_2}_0 v f'_A(v)dv \in \left ( \beta \int^{\tilde w_2}_0 v f_A(v)dv-\frac{\epsilon}{8},  \beta \int^{\tilde w_2}_0 v f_A(v)dv+\frac{\epsilon}{8}\right ).\label{inequality2}
\end{align}

Next, again by Portmanteu Theorem, it follows that
for any $\delta< w_2$ (where  $w_2$ is the wage paid by firm 2 when the distributions of worker productivities are given by $(F_A,F_B)$),

\beta \int_0^{w_2-\delta} v f'_A(v)dv+\int_0^{w_2-\delta} v f'_B(v)dv \to \beta \int_0^{w_2-\delta} v f_A(v)dv+\int_0^{w_2-\delta} v f_B(v)dv

\hspace{-23mm}\beta \int_{w_2-\delta}^1 (v-(w_2-\delta)) f'_A(v)dv+\int_{w_2-\delta}^1 (v-(w_2-\delta)) f'_B(v)dv \to \beta \int_{w_2-\delta}^1 (v-(w_2-\delta)) f_A(v)dv+\int_{w_2-\delta}^1 (v-(w_2-\delta)) f_B(v)dv,

as $(F'_A,F'_B)$  converges in weak$^*$ topology to $(F_A,F_B)$.
Because we know that

\beta \int_0^{w_2-\delta} v f_A(v)dv+\int_0^{w_2-\delta} v f_B(v)dv
<\beta \int_{w_2-\delta}^1 (v-(w_2-\delta)) f_A(v)dv+\int_{w_2-\delta}^1 (v-(w_2-\delta)) f_B(v)dv$$

from the construction of outcome $O$ at $(F_A,F_B)$, it follows that
there exists a neighborhood of $(F_A,F_B)$ such that, for any pair of distributions $(F'_A,F_B')$  in that neighborhood, we have

\beta \int_0^{w_2-\delta} v f'_A(v)dv+\int_0^{w_2-\delta} v f'_B(v)dv <\beta \int_{w_2-\delta}^1 (v-(w_2-\delta)) f'_A(v)dv+\int_{w_2-\delta}^1 (v-(w_2-\delta)) f'_B(v)dv,

and, with an analogous argument, that

\beta \int_0^{w_2+\delta} v f'_A(v)dv+\int_0^{w_2+\delta} v f'_B(v)dv >\beta \int_{w_2+\delta}^1 (v-(w_2+\delta)) f'_A(v)dv+\int_{w_2+\delta}^1 (v-(w_2+\delta)) f'_B(v)dv.

These inequalities imply that if firm 2 pays $w'_2$ to all workers it employs and firm 1 pays $w'1=0$ to all workers it employs, both given distributions $(F'_A,F'_B)$, such that the Equal Profit Condition is satisfied, then $w'_2 \in (w_2-\delta,w_2+\delta)$.
Moreover, by monotonicity of the expression $
\beta \int_0^{w_2'} v f'_A(v)dv$ in $w_2'$, it follows that
\beta \int_0^{w_2'} v f'_A(v)dv  \in \left ( \beta \int_0^{w_2-\delta} v f'_A(v)dv, \beta \int_0^{w_2+\delta} v f'_A(v)dv \right )$ for any distribution $(F'_A,F'_B)$ in the  neighborhood. Take a sufficiently small $\delta>0$ such that

\begin{align}
\beta \int_0^{w_2-\delta} v f_A(v)dv>\beta \int_0^{w_2} v f_A(v)dv-\frac{\epsilon}{8} \text{ and } \beta \int_0^{w_2+\delta} v f_A(v)dv <\beta \int_0^{w_2} v f_A(v)dv+\frac{\epsilon}{8}.\label{lemmna-inequality}
\end{align}

Moreover, we apply (<ref>) to $\tilde w_2=w_2-\delta$ and $\tilde w_2=w_2+\delta$ to obtain

\begin{align}
\beta \int^{w_2-\delta}_0 v f'_A(v)dv \in \left ( \beta \int^{w_2-\delta}_0 v f_A(v)dv-\frac{\epsilon}{8},  \beta \int^{w_2-\delta}_0 v f_A(v)dv+\frac{\epsilon}{8}\right ),\notag \\
\beta \int^{w_2+\delta}_0 v f'_A(v)dv \in \left ( \beta \int^{w_2+\delta}_0 v f_A(v)dv-\frac{\epsilon}{8},  \beta \int^{w_2+\delta}_0 v f_A(v)dv+\frac{\epsilon}{8}\right ) \label{inequality3}. \end{align}

Thus, by (<ref>) and (<ref>) it follows that $\pi'_{1,A} \in (\pi_{1,A}-\frac{\epsilon}{4},\pi_{1,A}+\frac{\epsilon}{4})$.
Similarly we can establish that $\pi'_{i,g}$ for $i \in \{1,2\},g \in \{A,B\}$, each satisfies $\pi'_{i,g} \in (\pi_{i,g}-\frac{\epsilon}{4},\pi_{i,g}+\frac{\epsilon}{4})$.
This and (<ref>) imply that

completing the proof of openness.

Second, we will show the denseness of the set of distributions $F_A,F_B$ such that $\pi_{1,A}+\pi_{2,A}>\pi_{1,B}+\pi_{2,B}$ or $\pi_{1,A}+\pi_{2,A}<\pi_{1,B}+\pi_{2,B}$. To do so, suppose for a given $F_A,~F_B$ satisfying our regularity conditions there is a core outcome that does not feature segregation, i.e. $\pi_{1,A}+\pi_{2,A}=\pi_{1,B}+\pi_{2,B}.$ We show that for any $\delta_1>0$ there exist distributions $F'_A,~F'_B$ satisfying our regularity conditions such that $\vert F'_A(v)-F_A(v)\vert \leq \delta_1$ and $\vert F'_B(v)-F_B(v)\vert \leq \delta_1$ for all $v\in[0,1]$ and for which there does not exist a non-segregation core outcome.[Note that our pointwise notion of “closeness” corresponds to that of the weak-* topology on the set of distributions satisfying our regularity conditions.] We show this by modifying the distributions such that the same wages by firms 1 and 2, $0$ and $w_2$, respectively, yield equal profit, but show that aggregate profit derived from $A-$group workers no longer equals that derived from $B-$group workers.

Take any $\delta\in(0,\min\{\underline f_A,\underline f_B,\delta_1\})$ and any $w^*\in(0,w_2)$ such that $w_2-w^*<1-w_2$. Define $F'_A$ and $F'_B$ via their respective densities:

\begin{align*}
f'_A(v) & := \begin{cases}
f_A(v) & \text{ if }  v <w^*,\\
f_A(v)-\frac{\delta}{\beta} & \text{ if }  v \in[w^*,w_2),\\
f_A(v)+\frac{\delta}{\beta} & \text{ if }  v \in[w_2,w_2+(w_2-w^*)),\\
f_A(v) & \text{ otherwise}
\end{cases},
\end{align*}

\begin{align*}
f'_B(v) & := \begin{cases}
f_B(v) & \text{ if }  v <w^*,\\
f_B(v)+\delta & \text{ if }  v \in[w^*,w_2),\\
f_B(v)-\delta & \text{ if }  v \in[w_2,w_2+(w_2-w^*)),\\
f_B(v) & \text{ otherwise}
\end{cases}.
\end{align*}

Noting there exists a $\beta$ measure of $A-$group workers, by construction of $F'_A$ and $F'_B$ it is the case that

\begin{align*}
\pi_{2,A}+\pi_{2,B}&= \beta \int_{w_2}^1 [v-w_2]f_A(v)dv+\int_{w_2}^1 [v-w_2]f_B(v)dv \notag \\
&= \beta \int_{w_2}^1 [v-w_2]f'_A(v)dv+\int_{w_2}^1 [v-w_2]f'_B(v)dv\\
&= \pi'_{2,A}+\pi'_{2,B} \notag
\end{align*}

\begin{align*}
\pi_{1,A}+\pi_{1,B}&= \beta \int_{0}^{w_2} vf_A(v)dv+\int_{0}^{w_2} vf_B(v)dv \notag \\
&= \beta \int_{0}^{w_2} vf'_A(v)dv+\int_{0}^{w_2} vf'_B(v)dv\\
&= \pi'_{1,A}+\pi'_{1,B} \notag
\end{align*}

where $\pi'_{i,g}$ represents firm $i$'s profit from $g-$group workers it hires given distribution $F'_g$. The above two equations imply that

\begin{equation*}
\pi'_i:=\pi'_{i,A}+\pi'_{i,B}=\pi^{O_i}_i \textrm{ for } i\in\{1,2\} %\label{same profit denseness}
\end{equation*}

It is also the case that

\begin{align*}
\pi_{1,A}+\pi_{2,A}-(\pi'_{1,A}+\pi'_{2,A})&= \beta \int_{w_2}^{w_2+(w_2-w^*)} [v-w_2]\frac{\delta}{\beta}dv+\beta\int_{w^*}^{w_2} v\frac{\delta}{\beta}dv \notag \\
&= \delta(w_2-w^*)w_2 \notag\\
\end{align*}

where the final inequality comes from the assumptions that $w_2>w^*>0$. Similarly it can be shown that

The above two equations imply that

The initial condition that $\pi_{1,A}+\pi_{2,A}=\pi_{1,B}+\pi_{2,B}$ further implies that

which completes the claim.

§.§ Proof of <Ref>

Proof of part <ref>. We first constructively show that there exist a continuum of core outcomes $O(\delta,\delta')$ with group-based segregation, parameterized by the following class of wage functions for $\delta, \delta' \in [0,1]$:

\begin{align*}
w_1(v) := \begin{cases}
v & \text{if } v<\delta, \\
\delta &  \text{otherwise.}
\end{cases}~~~~~  w_2(v) := \begin{cases}
0 & \text{if }  v<\delta', \\
v &  \text{otherwise.}
\end{cases}
\end{align*}

with firms 1 and 2 hiring all $A-$ and $B-$group workers, respectively.
We show that these wage functions permit a core outcome if $\delta \ge \delta'$ and

\begin{align}
\beta \stackrel[\delta]{1}{\int}\left(v-\delta \right) f_A(v)dv=\stackrel[0]{\delta'}{\int} v f_B(v)dv. \label{equal-profit-condition}
\end{align}

Note that the left-hand side of (<ref>) is $\pi^{O(\delta,\delta')}_1$, while the right-hand side $\pi^{O(\delta,\delta')}_2$.

To see that the above wage functions permit a core outcome $O(\delta,\delta')$, first note that there is no firm $i$ and outcome $\tilde O_i$ with group-based segregation that blocks $O(\delta,\delta')$. Suppose without loss of generality that in outcome $\tilde{O}_i$, firm $i$ hires a positive measure of $A-$group workers only with associated wage function $\tilde w_i(\cdot)$. By the definition of block, for almost every $v$, $\tilde f^i_A(v)>0$ only if $\tilde w_i(v)\geq w_1(v)$, implying that $\pi^{\tilde O_i}_i\leq \pi^{O(\delta,\delta')}_1=\pi^{O(\delta,\delta')}_2$. Therefore, $\tilde {O}_i$ is not a block.

Thus, consider a firm $i$ and a potential blocking outcome $\tilde O_i$ in which $i$ employs positive measures of both $A-$ and $B-$group workers. Under group-based EPSW, such a  firm pays a common wage $w$ to almost all workers. There are two exhaustive cases:

* Consider $w \le \delta$. By the definition of block, firm $i$ does not benefit from hiring $A-$group workers, as almost all $A-$group workers with $v\geq w$ are paid $w_1(v)=\min\{v,\delta\} \ge w$ in outcome $O(\delta,\delta').$ Therefore, $\pi^{\tilde O_i}_i$ is upper bounded by the case in which firm $i$ hires only $B-$group workers. It has been shown in the preceding paragraph, however, that there exists no such block.

* Consider $w > \delta$. By the definition of block, firm $i$ does not benefit from hiring $B-$group workers, as almost all $B-$group workers with $v\geq w$ are paid $w_2(v)=v$ in outcome $O(\delta,\delta').$ Therefore, $\pi^{\tilde O_i}_i$ is upper bounded by the case in which firm $i$ hires only $A-$group workers. It has been shown in the preceding paragraph, however, that there exists no such block.

The preceding arguments demonstrate that any outcome $O(\delta,\delta')$ with $\delta\geq \delta'$ and $\pi^{O(\delta,\delta')}_1=\pi^{O(\delta,\delta')}_2$ is a core outcome. Next, we show that there is a continuum of pairs $\delta,\delta'$ that satisfy these conditions. To see this, note that if $\delta=\delta'=1$, then the right-hand side of (<ref>) is strictly positive while the left-hand side is zero, and hence the former is strictly larger than the latter. Because of continuity of the left-hand side in $\delta$, the left-hand side is strictly smaller than the right-hand side for any $\delta$ that is sufficiently close to $1$. Now, noting continuity of the right-hand side in $\delta'$ and the fact that it is equal to zero for $\delta'=0$, by intermediate value theorem, it follows that there exists $\delta'$ such that  (<ref>) holds with equality. This concludes the proof of part 1.

Proof of part <ref>.  Recall that $AW^O_A$ and $AW^O_B$ denote the average wages for  $A-$ and $B-$group workers, respectively.
In addition, denote $TS_A$ and $TS_B$ to be the total surpluses created by $A-$ and $B-$group workers (together with the firms hiring them), respectively, i.e.,

\begin{align*}
TS_A & := \beta \stackrel[0]{1}{\int} v f_A(v)dv =\beta \mathbb{E}_A(v), \\
TS_B & :=\stackrel[0]{1}{\int} v f_B(v)dv=\mathbb{E}_B(v).
\end{align*}

Then, note that $\pi^O_1=TS_A - \beta AW^O_A$ and $\pi^O_2=TS_B - AW^O_B$. The Equal Profit Condition tells us that

\begin{align}
& TS_A - \beta AW^O_A  = TS_B - AW^O_B  \notag \\
\Leftrightarrow &
TS_A - TS_B=\beta AW^O_A - AW^O_B \notag \\
\Leftrightarrow &
\label{eq:gap_widens}
TS_A - TS_B + (1-\beta)AW^O_A= AW^O_A - AW^O_B.
\end{align}

Proof of the “only if” part.
Suppose that $AW^O_A - AW^O_B \le AW^{O'}_A - AW^{O'}_B$ for core outcomes $O$ and $O'$. Then (<ref>) implies that $TS_A - TS_B+ (1-\beta)AW^O_A \le TS_A - TS_B+ (1-\beta)AW^{O'}_A$. Note that $TS_A - TS_B$ is a constant. Since $\beta>1$, this implies that $AW^O_A \ge AW^{O'}_A$. Since firm 1 hires almost all $A-$group workers in both $O$ and $O'$, this implies that $\pi^O_1 \le \pi^{O'}_1$. Then  $\pi^O_2 = \pi^O_1 \le \pi^{O'}_1 = \pi^{O'}_2$, where the equalities follow from the Equal Profit Condition.

Proof of the “if” part.
Suppose $\pi^{O}_1 \le \pi^{O'}_1$ for two core outcomes $O$ and $O'$ (which implies $\pi^O_2 = \pi^O_1 \le \pi^{O'}_1 = \pi^{O'}_2$ by the Equal Profit Condition). Since firm 1 hires almost all $A-$group workers in both $O$ and $O'$, this implies that $AW^{O}_A \ge AW^{O'}_A$. Since $\beta>1$, this
implies that $TS_A - TS_B+ (1-\beta)AW^O_A \le TS_A - TS_B+ (1-\beta)AW^{O'}_A$. Thus, by (<ref>),  $ AW^{O}_A - AW^{O}_B \le AW^{O'}_A - AW^{O'}_B.$

Proof of Part <ref>. Consider the core outcome in which $w_1(v)=w_2(v)=v$ for all $v \in [0,1]$ and all $A-$ and $B-$group workers are hired by firms 1 and 2, respectively. The wage gap in this core outcome is the same as in any core outcome without EPSW, and both firms' profits are zero. Now, it is straightforward to see that in any non-equivalent core outcome firm profits are strictly positive. Therefore, applying the conclusion of part <ref> completes the claim.

§.§ Proof of <Ref>

Following Proposition <ref>, without loss of generality, we focus on core outcomes such that $f^1_A(v)=f_A(v)$ and $f^2_B(v)=f_B(v)$ for all $v$. We show that there exists $\beta^*\in[1,\infty)$ such that for all $\beta>\beta^*$, any core outcome $O$ must satisfy $w_1(v)>v-\delta$ for all $v$.

To show this, suppose not for the sake of contradiction. Then for any $\beta^*$ there exists $\beta>\beta^*$, a core outcome $O$, and some $v_\beta$ such that
$w_1(v_\beta)\leq v_\beta-\delta.$  Noting that
$w_1(v)$ is non-decreasing in $v$,
it must be the case that

\begin{equation}\label{monotonicty_and_45_degree}
w_1(v)\leq w_1(v_\beta) \le v_\beta-\delta \mbox{ for  all } v<v_\beta.
\end{equation}

Therefore, we can bound the profit of firm 1 from below as follows:

\begin{align}\label{triangle_of_profit}
\pi^O_1&= \beta\stackrel[0]{1}{\int}\left(v-w_1(v) \right)f_A(v) dv \ge \beta\stackrel[v_\beta-\delta]{v_\beta}{\int}\left(v-w_1(v) \right)f_A(v) dv \nonumber \\
&\ge \beta\stackrel[v_\beta-\delta]{v_\beta}{\int}\left(v-(v_\beta-\delta) \right) f_A(v) dv \ge \beta\stackrel[v_\beta-\delta]{v_\beta}{\int}\left(v-(v_\beta-\delta) \right)\underline f_A dv \\
&=\frac{\beta}{2}\delta^2\underline f_A \nonumber,
\end{align}

where the first inequality comes from the fact that $w_1(v)\leq v$ for all $v$, the second inequality comes from (<ref>), and the third inequality comes from the fact that $f_A(v)$ is bounded below by $\underline f_A>0$.  But the right-most term in  (<ref>) which lower bounds $\pi^O_1$ grows arbitrarily large as  $\beta\to\infty$. Therefore, for any sufficiently large $\beta$, it follows that

\begin{align*}
\pi^O_1\geq\frac{\beta}{2}\delta^2\underline f_A> \bar{f}_B>\stackrel[0]{1}{\int}\left(v-w_2(v) \right)\bar{f}_B dv\geq  \stackrel[0]{1}{\int}\left(v-w_2(v) \right) f_B(v) dv=\pi^O_2.   \end{align*}

But this is a contradiction with the Equal Profit Condition, as we have shown $\pi^O_1>\pi^O_2.$

§.§ Proof of <Ref>

To show the “only if” part, we first note that it follows from previous arguments that any core outcome satisfies Individual Rationality and the Equal Profit Condition.

To show that the No Desegregation Condition is satisfied, suppose for contradiction that (<ref>) is violated at some $\epsilon \in [0,1]$.
Because the right-hand side of (<ref>) is zero for $\epsilon=1,$ in the remainder we assume that $\epsilon<1$. Define

Let $\tilde O_1$ be an outcome such that,

\begin{align*}
\tilde f_A^1(v) & := \begin{cases}
f_A(v) & \text{ if } v\in [\epsilon+\delta',\hat w^{-1}_1(\epsilon+\delta)],  \\
0 & \text{ otherwise.}
\end{cases}~~~~~ \tilde w_1^A(v) := \begin{cases}
\epsilon+\delta' & \text{if } \tilde f_A^1(v)>0, \\
0 &  \text{otherwise.}
\end{cases}
\end{align*}

\begin{align*}
\tilde f_B^1(v) & := \begin{cases}
f_B(v) & \text{ if } v\in [\epsilon+\delta',\sup\{v' \vert w_2(v')<\epsilon+\delta'\}],  \\
0 & \text{ otherwise.}
\end{cases}~~~~~ \tilde w_1^B(v) := \begin{cases}
\epsilon+\delta' & \text{if } \tilde f_B^1(v)>0, \\
0 &  \text{otherwise.}
\end{cases}
\end{align*}

There exists  $\delta'>0$ such that,

\begin{align*}
\pi^{\tilde O_1}_1&=\beta\stackrel[\epsilon+\delta']{w_{1}^{-1}(\epsilon+\delta')}{\int}\left(v-\epsilon-\delta'\right)f_A(v)dv+\stackrel[ \epsilon+\delta' ]{\sup\{v' \vert w_2(v')<\epsilon+\delta'\}}{\int}\left(v-\epsilon-\delta'\right)f_B(v)dv\\
\end{align*}

where the inequality holds due to the following arguments. The  lower limit of the integrals  are continuous in $\epsilon$ and the integrand $(v-\epsilon)f_g(v)$ for $g\in\{A,B\}$ are continuous by assumption, the latter because $F_g$, $g\in\{A,B\}$ has no mass points. Also, the upper limit of the integrals, respectively, are weakly higher in the first line than in the second line: $w_{1}^{-1}(\epsilon+\delta') \ge w_{1}^{-1}(\epsilon)$ by monotonicity of $w_1^{-1}$ and $\sup\{v' \vert w_2(v')<\epsilon+\delta'\} \ge w^{-1}_2(\epsilon)$ by construction. Recalling that $\delta>0$ and that $\delta'>0$ is arbitrary, the desired conclusion follows.

$\tilde O_1$ blocks $O$ because firm 1's profit increases by assumption, for all $(v,A)$ such that $\tilde f_A^1(v)>0$, Condition 1 of the definition of block is satisfied, for all $(v,B)$ such that $\tilde f_B^1(v)>0$, Condition 3 of the definition of block is satisfied, and for all other $(g,v)$, $g\in\{A,B\}$, Condition 4 of the definition of block is satisfied. This contradicts the initial assumption that $O$ is a core outcome.

To show the “if” direction, suppose the stated conditions hold for an outcome $O$, and suppose for contradiction that there is a firm $i$ and a blocking outcome $\tilde O_i$. We take this firm $i$ to be 1, and the argument for firm 2 is similar.

We consider the following three exhaustive cases:

* $\tilde f_B^1(v)=0$ for almost all $v$.
For any $v$, either $\tilde f^1_A(v)=0$ in which case firm 1 makes 0 infinitesimal profit from $v$, or $\tilde f^1_A(v)>0$. In the latter case, since $f^1_A(v)=f_A(v)$ and $f^1_B(v)=0$ for almost all $v$, it must be that for $A$ and almost all $v$, Conditions 3 and 4 of the definition of block cannot be satisfied, while Conditions 1 and 2 of the definition of block become identical, implying $\tilde w^A_1(v)\geq w_1(v)$ for almost all $v$ with $\tilde f_A^1(v)>0$. Given that $w_1(v)\leq v$ for almost all $v$ by assumption, it must be that:

\begin{align*}
\pi_1^O = \beta \stackrel[0]{1}{\int}\left(v-w_1(v)\right)f_A(v)dv
\ge  \beta \stackrel[0]{1}{\int}\left(v-w_1(v)\right)\tilde f^1_A(v)dv
\ge \beta \stackrel[0]{1}{\int}\left(v-\tilde w^A_1(v)\right)\tilde f^1_A(v)dv
=  \pi_1^{\tilde O_1},
\end{align*}

where the first equality comes from the fact that $f^1_A(v)=f_A(v)$ for almost all $v$ by assumption, the first inequality follows since $ w_1(v)\leq v$ for almost all $v$ by assumption and since $\tilde f^1_A(v)\leq f_A(v)$ for all $v$, and the second inequality follows from the fact $\tilde w_1^A(v) \ge w_1(v)$ for all $v$ such that $\tilde f^1_A(v)>0$ by Condition 2 of the definition of block. Therefore, the above inequalities contradict that $\pi^{\tilde O_1}_1> \pi^O_1.$

* $\tilde f_A^1(v)=0$ for almost all $v$. For any $v$, either $\tilde f^1_B(v)=0$ in which case firm 1 makes 0 infinitesimal profit from $v$, or $\tilde f^1_B(v)>0$. In the latter case, since $f^2_B(v)=f_B(v)$ for almost all $v$, it must be that for $B$ and almost all $v$, Conditions 2 and 4 of the definition of block cannot be satisfied, while Conditions 1 and 3 of the definition of block become identical, implying $\tilde w^B_1(v)> w_2(v)$ for almost all $v$ with $\tilde f_B^1(v)>0$. Given that $w_2(v)\leq v$ for almost all $v$ by assumption, it must be that:

\begin{align*}
\pi_2^O = \stackrel[0]{1}{\int}\left(v-w_2(v)\right)f_B(v)dv
\ge  \stackrel[0]{1}{\int}\left(v-w_2(v)\right)\tilde f^1_B(v)dv
\ge  \stackrel[0]{1}{\int}\left(v-\tilde w^B_1(v)\right)\tilde f^1_B(v)dv
=  \pi_1^{\tilde O_1},
\end{align*}

where the first equality comes from the fact that $f^2_B(v)=f_B(v)$ for almost all $v$ (Proposition <ref>), the first inequality follows since $ w_2(v)\leq v$ for almost all $v$ by assumption and since $\tilde f^1_B(v)\leq f_B(v)$ for all $v$, and the second inequality follows from the fact $\tilde w_1^B(v)>w_2(v)$ for all $v$ such that $\tilde f^1_B(v)>0$ by Condition 3 of the definition of block. The Equal Profit Condition guarantees that $ \pi_2^O= \pi_1^O$, and therefore, the above inequalities contradict that $\pi^{\tilde O_1}_1> \pi^O_1.$

* There exist positive Lebesgue measure sets $V_A$ and $V_B$ such that $\tilde f_A^1(v)>0$ for all $v\in V_A$ and $\tilde f_B^1(v)>0$ for all $v\in V_B$. Group-based EPSW implies that firm 1 must pay a common wage $\epsilon$ in $\tilde O_1$ to almost every $(g,v)$ such that $\tilde f_g^1(v)>0$. Therefore,

\begin{align*}
\pi_1^{\tilde O_1} \leq & \beta\stackrel[0]{w_{1}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)\tilde f^1_A(v)dv+\stackrel[0]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)\tilde f^1_B(v)dv \\
\leq & \beta\stackrel[\epsilon]{w_{1}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)\tilde f^1_A(v)dv+\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)\tilde f^1_B(v)dv \\
\leq & \beta\stackrel[\epsilon]{w_{1}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv \\ \leq & \pi^O_2,
\end{align*}

where the first inequality follows from two arguments:

* Consider the first integral. Since $f^1_A(v)=f_A(v)$ for almost all $v$ by assumption, for $A$ and almost all $v\in V_A$, Conditions 3 and 4 of the definition of block cannot be satisfied, while Conditions 1 and 2 of the definition of block become identical, implying $\tilde w^A_1(v)\geq w_1(v)$ for almost all $v\in V_A$. By definition of $w_1^{-1}(\cdot)$, it therefore must be the case that $\tilde f^1_A(v)=0$ for almost all $v$ such that $w_1^{-1}(\epsilon)<v$.

* Consider the second integral. Since $f^2_B(v)=f_B(v)$ for almost all $v$ by assumption, for $B$ and almost all $v\in V_B$, Conditions 2 and 4 of the definition of block cannot be satisfied, while Conditions 1 and 3 of the definition of block become identical, implying $\tilde w^B_1(v)> w_2(v)$ for almost all $v\in V_B$. By definition of $w_2^{-1}(\cdot)$, it therefore must be the case that $\tilde f^1_B(v)=0$ for almost all $v$ such that $w_2^{-1}(\epsilon)<v$.

The second inequality follows from the fact that $v-\epsilon<0$ for all $v<\epsilon$ and the restriction that $\tilde f^1_A(v)\geq0$ for all $v$ and $\tilde f^1_B(v)\geq0$ for all $v$. The third inequality follows from the fact that $v-\epsilon \geq 0$ over the range of integration, and that $\tilde f^1_A(v)\leq f_A(v)$ and $\tilde f^1_B(v)\leq f_B(v)$. The final inequality follows from the No Desegregation Condition.

The Equal Profit Condition guarantees that $ \pi_2^O= \pi_1^O$, and therefore, the above inequalities contradict that $\pi^{\tilde O_1}_1> \pi^O_1.$

§.§ Proof of <Ref>

Before beginning the proof, we define two terms.

A function $g:[0,1]\to[0,1]$ has an upward jump at $x\in[0,1]$ if either i) $\exists \delta>0$ such that $ \forall \eta>0$, there exists $x'\in (x-\eta,x)\cap[0,1]$ for which $g(x)> g(x') + \delta$, or ii) $\exists \delta>0$ such that $ \forall \eta>0$, there exists $x'\in (x,x+\eta)\cap[0,1]$ for which $g(x')> g(x) + \delta$.

A function $g:[0,1]\to[0,1]$ is upward continuous at $x\in[0,1]$ if
it does not have an upward jump at $x$. $g$ is upward continuous if it is upward continuous at $x$ for all $x\in[0,1]$.

We now proceed with the proof of <Ref>.

We begin the proof by showing several claims.

Take any $\epsilon \in [0,1]$ for which $\phi(\epsilon)<1.$ Then   (<ref>) holds with equality at $\epsilon$ and $\tilde v=\phi(\epsilon)$.

Because the right-hand side of (<ref>) is continuous  in $\tilde v$ while the left-hand side is constant in $\tilde v$, the definition of $\phi(\epsilon)$ implies that (<ref>)
holds, with either a strict inequality or equality, at $\epsilon$ and $\tilde v = \phi(\epsilon).$  Suppose for contradiction that  (<ref>) holds as a strict inequality at $\epsilon$ and $\tilde v = \phi(\epsilon)$. Then, because the right-hand side of the inequality is continuous
in $\tilde v$ and $\phi(\epsilon)<1$ (while, again, the left-hand side is constant in $\tilde v$), there exists $v' \in (\phi(\epsilon),1]$ such that the inequality holds for $\tilde v =v'$. This is a contradiction to the definition of $\phi(\epsilon).$

Function $\phi(\cdot)$ is upward continuous.

We consider for contradiction that $\phi(\cdot)$ has an upward jump of type (ii) at value $\epsilon<1$, and the case of a type (i) upward jump is similar and therefore omitted. Under this premise, there exists $\delta>0$ such that for all $\eta>0$ there exists $\epsilon'\in(\epsilon,\epsilon+\eta)\cap [0,1]$ for which $\phi(\epsilon')>\phi(\epsilon)+\delta.$ Since $\phi(\cdot)$ is a function from $[0,1]$ to $[0,1]$ and $\phi(\epsilon')>\phi(\epsilon)+\delta,$ it follows that $\phi(\epsilon) <1.$ <Ref> implies that (<ref>) holds with equality at $\epsilon$ and $\tilde v=\phi(\epsilon)$.

First consider the case in which $\epsilon=\phi(\epsilon)$. Because the definition of an upward jump specifies the quantifier “for all $\eta>0$” it suffices to assume that $\delta>\eta$.
Comparing (<ref>) at $\epsilon$ and $\epsilon'$ respectively implies that

\begin{align*}
0 \leq & \stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv-\beta\stackrel[\epsilon']{\phi(\epsilon')}{\int}\left(v-\epsilon'\right)f_A(v)dv-\stackrel[\epsilon']{w_{2}^{-1}(\epsilon')}{\int}\left(v-\epsilon'\right)f_B(v)dv\\
\leq & \stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv-\beta\stackrel[\epsilon+\eta]{\phi(\epsilon')}{\int}\left(v-\epsilon-\eta\right)f_A(v)dv-\stackrel[\epsilon+\eta]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon-\eta\right)f_B\left( v\right)dv\\
= & \stackrel[\epsilon]{\epsilon+\eta}{\int}\left(v-\epsilon\right) f_B(v)dv+\stackrel[\epsilon+\eta]{w_2^{-1}(\epsilon)}{\int}\eta f_B(v)dv-\beta\stackrel[\epsilon+\eta]{\phi(\epsilon')}{\int}\left(v-\epsilon-\eta\right)f_A(v)dv\\
\leq & \eta \cdot \bar f_B+\eta\cdot \bar f_B-\beta\stackrel[\epsilon+\eta]{\phi(\epsilon')}{\int}\left(v-\epsilon-\eta\right)\underline f_A dv\\
= & \eta \cdot \bar f_B+\eta\cdot \bar f_B-\frac{1}{2}\underline f_A\cdot \beta\cdot(\phi(\epsilon')-\epsilon-\eta)^2
\end{align*}

where the second inequality follows from the fact that $w_2^{-1}(\cdot)$ is non-decreasing and all integrands are weakly positive over their respective regions of integration. Rearranging terms implies that it must be that

$$ \frac{4}{\beta}\frac{\bar f_B}{\underline f_A}\geq \frac{(\phi(\epsilon')-\epsilon-\eta)^2}{\eta}$$

Therefore, if $\phi(\epsilon')-\phi(\epsilon)=\phi(\epsilon')-\epsilon>\delta$ then the numerator on the right-hand side of the previous expression is positive, and the entire right-hand side grows unboundedly large as $\eta\to 0$, which is a contradiction with the inequality holding.

Second, consider the case in which $\epsilon<\phi(\epsilon)$. Because the  definition of an upward jump specifies the quantifier “for all $\eta>0$” it suffices to assume that $\delta>\eta$.
Comparing (<ref>) at $\epsilon$ and $\epsilon'$ respectively implies that

\begin{align*}
0 \leq & \beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv-\beta\stackrel[\epsilon']{\phi(\epsilon')}{\int}\left(v-\epsilon'\right)f_A(v)dv-\stackrel[\epsilon']{w_{2}^{-1}(\epsilon')}{\int}\left(v-\epsilon'\right)f_B(v)dv\\
\leq & \beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv-\beta\stackrel[\epsilon+\eta]{\phi(\epsilon')}{\int}\left(v-\epsilon-\eta\right)f_A(v)dv-\stackrel[\epsilon+\eta]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon-\eta\right)f_B\left( v\right)dv\\
= & \beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+ \stackrel[\epsilon]{\epsilon+\eta}{\int}\left(v-\epsilon\right) f_B(v)dv+\stackrel[\epsilon+\eta]{w_2^{-1}(\epsilon)}{\int}\eta f_B(v)dv-\beta\stackrel[\epsilon+\eta]{\phi(\epsilon')}{\int}\left(v-\epsilon-\eta\right)f_A(v)dv\\
= & \beta\stackrel[\epsilon]{\epsilon+\eta}{\int}\left(v-\epsilon\right)f_A(v)dv+ \stackrel[\epsilon]{\epsilon+\eta}{\int}\left(v-\epsilon\right) f_B(v)dv+\stackrel[\epsilon+\eta]{w_2^{-1}(\epsilon)}{\int}\eta f_B(v)dv-\beta\stackrel[\phi(\epsilon)]{\phi(\epsilon')}{\int}\left(v-\epsilon\right)f_A(v)dv+\beta\stackrel[\epsilon+\eta]{\phi(\epsilon')}{\int}\eta f_A(v)dv\\
\le  &
\beta\eta\bar f_A
+\eta \cdot \bar f_B+\eta\cdot \bar f_B-\beta[\phi(\epsilon')-\phi(\epsilon)]\cdot [\phi(\epsilon)-\epsilon]\underline f_A +\beta \eta \bar f_A
\\
\le  &
\beta\eta\bar f_A
+\eta \cdot \bar f_B+\eta\cdot \bar f_B-\beta \cdot \delta \cdot [\phi(\epsilon)-\epsilon]\underline f_A +\beta \eta \bar f_A,
\end{align*}

where the transition from the third to the forth line follows from the identity    $
\int_{a}^b g(x)dx -    \int_{c}^d g(x)dx =     \int_{a}^c g(x)dx -    \int_{b}^d g(x)dx,$ and that from the second-last to the last line follows from the assumption that $\phi(\epsilon')-\phi(\epsilon)>\delta$ (other manipulations are either straightforward or analogous to the first case,  and hence the illustrations are omitted).
However, for any sufficiently small $\eta,$ the last expression is negative because all terms except for $-\beta \cdot \delta \cdot [\phi(\epsilon)-\epsilon]\underline f_A<0$ converges to zero, a contradiction.

$\phi(\epsilon) \ge \hat w_1^{-1}(\epsilon)$ for all $\epsilon$.

By the definition of $\hat{w}_1^{-1}(\cdot)$ it suffices to show that for any $\epsilon$, if $\hat{w}_1(v)\leq \epsilon$ then $v\leq \phi(\epsilon).$ Suppose for contradiction that $\hat{w}_1(v)\leq \epsilon$ but that $v> \phi(\epsilon).$ Then by the upward continuity of $\phi(\cdot)$ (see <Ref>), there is some $\epsilon'>\epsilon$ such that $\phi(\epsilon')<v$. Then this implies, by the definition of $\hat{w}_1$, that $\hat{w}_1(v)\geq \epsilon'>\epsilon$. But this contradicts our ongoing assumption that $\hat{w}_1(v)\leq \epsilon$.

$\hat{w}_1^{-1}(\epsilon)=\phi(\epsilon)$ for all   $\epsilon>0$ such that $\phi(\epsilon') \ge \phi(\epsilon)$ for all $\epsilon'>\epsilon.$

Let $\epsilon>0$ such that $\phi(\epsilon') \ge \phi(\epsilon)$ for all $\epsilon'>\epsilon.$ Let $v:=\phi(\epsilon)$. Note that $\hat w_1(v')> \epsilon$ for every $v'>v$ because $\phi(\cdot)$ is upward continuous (see <Ref>). Also note that $\hat w_1(v') \le \epsilon$ for every $v'<v$ because $\phi(\epsilon') \ge \phi(\epsilon)=v>v'$ for every $\epsilon'>\epsilon$ by assumption. Now, consider
$$\hat w^{-1}_1(\epsilon):=\sup \{v'| \hat w_1(v') \le \epsilon\}.$$
From the two previous inequalities, we conclude that $\hat w^{-1}_1(\epsilon)=\phi(\epsilon)$.

Let $w_1:[0,1]\to[0,1]$ be a non-decreasing function that satisfies (<ref>) for every $\epsilon\in[0,1]$. Then $w_1(v)\geq \hat{w}_1(v)$ for almost all $v\in[0,1].$

We first show that for any $v$ it is either the case that $w_1(v)\geq \hat{w}_1(v)$ or that $w_1(\cdot)$ is discontinuous from the right at $v$, i.e. $w_1(v) \neq \lim_{v' \to v+} w_1(v')$. Suppose the negation for contradiction: $w_1(v)<\hat{w}_1(v)$ for some $v\in[0,1]$ and $w_1(\cdot)$ is continuous from the right at $v$. Then by definition of $\hat{w}_1(\cdot)$ there exists $\epsilon^{*}$ such that $w_1(v)<\epsilon^{*}$ and $\phi(\epsilon^{*})\leq v.$ Since (<ref>) holds for $\epsilon^{*}$ and  $w_1^{-1}(\epsilon^{*})$ by assumption, it must be that $w_1^{-1}(\epsilon^{*})\leq\phi(\epsilon^{*})\leq v$. Since $w_1(v)<\epsilon^{*},$
the assumption that $w_1(v) = \lim_{v' \to v+} w_1(v')$  implies that  there exists $\delta>0$ such that $w_1(v+\delta)<\epsilon^{*}$. But then that implies that $w_1^{-1}(\epsilon^{*})>v$ which contradicts $w_1^{-1}(\epsilon^{*})\leq v$.

Because $ w_1$ is non-decreasing, there are at most countably many values $v$ such that $w_1(v) \neq \lim_{v' \to v+} w_1(v')$.[This claim can be shown as follows. Suppose that $w_1(v) \neq \lim_{v' \to v+}  w_1(v')$. Then, because $ w_1$ is non-decreasing, we have $w_1(v) < \lim_{v' \to v+} w_1(v')$. Then,
the set of rational numbers is dense in the set of real numbers, there exists at least one rational number in $(w_1(v), \lim_{v' \to v+}w_1(v'))$; denote one such rational number as $\epsilon_v$. Moreover, because $w_1$ is non-decreasing, $v<\bar v$ implies $\lim_{v' \to v+} w_1(v') \le w_1(\bar v),$ and hence
$\epsilon_v \neq \epsilon_{\bar v}$. Therefore, we have established that there is an injective function from the sets of all points $v$ such that $w_1(v) \neq \lim_{v' \to v+} w_1(v')$ to the set of rational numbers, which implies that the cardinality of the former set is at most as large as the cardinality of the latter.]
By the previous paragraph, this implies that $w_1(v)\geq\hat{w}_1(v)$ for almost all $v\in[0,1].$

For any $w_2(\cdot),$ $\hat{w}_1(v)\leq v$ for all $v\in[0,1).$

Fix any $v\in[0,1]$. If the set $\{\epsilon\vert \phi(\epsilon)\leq v\}$ is empty then $\hat{w}_1(v)=0\leq v.$ Therefore, we focus on the case in which the set $\{\epsilon\vert \phi(\epsilon)\leq v\}$ is non-empty.

For each $\epsilon>0$, it is the case that $\tilde v =\epsilon$ satisfies (<ref>). To see this, note the first term of the right-hand side equals zero if $\tilde v =\epsilon$. Then comparing the left-hand side to the second term of the right-hand side, the inequality holds because: 1) the region of integration is larger for the left-hand side ($1\geq w_2^{-1}(\epsilon)$ for all $\epsilon$ by definition of $w_2^{-1}(\cdot)$, and $0<\epsilon$),
and 2) the left-hand side integrand is weakly bigger than the right-hand side integrand since $\epsilon\geq w_2(v)$ for all $v<w_2^{-1}(\epsilon)$ by definition of $w_2(\cdot)$ and $w_2^{-1}(\epsilon)$.

Since $\phi(\epsilon)$ is defined as the supremum value of $\tilde v$ for which (<ref>) holds for any $\epsilon$, it must be that $\phi(\epsilon)\geq\epsilon$ for all $\epsilon\in[0,1]$. By the definition of $\hat{w}_1$, this implies that $\hat{w}_1(v)\leq v$: if $\hat{w}_1(v)=\epsilon^*> v$ then $\epsilon^*:=\sup\{\epsilon\vert\phi(\epsilon)\leq v\}$ implies that $\phi(\epsilon^*)\leq v <\epsilon^*$ which contradicts $\phi(\epsilon)\geq\epsilon$ for all $\epsilon\in[0,1]$.

We now return to the proof of the Lemma.

Proof of the “only if” part.
Suppose $\hat{\pi}_1<\pi^{O_2}_2$. By <Ref>, for any monotonic wage function $w_1(\cdot)$ satisfying the No Desegregation Condition, $\pi_1:=\beta \stackrel[0]{1}{\int}\left(v-w_{1}(v)\right)f_A(v)dv\leq \hat{\pi}_1<\pi^{O_2}_2$. Therefore, it is impossible to satisfy both the No Desegregation Condition and the Equal Profit Condition simultaneously. By  <Ref>, this implies there  exists no core outcome in which firm 2 pays wage schedule $w_2(\cdot)$.

Proof of the “if” part.
Suppose $\hat{\pi}_1\geq\pi^{O_2}_2$.
Consider a wage function $w^x_1(\cdot)$ for $x\in[0,1]$ where

\begin{align*}
w^x_1(v) & := \begin{cases}
\hat{w}_1(v) & \text{if }  v<x  \\
v & \text{otherwise.}
\end{cases}
\end{align*}

First, we claim that for any $x$, $w^x_1(\cdot)$ is monotonic.  This follows from the monotonicity of $\hat{w}_1(\cdot)$, and the fact that $\hat{w}_1(v)<x$ for all $v<x$ (<Ref>).

Second, we claim that for any $x$, $w^x_1(\cdot)$ satisfies Individual Rationality. This follows from the facts that $\hat{w}_1(v)\leq v$ for all $v<x$ (<Ref>), and that $w^x_1(v)=v$ for all $v\geq x$.

Third, we claim that for any $x$, $w^x_1(\cdot)$ satisfies the No Desegregation Condition. To see this, first note  that $\hat{w}_1(\cdot)$ satisfies the No Desegregation Condition (Lemma <ref> and Claim <ref>). This implies that any $w_1(\cdot)$ such that $w_1(v)\geq \hat{w}_1(v)$ for all $v$ satisfies the No Desegregation Condition as, $w^{-1}_1(\epsilon)\leq \hat{w}^{-1}_1(\epsilon)$ for all $\epsilon\in[0,1]$ and therefore the first integral on the right-hand side of the No Desegregation Condition is weakly smaller given wage schedule $w_1(\cdot)$.

Finally, we claim that for some $x^*\in[0,1]$, $\pi^{x^*}_1:=\beta \stackrel[0]{1}{\int}\left(v-w^{x^*}_{1}(v)\right)f_A(v)dv=\pi^{O_2}_2$. To see this, note that $\pi^x_1=\beta \stackrel[0]{x}{\int}\left(v-\hat{w}_{1}(v)\right)f_A(v)dv$. Then $\pi^x_1$ can be viewed as a one-variable, weakly increasing function of $x$. By the assumption that $F_A$ is atomless,  $\pi^x_1$ is continuous in $x$. Since $\hat{\pi}_1=\pi^1_1\geq\pi^{O_2}_2$ and  $\pi_1^0=0 \le \pi^{O_2}_2$, the intermediate value theorem applies, implying that there is some $x^*\in[0,1]$ such that $\pi^{x^*}_1=\pi^{O_2}_2$. We have shown all the conditions in  <Ref>, so we conclude that $O=(O_1,O_2)$, where $O_1$ is given by $f^1_A(v)=f_A(v),w^A_1(v)=w^{x^*}_1(v), f^1_B(v)=0,$ and $w^B_1(v)=0$ for all $v$, is a core outcome.

We observe that $\hat w_1^{-1}(\cdot)$ is the largest monotonic function that is everywhere below $\phi(\cdot)$. Formally,

If function $h:[0,1] \to [0,1]$ is monotone nondecreasing and $h(\epsilon) \le \phi(\epsilon)$ for all $\epsilon \in [0,1]$, then $h(\epsilon) \le \hat w_1^{-1}(\epsilon)$ for all $\epsilon \in [0,1]$.

Suppose for contradiction that there exists $\epsilon$ such that $h(\epsilon) > \hat w_1^{-1}(\epsilon)$. Since $h(\epsilon) \le \phi(\epsilon)$ by assumption, it follows that $\hat w_1^{-1}(\epsilon)<\phi(\epsilon)$.

First, suppose that the set $\{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}$ is nonempty. In that case, define $\epsilon^*:=\sup \{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}.
Because $\phi(\cdot)$ is upward continuous, by <Ref> it follows that $\phi(\epsilon^*)=\hat w^{-1}_1(\epsilon^*)$.[The proof of this claim is as follows. By <Ref>, it suffices to show that $\epsilon^*$ is in the set $\{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}$. To show this, assume for contradiction that there exists $\hat \epsilon>\epsilon^*$ such that $\phi(\hat \epsilon) < \phi(\epsilon^*)$. Then, by  upward continuity of $\phi(\cdot)$, there exists $\delta>0$ such that $\phi(\epsilon^{**})>\phi(\hat \epsilon)$ for all $\epsilon^{**} \in (\epsilon^*-\delta,\epsilon^*)$. However, this is a contradiction to the definition of $\epsilon^*$.] Because $\hat w^{-1}_1(\cdot)$ is nondecreasing, $\hat w^{-1}_1(\epsilon^*) \le \hat w^{-1}_1(\epsilon)$. Combined with the assumption that $h(\epsilon) > \hat w_1^{-1}(\epsilon)$, it follows that $h(\epsilon) > \hat w_1^{-1}(\epsilon^*)=\phi(\epsilon^*)$. Since this is a contradiction to $h(\epsilon) \le \phi(\epsilon)$ if $\epsilon^*=\epsilon,$ we proceed with the assumption that $\epsilon^*<\epsilon.$
By definition of $\epsilon^*$, it follows that there exists $\bar \epsilon>\epsilon^*$ such that $\phi(\bar \epsilon) \le \phi(\epsilon^*)$.[The proof of this claim is as follows.
Suppose for contradiction that $\phi(\bar \epsilon)>\phi(\epsilon^*)$ for all $\bar \epsilon>\epsilon^*$. Then, by upward continuity of $\phi(\cdot)$, $\epsilon^*$ would not be the supremum of the set $\{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}$, a contradiction.]  Moreover, we observe that $\epsilon\leq \bar \epsilon$.[ Suppose for contradiction that $\epsilon> \bar \epsilon.$ Recall that $\phi(\bar \epsilon)\leq \phi(\epsilon^*).$ Since $\epsilon^*\leq \bar \epsilon<\epsilon$, then this is a contradiction with the fact that $\epsilon^*$ is the supreumum of the set $\{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}$.]
Because $h(\cdot)$ is nondecreasing, it follows that $h(\bar  \epsilon) \ge h(\epsilon)>\phi(\epsilon^*) \ge \phi(\bar \epsilon)$, a contradiction.

Next, suppose that the set $\{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}$ is empty. Let $\underline \epsilon$ be a minimizer of $\phi(\cdot)$ over $[0,1]$.[The claim that a minimizer of $\phi(\cdot)$ over $[0,1]$ exists is proven as follows. Let $\underline v:=\inf_{\tilde \epsilon \in [0,1]}\phi(\tilde \epsilon).$ By definition, there exists a sequence $(\epsilon_n)_{n=1}^\infty$ such that $\underline \epsilon:=\lim_{n \to \infty}\epsilon_n$ exists and $\lim_{n \to \infty}\phi(\epsilon_n)=\underline v$. We will show that $\phi(\underline \epsilon)=\underline v$. To show this, suppose for contradiction that $\phi(\underline \epsilon) \neq \underline v$. Because $\underline v$ is the infimum of $\phi$, this implies that $\phi(\underline \epsilon) > \underline v$.
Suppose that there is a subsequence $(\epsilon_{n_k})_k$ of $(\epsilon_n)_n$ such that $\epsilon_{n_k} <\underline \epsilon$ for each $k$.
Now, pick an arbitrary $\delta \in \left (0,\frac{\phi(\underline \epsilon)-\underline v}{2} \right)$: Note that $\phi(\underline \epsilon)-\underline v>0$. Because $\underline \epsilon=\lim_{n \to \infty}\epsilon_n$ and $\lim_{n \to \infty}\phi(\epsilon_n)=\underline v$, for any $\eta>0$ there exists $k$ such that $\epsilon_{n_k}>\underline \epsilon- \eta$ and $\phi(\epsilon_{n_k})+\delta<\underline v + 2
\delta<\phi(\underline \epsilon)$.
But this is a contradiction to upward continuity of $\phi.$
Now, suppose that there is no subsequence $(\epsilon_{n_k})_k$ of $(\epsilon_n)_n$ such that $\epsilon_{n_k} <\underline \epsilon$ for each $k$.
Then there is a subsequence $(\epsilon_{n_k})_k$ of $(\epsilon_n)_n$ such that $\epsilon_{n_k} \ge \underline \epsilon$ for each $k$. But this is a contradiction to the fact that, by inspection of the No Desegregation Condition, $\lim \inf_{\tilde \epsilon \to \underline \epsilon+0} \phi(\tilde \epsilon) \ge \phi(\underline \epsilon).$]
Note that it cannot be that $\underline \epsilon \le \epsilon$, because then
$\underline \epsilon \in \{\tilde \epsilon \in [0,\epsilon]
| \phi(\hat \epsilon)\ge \phi(\tilde \epsilon) \text{ for every $\hat \epsilon>\tilde \epsilon$}\}$, a contradiction to the assumption made in the present case. Thus, assume $\underline \epsilon > \epsilon$.
From the constructions of $\hat{w}_1(\cdot)$ and $\hat{w}_1^{-1}(\cdot)$ on page 16, it follows that $\hat{w}_1^{-1}(\epsilon)=\phi(\underline \epsilon)$. But this is a contradiction to monotonicity of $h(\cdot)$ because $h(\epsilon)>\hat{w}^{-1}(\epsilon)=\phi(\underline \epsilon) \ge h(\underline \epsilon)$, where the last inequality follows from the hypothesis of the claim. This completes the proof.

§.§ Proof of <Ref>

First, consider the case where $w_2(v)=v$ for almost all $v$. The conclusions of the proposition hold trivially with $\beta^*=0$ in this case because $w_1(v)=w_2(v)=v$ for all $v$ can be supported in a core outcome for every $\beta$.
Thus, in the following, we exclude the case in which  $w_2(v) = v$ for almost all $v$.

From Lemma <ref> note that, to tell whether there exists a core outcome in which firm 2 chooses payment schedule $w_2(\cdot)$ (for  $B-$group workers), it suffices  to check whether

\begin{equation}
\hat{\pi}_1 \ge \pi^{O_2}_2 \label{No Desegregation Condition_w_2=0}.
\end{equation}

There exists $\beta^* \in [0,\infty)$ such that (<ref>) holds for all $\beta>\beta^*$.

Let $\epsilon^*$ be the supremum of $v$ such that $w_2(v')=v'$ for almost all $v'<v$, that is, $\epsilon^*:=\{v \in [0,1] | w_2(v')=v'  \text{ for almost all } v' \in [0,v)\}$. Note that we have already excluded the function $w_2(\cdot)$ such that $w_2(v)=v$ for almost all $v \in [0,1]$ from consideration, so it follows that $\epsilon^*<1$. We proceed by proving the following three claims.

$\phi(\cdot)$ uniformly converges to the identity function as $\beta \to \infty$.

By (<ref>) and <Ref>, we have

\begin{equation*}
\stackrel[0]{1}{\int}\left(v-w_{2}(v)\right)f_B(v)dv \ge  \beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv.
\end{equation*}

The right-hand side of this inequality can be bounded from below as follows:

\begin{align*}
\beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv & \ge \beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)\underline{f}_Adv \\
& = \underline{f}_A \beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right) dv \\
& = \frac{\underline{f}_A \beta}{2}  (\phi(\epsilon)-\epsilon))^2.
\end{align*}

This and the fact that $\phi(\epsilon) \ge \epsilon$ for all $\epsilon$ imply the following bound:

\begin{align}
0 \le   \phi(\epsilon)-\epsilon \le \left [ \frac{2}{\underline{f}_A \beta}\stackrel[0]{1}{\int}\left(v-w_{2}(v)\right)f_B(v)dv \right ]^{1/2}.\label{inequality15}
\end{align}

Because the right-hand side of (<ref>) does not depend on $\epsilon$ and converges to zero as $\beta \to \infty$, we have established uniform convergence of $\phi(\cdot)$ to the identity function as $\beta \to \infty$, as desired.

For any $n>0$ and any $0<\underline{\epsilon}<\bar{\epsilon}<1$  there exists $\beta^*$ such that for all $\beta>\beta^*$ there exists a sequence $\epsilon_1,...,\epsilon_n$ satisfying:

* Intervals $[\epsilon_k,\hat{w}_1^{-1}(\epsilon_k))$ and $[\epsilon_{\ell},\hat{w}_1^{-1}(\epsilon_{\ell}))$ are disjoint for any $k\neq \ell$,

* $\phi(\epsilon_k)=\hat{w}_1^{-1}(\epsilon_k)$, and

* $\underline{\epsilon}<\epsilon_k<\bar{\epsilon}$ for all $k$.

Define $v_1=\underline{\epsilon}+ 2\gamma, v_2=\underline{\epsilon}+4\gamma,..., v_n=\underline{\epsilon}+ 2n\gamma$, where $\gamma>0$ is sufficiently small such that $v_n <\bar{\epsilon}$. Let $\beta$ be large enough so that

\begin{equation}
\phi(\epsilon) \in [\epsilon,\epsilon+\gamma), \label{phi-range}
\end{equation}

for every $\epsilon \in [0,1]$ (note that such $\beta$ exists by <Ref>).

For each $k\in\{1,2,...,n\}$, let $\epsilon_k:=\hat{w}_1(v_k)$.  Then $\epsilon_k=\hat w_1(v_k) \in [v_k-\gamma, v_k]=[\underline{\epsilon}+(2k-1)\gamma, \underline{\epsilon}+ 2k \gamma]$, where the first  relation follows from the definition of $\epsilon_k$, the second relation follows from the definition of $\hat{w}_1(\cdot)$ because:

* $\phi(\epsilon)\in [\epsilon, \epsilon+\gamma)$ for any $\epsilon$ by (<ref>), so if $\epsilon<v_k-\gamma$ then $\phi(\epsilon) < v_k$. Because $\hat w_1(v_k)$ is defined as the supremum of all $\epsilon$ such that $\phi(\epsilon)\leq v_k$, we conclude that  $\hat w_1(v_k) \ge v_k-\gamma$.

* $\hat w_1(v_k) \le v_k$ by <Ref>.

Moreover,  $\hat{w}_1^{-1}(\epsilon_k) < \epsilon_k +\gamma \le \underline{\epsilon}+(2k+1) \gamma$, where the first inequality follows from (<ref>) and that $\phi(\epsilon)\geq \hat{w}_1^{-1}(\epsilon)$ by Claim <ref>, and the second inequality follows from our earlier assertion that $\epsilon_k \le \underline{\epsilon}+ 2k \gamma$.
We also know that $\epsilon_k\geq \underline{\epsilon}+(2k-1)\gamma$, so $\hat{w}_1^{-1}(\epsilon_k) \geq \underline{\epsilon}+(2k-1)\gamma$. Thus, it follows that
$\hat{w}_1^{-1}(\epsilon_k)\in [\underline{\epsilon}+(2k-1)\gamma, \underline{\epsilon}+(2k+1) \gamma]$. Recalling that $\epsilon_k \in [\underline{\epsilon}+(2k-1)\gamma, \underline{\epsilon}+ 2k \gamma]$,
the  intervals $[\epsilon_k,\hat{w}^{-1}(\epsilon_k))$ are all disjoint, satisfying requirement 1 of the claim.

We can also see that $\phi(\epsilon')>\phi(\epsilon_k)$ for each $k$ and all $\epsilon'>\epsilon_k$, because the definition $\epsilon_k=\hat{w}_1(v_k)$ and the upward continuity of $\phi(\cdot)$ (see <Ref>) imply that $\phi(\epsilon_k)=v_k$[To see this, suppose for contradiction that $\phi(\epsilon_k) \neq v_k$. If $\phi(\epsilon_k) < v_k$, then since $\phi(\epsilon_k')>v_k$ for all $\epsilon_k'>\epsilon_k$ by definition of $\epsilon_k=\hat{w}_1(v_k)$, $\phi(\cdot)$ has an upward jump of type (ii) at $\epsilon_k'$. If $\phi(\epsilon_k) > v_k$, then since for any $\eta>0$ there exists $x \in (\epsilon_k-\eta,\epsilon_k)$ such that $\phi(x) \le v_k$, $\phi(\cdot)$ has an upward jump of type (i) at $\epsilon_k.$ In both cases, it is a contradiction to upward continuity of $\phi(\cdot).$] and that there is no $\epsilon'>\epsilon_k$ such that $\phi(\epsilon')\le v_k$. By  Claim <ref>, this implies that $\phi(\epsilon_k)=\hat{w}^{-1}(\epsilon_k)$ for all $k$, satisfying requirement 2 of the claim.

Finally, the property that $\epsilon_k \in [v_k-\gamma, v_k]$ for every $k$ and that $\underline{\epsilon}+2\gamma=v_1<v_2<\dots<v_n<\bar{\epsilon}$ imply condition 3 of the claim holds.

For any $\epsilon$ at which the No Desegregation Condition (i.e. (<ref>)) holds with equality, we can rearrange terms to obtain

\begin{equation}
\stackrel[0]{\epsilon}{\int}\left(v-w_{2}(v)\right)f_B(v)dv + \stackrel[w_{2}^{-1}(\epsilon)]{1}{\int}\left(v-w_{2}(v)\right)f_B(v)dv +  \stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(\epsilon-w_{2}(v)\right)f_B(v)dv=  \beta\stackrel[\epsilon]{\hat w_{1}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv.
\label{temp_No Desegregation Condition}\end{equation}

For any $\underline \epsilon$ and any $\bar \epsilon$ such that $\epsilon^*<\underline{\epsilon}<\bar{\epsilon}<1$, there exists $n$ for which the left-hand side of (<ref>) is strictly larger than $\pi^{O_2}_2/n$ for all $\epsilon\in(\underline \epsilon, \bar{\epsilon}).$

By the definition of $\epsilon^*$,  $\stackrel[0]{\underline \epsilon}{\int}\left(v-w_{2}(v)\right)f_B(v)dv:=\delta>0$. Moreover, because $v-w_{2}(v) \ge 0$ for every $v \in [0,1]$ by assumption, it follows that

$$\stackrel[0]{\epsilon}{\int}\left(v-w_{2}(v)\right)f_B(v)dv \ge \stackrel[0]{\underline \epsilon}{\int}\left(v-w_{2}(v)\right)f_B(v)dv=\delta,$$

for any $\epsilon \in (\underline \epsilon,\bar \epsilon)$. Moreover, each term of the left-hand side of (<ref>) is nonnegative, so the left-hand side of (<ref>) is at least $\delta>0$ for all $\epsilon \in (\underline \epsilon,\bar \epsilon)$. Therefore, for any $n>\pi^{O_2}_2/\delta$, the left-hand side of (<ref>) is strictly larger than $\pi^{O_2}_2/n$ for  all $\epsilon \in (\underline \epsilon,\bar \epsilon)$, as desired.

We now return to the proof of <Ref> to establish that $\hat{\pi}_1 > \pi^{O_2}_2$ for sufficiently large $\beta$. To do so, take any $\underline \epsilon$ and any $\bar \epsilon$ such that $\epsilon^*<\underline{\epsilon}<\bar{\epsilon}<1$.
By Claim <ref> and (<ref>),  for any sufficiently large $n$ and any $\epsilon\in(\underline \epsilon, \bar \epsilon)$ we have

\begin{align*}
\pi^{O_2}_2 < \beta n \stackrel[\epsilon]{\hat{w}_{1}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv.
\end{align*}

Further noting that any $v \in [\epsilon, \hat w^{-1}_1(\epsilon)]$ satisfies  $\epsilon \ge \hat{w}_1(v)$, we can see that

\begin{align}
\pi^{O_2}_2 < \beta n \stackrel[\epsilon]{\hat{w}_{1}^{-1}(\epsilon)}{\int}\left(v-\hat{w}_1(v)
\right)f_A(v)dv.\label{equation-13.75}
\end{align}

Consider any $\beta^*$, $\beta>\beta^*$ and a sequence $\epsilon_1,\epsilon_2,\dots,\epsilon_n$ with $\underline{\epsilon}<\epsilon_k<\bar{\epsilon}$ for all $k\in\{1,...,n\}$ that satisfy the conditions in Claim <ref>. Let

k^* \in \arg\min_{k \in \{1,\dots,n\}} \stackrel[\epsilon_k]{\hat{w}_{1}^{-1}(\epsilon_k)}{\int} \left(v-\hat{w}_{1}(v)\right)f_A(v)dv.

Then, by definition of $k^*$,

\begin{align}
\stackrel[\epsilon_{k^*}]{\hat{w}_{1}^{-1}(\epsilon_{k^*})}{\int} \left(v-\hat{w}_{1}(v)\right)f_A(v)dv \le \stackrel[\epsilon_k]{\hat{w}_{1}^{-1}(\epsilon_k)}{\int} \left(v-\hat{w}_{1}(v)\right)f_A(v)dv,\label{interval-comparison}
\end{align}

for every $k \in \{1,\dots,n\}$. Therefore,

\begin{eqnarray*}
\beta n \stackrel[\epsilon_{k^*}]{\hat{w}_{1}^{-1}(\epsilon_{k^*})}{\int}  \left(v-\hat{w}_{1}(v)\right)f_A(v)dv & \le &\beta \sum_{k \in \{1,\dots,n\}} \stackrel[\epsilon_k]{\hat{w}_{1}^{-1}(\epsilon_k)}{\int} \left(v-\hat{w}_{1}(v)\right)f_A(v)dv \\
& \le & \beta
\stackrel[0]{1}{\int}\left(v-\hat{w}_{1}(v)\right)f_A(v)dv\\
& = & \hat \pi_1,
\end{eqnarray*}

where the first inequality follows from (<ref>),
the second inequality follows from the facts that $v-\hat{w}_{1}(v) \ge 0$ for every $v$
and that all the intervals of the form $[\epsilon_{k},\hat{w}_{1}^{-1}(\epsilon_k))$ are disjoint subsets of $[0,1]$, and the last equality follows from the definition  of $\hat \pi_1$. This and (<ref>) taken at $\epsilon=\epsilon_{k^*}$ imply $\hat \pi_1 > \pi^{O_2}_2$, as desired.
Application of <Ref> completes the proof.

§.§ Proof of <Ref>

In the main body, <Ref> states

Suppose there is a group-based EPSW and fix $F_A$ and $F_B$ arbitrarily. There exists $\beta^* \in (0,\infty)$ such that
in the market $(\beta, F_A, F_B)$ with any $\beta>\beta^*$ there exists a core outcome in which $w_2(v)=0$ for all $v$
(i.e. the wage of each $B-$group worker is $0$) while
in the market $(\beta, F_A, F_B)$ with any $\beta<\beta^*$
there exists no core outcome in which $w_2(v)=0$ for all $v$.
It also claims that this statement can be generalized. We formally state and prove the following result, which subsumes the case with $w_2(v)=0$ for all $v$.

Let $w_2(\cdot)$ be a continuous  wage function such that $w_2(v) \le v$ for all $v \in [0,1)$,  $w^{-1}_2(\cdot)$ is absolutely continuous,  and $\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv$ is non-increasing in $\epsilon$. Also, let $f_A(\cdot)$ be continuous. Then there exists $\beta^*\in [0,\infty)$ such that
in the market $(\beta, F_A, F_B)$ with any $\beta>\beta^*$
there exists a core outcome in which the wage schedule of $B-$group workers is given by $w_2(\cdot)$ while
in the market $(\beta, F_A, F_B)$ with any $\beta<\beta^*$
there exists no core outcome in which the wage schedule of $B-$group workers is given by $w_2(\cdot)$.

First, consider the case where $w_2(v)=v$ for almost all $v$. The conclusion of the proposition holds trivially with $\beta^*=1$ in this case because $w_1(v)=w_2(v)=v$ for all $v$ can be supported in a core outcome for every $\beta$.
Thus, in the following, we exclude the case in which  $w_2(v) = v$ for almost all $v$.

Fix $w_2$ satisfying the conditions in the proposition.
From  <Ref>, there exists $\beta'$ such that there exists a core outcome in which the wage schedule of $B-$group workers is given by $w_2$ for all $\beta>\beta'.$ Let $\bar \beta>\beta'$, and consider the set $[1,\bar \beta]$. In the remainder of the proof, we will show that there exists $\beta^*\in [1,\bar \beta)$ such that for all $\beta \in(\beta^*, \bar \beta]$  there exists a core outcome in which the wage schedule of $B-$group workers is given by $w_2(\cdot)$ while for all $\beta\in[1,\beta^*)$ there exists no core outcome in which the wage schedule of $B-$group workers is given by $w_2$ (which implies that $\beta^* \le \beta'$): This will complete the proof because <Ref> implies that the same conclusion holds for all $\beta \in [\bar \beta, \infty)$.

With the additional assumptions made for this part, we first establish two technical properties: first,  <Ref> establishes that $\hat{w}_1^{-1}(\epsilon)=\phi(\epsilon)$  for all $\epsilon$, and second, <Ref> implies that $\hat{w}_1^{-1}(\epsilon)$ is differentiable in $\beta$ and $\epsilon$. Building on these results, the proof of the “threshold structure” is provided in <Ref>.

If $\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv$ is non-increasing in $\epsilon$, then

* There exists some $E \in [0,1]$ such that $\phi(\cdot)$ is strictly increasing in $\epsilon$ for all $\epsilon<E$ and $\phi(\epsilon)=1$ for all $\epsilon\geq E$, and

* $\hat{w}_1^{-1}(\epsilon)=\phi(\epsilon)$ for all $\epsilon\in[0,1]$.

Let $0\leq\epsilon_1<\epsilon_2\leq 1.$ First, suppose that $\tilde v_1:=\phi(\epsilon_1)=1.$ By inspection, we note the following properties from (<ref>):  the left-hand side does not depend on $\epsilon,$ the second integral on the right-hand side is non-increasing in $\epsilon$ by assumption, and the first term on the right-hand side is strictly decreasing in $\epsilon$ when fixing $\tilde v_1$ as the upper limit of integration, it becomes clear that the inequality is satisfied at $\epsilon_2$ and $\tilde v_1=1.$ Therefore, $\phi(\epsilon_2)=1$, which demonstrates that there exists $E$ such that $\phi(\epsilon)=1$ for any $\epsilon>E$.

Second, suppose $\phi(\epsilon_1)<1$. Then, $\tilde v_1:=\phi(\epsilon_1)$ satisfies (<ref>) at $\epsilon_1$ with equality. Define $\hat v:=\max\{\tilde v_1,\epsilon_2\}.$ We consider the following exhaustive cases, which jointly show that $\phi(\epsilon_1)<\phi(\epsilon_2),$ as desired.

* Suppose that $\tilde v_1\geq\epsilon_2$, which implies $\hat v=\tilde v_1$. By inspection, we note the following properties from (<ref>): the left-hand side does not depend on $\epsilon,$ the second integral on the right-hand side is non-increasing in $\epsilon$ by assumption, and the first term on the right-hand side is strictly decreasing in $\epsilon$ when fixing $\tilde v_1$ as the upper limit of integration. Therefore, $\hat v$ satisfies  (<ref>) at $\epsilon_2$ with a strict inequality. Noting that the right-hand side of (<ref>) is continuous in the variable $\tilde v$ for all $\epsilon\in[0,1]$, there exists $\bar v>\hat v$ that satisfies (<ref>) at $\epsilon_2$. Hence, we obtain   $\phi(\epsilon_2)>\phi(\epsilon_1)$, as desired.

* Suppose that $\tilde v_1 < \epsilon_2$. The proof of <Ref> reveals that $\phi(\epsilon)\geq \epsilon$ for all $\epsilon\in[0,1].$ Therefore, $\phi(\epsilon_2)\geq \epsilon_2>\tilde v_1= \phi(\epsilon_1),$ establishing that $\phi(\epsilon_2)>\phi(\epsilon_1)$ as desired.

By Part <ref>, it follows that $\phi(\epsilon') \ge \phi(\epsilon)$ for any $\epsilon'>\epsilon$. By applying <Ref>, we complete the proof.

Second, we establish relevant differentiability properties of $\phi(\cdot)$. Note that we have treated $\phi(\cdot)$ as a function of $\epsilon,$ but it also depends on $\beta$. Thus, in the following result, we regard $\phi(\cdot)$ as  a function of $\epsilon$ and $\beta$.

Suppose that $w^{-1}_2(\cdot)$ is absolutely continuous and $\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv$ is non-increasing in $\epsilon$.

Fix $\beta\geq 1$. For almost all $\epsilon\in(0,1)$
, $\frac{\partial \phi(\cdot)}{\partial\epsilon}$ exists.

Fix $\epsilon\in(0,1)$.
For almost all $\beta\geq 1$, $\frac{\partial \phi(\cdot)}{\partial\beta}$ exists.

*  Let $f_A(\cdot)$ be continuous. Then for  almost all $\epsilon \in(0,E)$ and almost all $\beta\geq 1$,  $\frac{\partial\hat{w}_1}{\partial \beta}(\hat{w}_1^{-1}(\epsilon))$ exists and is equal to $-\frac{\partial\hat{w}_1}{\partial v}(\hat{w}_1^{-1}(\epsilon))\frac{\partial\hat{w}^{-1}_1}{\partial \beta}(\epsilon)$.

*  There exists constant $c>0$ such that for every $\beta<\bar\beta$, $\hat{w}_1(\cdot)$ is Lipschitz continuous in $v\in[0,1)$ with constant $1/c$.

First, consider the case in which $w_2(v)=v$ for all $v \in [0,1].$ Then, $\phi(\epsilon)=\epsilon$ for all $\epsilon \in [0,1]$, so the conclusion holds.

Second, suppose it is not the case that $w_2(v)=v$ for all $v$. Fix $\underline \epsilon >0$. It suffices to show that for almost all $\epsilon\in[\underline \epsilon,1]$
, $\frac{\partial \phi(\cdot)}{\partial\epsilon}$ exists. To do so we establish the following claim:

Suppose it is not the case that $w_2(v)=v$ for all $v$. Then there exists $\delta>0$ such that, for every $\epsilon \in [\underline \epsilon,1]$ with $\phi(\epsilon)<1$, we have $\phi(\epsilon)-\epsilon>\delta.$

For any $\epsilon$ with $\phi(\epsilon)<1$,  (<ref>) holds with equality at $\tilde v=\phi(\epsilon)$, so
we have

\begin{equation} \label{equality-to-be-evaluated}
\beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv=\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv-\stackrel[\epsilon]{w^{-1}_2(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv.
\end{equation}

The left-hand side of (<ref>) can be bounded from above by

\begin{equation}
\beta\stackrel[0]{1}{\int}\left(\phi(\epsilon)-\epsilon\right) f_A(v)dv = \beta (\phi(\epsilon)-\epsilon). \label{eq:left-hand side}\end{equation}

The assumption that $\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv$ is non-increasing in $\epsilon$ implies that the right-hand side of (<ref>) is bounded from below by

\begin{equation}
\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv-\stackrel[0]{w^{-1}_2(0)}{\int}\left(v-0\right)f_B(v)dv =\stackrel[w^{-1}_2(0)]{1}{\int} [v-w_2(v)] f_B(v)dv, \label{new-equation}
\end{equation}

where the equality holds because $w_2(v)=0$ for all $v<w^{-1}_2(0)$. If the right-hand side of (<ref>) is strictly positive, then we have established the desired conclusion, because (<ref>), (<ref>), and  (<ref>) imply that $\phi(\epsilon)-\epsilon\geq\frac{1}{\beta}\stackrel[w^{-1}_2(0)]{1}{\int} [v-w_2(v)] f_B(v)dv>0$ for all $\epsilon$.

So, we now assume that the right-hand side of (<ref>) is zero. This implies that there exists $v^* \in [0,1]$ such that $w_2(\cdot)$ is equal (almost everywhere) to

\begin{equation*}
0 & v < v^*\\
v & v \ge v^*.
\end{cases}
\end{equation*}

Recall that we have assumed it is not the case that $w_2(v)=v$ for all $v$. Therefore, $v^*>0$. Now, assume $\underline \epsilon<v^*$ without loss of generality. For any $\epsilon \in [\underline \epsilon, v^*)$, the right-hand side of (<ref>) is equal to

\begin{equation}
\stackrel[0]{v^*}{\int} [v-0] f_B(v)dv-\stackrel[\epsilon]{v^*}{\int}\left(v-\epsilon\right)f_B(v)dv=\stackrel[0]{\epsilon}{\int} v f_B(v)dv+\stackrel[\epsilon]{v^*}{\int}\epsilon f_B(v)dv \geq \stackrel[0]{\underline \epsilon}{\int} v \underline f_B dv = \frac{\underline \epsilon^2}{2} \underline f_B>0, \label{eq:right-hand side}
\end{equation}

where the first inequality holds because $\stackrel[\epsilon]{v^*}{\int}\epsilon f_B(v)dv \ge 0$, $\epsilon\geq\underline \epsilon,$ and $f_B(v) \ge \underline f_B$ for all $v.$
For any $\epsilon \in [v^*,1]$, the right-hand side of (<ref>) is equal to

\begin{equation}
\stackrel[0]{v^*}{\int} [v-0] f_B(v)dv\geq \frac{(v^*)^2}{2}\underline f_B>0, \label{eq:right-hand side_alt}
\end{equation}

where the first inequality follows because $f_B(v) \ge \underline f_B$ for all $v$.
\begin{align}\label{eq:delta_beta}
\delta_\beta:=\frac{1}{2\beta}\min \left \{\frac{\underline\epsilon^2}{2}\underline f_B,\frac{(v^*)^2}{2}\underline f_B \right \}= \frac{\underline\epsilon^2 \underline f_B}{4\beta},
\end{align}

where the last equality follows from $v^*>\underline{\epsilon}$.
Combining (<ref>), (<ref>), and (<ref>) we have that for all $\epsilon>\underline \epsilon$, $\phi(\epsilon)-\epsilon>\delta_\beta$ as desired.

We now return to the Proof of Part <ref> of <Ref>. First, consider the case in which $\phi(\epsilon)=1$. Then for any $\Delta \epsilon>0$, $\phi(\epsilon + \Delta \epsilon)=1$, so $\phi(\epsilon + \Delta \epsilon)-\phi(\epsilon)=0$.

Next, consider the case in which $\phi(\epsilon)<1$. By <Ref>, let $\delta_\beta$ be defined as in (<ref>).
We reproduce (<ref>) (No Desegregation Condition) holding with equality (note that $\phi(\epsilon)<1$ implies that No Desegregation Condition holds with equality) here:

\begin{equation}
\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv=\beta\stackrel[\epsilon]{\phi(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w^{-1}_2(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv.\label{eq:ndc-equality-modified-w20}
\end{equation}

For any $\Delta \epsilon \in (0, 1-\epsilon)$, the No Desegregation Condition (with inequality) must also hold at $\epsilon+\Delta \epsilon$, so

\begin{equation}
\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv\geq\beta\stackrel[\epsilon+\Delta\epsilon]{\phi(\epsilon+\Delta\epsilon)}{\int}\left(v-(\epsilon+\Delta\epsilon)\right)f_A(v)dv+\stackrel[\epsilon+\Delta\epsilon]{w^{-1}_2(\epsilon+\Delta\epsilon)}{\int}\left(v-(\epsilon+\Delta\epsilon) \right)f_B(v)dv.\label{eq:ndc-equality-modified-w20-2}
\end{equation}

Suppose that $\Delta \epsilon<\delta_\beta/2$.
We continue the proof by subtracting (<ref>) from (<ref>). Note that $\phi(\epsilon)-\epsilon>\delta_\beta$, as we have previously shown, implies that $\phi(\epsilon)>\epsilon+\delta_\beta>\epsilon+\Delta\epsilon.$ Note that we do not know the sign of $w^{-1}_2(\epsilon)-(\epsilon+\Delta\epsilon)$ but our approach moving forward will not depend on that sign.

\begin{align}
& 0  \ge - \beta\stackrel[\epsilon]{\epsilon + \Delta \epsilon}{\int}\left(v-\epsilon \right)f_A(v)dv
- \beta\stackrel[\epsilon + \Delta \epsilon]{\phi(\epsilon)}{\int} \Delta \epsilon f_A(v)dv
+ \beta\stackrel[\phi(\epsilon)]{\phi(\epsilon+\Delta \epsilon)}{\int}\left(v-(\epsilon+\Delta \epsilon)\right)f_A(v)dv \notag \\
& - \stackrel[\epsilon]{\epsilon+\Delta \epsilon}{\int}\left(v-\epsilon \right)f_B(v)dv
- \stackrel[\epsilon+\Delta \epsilon]{w^{-1}_2(\epsilon)}{\int}\Delta \epsilon f_B(v)dv  + \stackrel[w^{-1}_2(\epsilon)]{w^{-1}_2(\epsilon+\Delta \epsilon)}{\int}\left(v-(\epsilon+\Delta \epsilon) \right ) f_B(v)dv
\notag \\
\Leftrightarrow &
\beta\stackrel[\phi(\epsilon)]{\phi(\epsilon+\Delta \epsilon)}{\int}\left(v-(\epsilon+\Delta \epsilon)\right)f_A(v)dv \le \notag \\ &\beta\stackrel[\epsilon]{\epsilon + \Delta \epsilon}{\int}\left(v-\epsilon \right)f_A(v)dv
+ \beta\stackrel[\epsilon + \Delta \epsilon]{\phi(\epsilon)}{\int} \Delta \epsilon f_A(v)dv
+ \stackrel[\epsilon]{\epsilon+\Delta \epsilon}{\int}\left(v-\epsilon \right)f_B(v)dv
+ \stackrel[\epsilon+\Delta \epsilon]{w^{-1}_2(\epsilon)}{\int}\Delta \epsilon f_B(v)dv \notag \\
& - \stackrel[w^{-1}_2(\epsilon)]{w^{-1}_2(\epsilon+\Delta \epsilon)}{\int}\left(v-(\epsilon+\Delta \epsilon) \right ) f_B(v)dv
\label{eq:ndc-equality-modified-w20-3}.
\end{align}

In the following, we bound each term in (<ref>) one by one:

* The left-hand side is bounded from below as follows: Recall our assumption that $\Delta \epsilon<\delta_\beta/2$.
Because $\phi(\epsilon)-\epsilon>\delta_\beta$ and $v \ge \phi(\epsilon)$ for $v$ in the range of the integration, one can bound the integrand as follows: $v-(\epsilon+\Delta \epsilon) \ge \phi(\epsilon)-\epsilon -\Delta \epsilon > \delta_\beta -\delta_\beta/2 = \delta_\beta/2.$ Therefore, we bound the left-hand side by

\begin{align*}  \beta\stackrel[\phi(\epsilon)]{\phi(\epsilon+\Delta \epsilon)}{\int}\left(v-(\epsilon+\Delta \epsilon)\right)f_A(v)dv
\ge & \beta\stackrel[\phi(\epsilon)]{\phi(\epsilon+\Delta \epsilon)}{\int}\frac{\delta_\beta}{2}f_A(v)dv  \ge \frac{\beta \delta_\beta \underline f_A}{2} \cdot [\phi(\epsilon+\Delta \epsilon)-\phi(\epsilon)]. \end{align*}

* We bound the first term of the right-hand side from above as follows:

\begin{align}
\beta\stackrel[\epsilon]{\epsilon + \Delta \epsilon}{\int}\left(v-\epsilon \right)f_A(v)dv \le \beta\stackrel[\epsilon]{\epsilon + \Delta \epsilon}{\int}1 \cdot f_A(v)dv \le \beta\Delta \epsilon \bar f_A.
\end{align}

* We bound the second term of the right-hand side as follows:

\begin{align}
\beta\stackrel[\epsilon + \Delta \epsilon]{\phi(\epsilon)}{\int}\Delta \epsilon f_A(v)dv \le \beta\stackrel[0]{1}{\int}\Delta \epsilon f_A(v)dv =\beta \Delta \epsilon
\end{align}

* We bound the third term of the right-hand side from above as follows:

\begin{align} \stackrel[\epsilon]{\epsilon+\Delta \epsilon}{\int}\left(v-\epsilon \right)f_B(v)dv \le \stackrel[\epsilon]{\epsilon+\Delta \epsilon}{\int}\Delta \epsilon f_B(v)dv \le \stackrel[0]{1}{\int}\Delta \epsilon f_B(v)dv =\Delta \epsilon.\end{align}

* We bound the fourth term of the right-hand side from above as follows:
\begin{align}
\stackrel[\epsilon+\Delta \epsilon]{w^{-1}_2(\epsilon)}{\int}\Delta \epsilon f_B(v)dv \le \stackrel[0]{1}{\int}\Delta \epsilon f_B(v)dv = \Delta \epsilon.\label{eq:ndc-equality-modified-w20-4} \end{align}

* We bound the fifth term of the right-hand side from above as follows:

\begin{align}
- \stackrel[w^{-1}_2(\epsilon)]{w^{-1}_2(\epsilon+\Delta \epsilon)}{\int}\left(v-(\epsilon+\Delta \epsilon) \right ) f_B(v)dv & \le - \stackrel[w^{-1}_2(\epsilon)]{w^{-1}_2(\epsilon+\Delta \epsilon)}{\int}\left(\epsilon-(\epsilon+\Delta \epsilon) \right ) f_B(v)dv \notag \\
& =
\stackrel[w^{-1}_2(\epsilon)]{w^{-1}_2(\epsilon+\Delta \epsilon)}{\int}\Delta \epsilon f_B(v)dv \notag \\
& \le  \stackrel[0]{1}{\int}\Delta \epsilon f_B(v)dv \notag \\
& \le \Delta \epsilon, \label{eq:fifth_term}
\end{align}

where the first inequality follows because $v \ge w^{-1}_2(\epsilon)$ for every $v$ in the range of the integration and $w^{-1}_2(\epsilon) \ge \epsilon$ for every $\epsilon$, so we have $v \ge \epsilon$.

Now, by (<ref>)–(<ref>), we obtain

\begin{align*}
&\frac{\beta \delta_\beta \underline f_A}{2} \cdot [\phi(\epsilon+\Delta \epsilon)-\phi(\epsilon)] \le \beta \Delta \epsilon \bar f_A + \beta \Delta \epsilon + \Delta \epsilon + \Delta \epsilon +\Delta \epsilon \notag \\
\Leftrightarrow &\phi(\epsilon+\Delta \epsilon)-\phi(\epsilon) \le    \frac{2}{\beta \delta_\beta \underline f_A} \left [ \beta \bar f_A + \beta + 3\right]\cdot \Delta \epsilon.
\end{align*}

Recall that the above bound has been obtained under the assumption that $\Delta \epsilon<\frac{\delta_\beta}{2}$. Now suppose that $\Delta\epsilon\geq \frac{\delta_\beta}{2}$. For any such $\Delta\epsilon$, we have $\phi(\epsilon+\Delta \epsilon)-\phi(\epsilon) \le 1 = \frac{\Delta \epsilon}{\Delta \epsilon} \le \frac{2}{\delta_\beta} \Delta \epsilon$. As shown in the Proof of <Ref>, $\phi(\cdot)$ is weakly increasing, so $\phi(\epsilon+\Delta\epsilon) - \phi(\epsilon) \ge 0$.

Combining the two cases, the function $\phi(\cdot)$ is Lipschitz continuous in $\epsilon$ with Lipschitz constant $\max\left \{\frac{2}{\beta \delta_\beta \underline f_A} \left [ \beta \bar f_A + \beta + 3\right],\frac{2}{\delta_\beta}\right\}=\frac{2}{\beta \delta_\beta} \left [ \beta \bar f_A + \beta + 3\right]$. By the Stepanov Theorem, this implies that $\frac{\partial\phi(\cdot)}{\partial \epsilon}$ exists for almost every $\epsilon\in [\underline \epsilon,1]$.

Let us temporarily write $\phi(\epsilon; \beta)$ to be explicit about the dependence of $\phi$ on $\beta$.
Fix $\underline \epsilon >0$. It suffices to show that for all $\epsilon\in[\underline \epsilon,1]$
, $\frac{\partial \phi(\epsilon;\beta)}{\partial\beta}$ exists.  Recall $\beta\geq 1$ and let $\Delta \beta>0$. We compare $\phi(\epsilon;\beta)$ to $\phi(\epsilon;\beta+\Delta \beta)$. By inspection of (<ref>), note that $\phi(\epsilon;\beta)$ is non-increasing in $\beta$.
Therefore, if $\phi(\epsilon;\beta+\Delta \beta)=1$, then $\phi(\epsilon;\beta)-\phi(\epsilon;\beta+\Delta \beta)=1-1=0,$ so there is nothing to show. So suppose $\phi(\epsilon;\beta+\Delta \beta)<1$. Then the No Desegregation Condition conditions for $\beta$ and $\beta+\Delta \beta$ are, respectively,

\begin{equation}
\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv \ge \beta\stackrel[\epsilon]{\phi(\epsilon;\beta)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w^{-1}_2(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv.\label{eq:ndc-equality-modified-w20-11}
\end{equation}

\begin{equation}
\stackrel[0]{1}{\int} [v-w_2(v)]  f_B(v)dv =  (\beta+\Delta \beta)\stackrel[\epsilon]{\phi(\epsilon;\beta+\Delta \beta)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w^{-1}_2(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv\label{eq:ndc-equality-modified-w20-12}.
\end{equation}

Subtracting (<ref>) from (<ref>) we obtain

\begin{align}
0 \ge   \beta\stackrel[\epsilon]{\phi(\epsilon;\beta)}{\int}\left(v-\epsilon\right)f_A(v)dv  -(\beta+\Delta \beta)\stackrel[\epsilon]{\phi(\epsilon;\beta+\Delta \beta)}{\int}\left(v-\epsilon\right)f_A(v)dv  \notag \\
\Leftrightarrow
\beta \stackrel[\phi(\epsilon;\beta+\Delta \beta)]{\phi(\epsilon;\beta)}{\int}\left(v-\epsilon\right)f_A(v)dv \le \Delta \beta \stackrel[\epsilon]{\phi(\epsilon;\beta+\Delta \beta)}{\int}\left(v-\epsilon\right)f_A(v)dv.
\label{eq:ndc-equality-modified-w20-13}
\end{align}

For each $\beta\geq 1$, define $\delta_\beta$ as in (<ref>). As shown in the proof of <Ref>, $\phi(\epsilon;\beta+\Delta \beta)-\epsilon>\delta_{\beta+\Delta\beta}$ for all $\epsilon \ge \underline \epsilon$. Therefore, the left-hand side of (<ref>) can be bounded from below by

\begin{align}\label{eq:57}
\beta \stackrel[\phi(\epsilon;\beta+\Delta \beta)]{\phi(\epsilon;\beta)}{\int}\left(v-\epsilon\right)f_A(v)dv \ge \beta \stackrel[\phi(\epsilon;\beta+\Delta \beta)]{\phi(\epsilon;\beta)}{\int} \delta_{\beta+\Delta\beta} f_A(v)dv \ge \beta \delta_{\beta+\Delta\beta} \underline f_A [\phi(\epsilon;\beta)-\phi(\epsilon;\beta+\Delta  \beta)].
\end{align}

Meanwhile, the right-hand side of (<ref>) can be bounded from above by

\begin{align}\label{eq:59.1}
\Delta \beta \stackrel[\epsilon]{\phi(\epsilon;\beta+\Delta \beta)}{\int}\left(v-\epsilon\right)f_A(v)dv \le \Delta \beta \stackrel[0]{1}{\int} 1 \cdot f_A(v)dv = \Delta \beta.
\end{align}

Therefore, from  (<ref>)-(<ref>) we conclude

\begin{align}\label{eq:stepanov}
\frac{\phi(\epsilon;\beta)-\phi(\epsilon;\beta+\Delta \beta)}{\Delta\beta} \le \frac{1}{\beta \delta_{\beta+\Delta\beta} \underline f_A}=\frac{1}{\beta \frac{\underline\epsilon^2 \underline f_B}{4(\beta+\Delta \beta)}\underline f_A}.
\end{align}

By (<ref>) and the fact that $\phi(\epsilon;\beta)-\phi(\epsilon;\beta+\Delta \beta) \ge 0$ for all $\Delta \beta>0,$ it follows that

\begin{align*}
\lim \sup _{\Delta \beta \to 0}\frac{\phi(\epsilon;\beta)-\phi(\epsilon;\beta+\Delta \beta)}{\Delta\beta} \in \left [0, \frac{1}{\frac{\underline\epsilon^2 \underline f_B}{4\beta}\underline f_A} \right ].
\end{align*}

By the Stepanov Theorem, this implies that $\frac{\partial\phi(\epsilon;\beta)}{\partial \beta}$ exists for almost every $\beta \in [1,\infty)$.

We begin this proof with a technical result.

$\frac{\partial \hat{w}_1(\hat w_1^{-1}(\epsilon))}{\partial v}$  is continuous in $\beta$ for almost all $\epsilon \in (0,E)$ and almost all $\beta\geq 1.$

We first note that $w_2^{-1}(\cdot)$ is absolutely continuous (in $\epsilon$) by assumption, and hence its partial derivative with respect to $\epsilon$ exists for almost all $\epsilon \in (0,1)$. Solving (<ref>) with equality, it follows that  $\beta\stackrel[\epsilon]{\hat w_{1}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{ w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv$ is constant in $\epsilon<E$, since $\hat{w}_1^{-1}(\epsilon)=\phi(\epsilon)$ by <Ref> for all $\epsilon<E$. This implies that for $\epsilon<E$

\[
\beta\left[(\hat{w}^{-1}_1(\epsilon)-\epsilon)f_A(\hat{w}^{-1}_1(\epsilon))\frac{\partial\hat{w}^{-1}_1(\epsilon)}{\partial \epsilon}-\stackrel[\epsilon]{\hat w_{1}^{-1}(\epsilon)}{\int}f_A(v)dv\right]+(w_{2}^{-1}(\epsilon)-\epsilon)f_B( w_2^{-1}(\epsilon))\frac{d w_{2}^{-1}(\epsilon)}{d \epsilon}-\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}f_B(v)dv=0.
\]

Simplifying yields

\[\frac{\partial\hat{w}_1^{-1}(\epsilon)}{\partial \epsilon}=\frac{F_B(w_{2}^{-1}(\epsilon))-F_B(\epsilon)-(w_{2}^{-1}(\epsilon)-\epsilon)f_B(w_2^{-1}(\epsilon))\frac{d w_{2}^{-1}(\epsilon)}{d \epsilon}+\beta[F_A(\hat{w}_1^{-1}(\epsilon))-F_A(\epsilon)]}{\beta[\hat{w}_1^{-1}(\epsilon)-\epsilon]f_A(\hat{w}_1^{-1}(\epsilon))},
\]

where we note that $\hat{w}_1^{-1}(\epsilon)-\epsilon >0$ for all $\epsilon \in (0,E)$.[The proof sketch is as follows. By <Ref> and (<ref>), we know that $\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv=\beta\stackrel[\epsilon]{\hat w_1^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_A(v)dv+\stackrel[\epsilon]{w^{-1}_2(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv>0$ for all $\epsilon<E$, where the inequality follows because we have excluded the case in which $w_2(v)=v$ for almost all $v\in[0,1]$. For contradiction, assume that there exists $\epsilon' \in (0,E)$ such that $\hat{w}_1^{-1}(\epsilon')=\epsilon'$. Then it must follow that
$\stackrel[\epsilon']{w^{-1}_2(\epsilon')}{\int}\left(v-\epsilon'\right)f_B(v)dv=\stackrel[0]{1}{\int} [v-w_2(v)] f_B(v)dv>0$. This equality is possible only if $w_2(v)=v$ for all $v <\epsilon'.$ Therefore, for this function $w_2(\cdot),$ $\stackrel[\epsilon'']{w^{-1}_2(\epsilon'')}{\int}\left(v-\epsilon''\right)f_B(v)dv=0$ for every $\epsilon''<\epsilon',$ a contradiction to the assumption that the term $\stackrel[\epsilon]{w_{2}^{-1}(\epsilon)}{\int}\left(v-\epsilon\right)f_B(v)dv$ is non-increasing in $\epsilon$.] Using this equality and the inverse function theorem tells us that

\begin{equation}
\frac{\partial \hat{w}_1(\hat{w}_1^{-1}(\epsilon))}{\partial v}=\frac{\beta[\hat{w}_1^{-1}(\epsilon)-\epsilon]f_A(\hat{w}_1^{-1}(\epsilon))}{F_B(w_{2}^{-1}(\epsilon))-F_B(\epsilon)-( w_{2}^{-1}(\epsilon)-\epsilon)f_B(w_2^{-1}(\epsilon))\frac{d w_{2}^{-1}(\epsilon)}{d \epsilon}+\beta[F_A(\hat{w}_1^{-1}(\epsilon))-F_A(\epsilon)]}\label{partial_epsilon_continuity}
\end{equation}

(<ref>) is continuous in $\beta$ because: 1) only the terms multiplied by $\beta$ are functions of $\beta$, 2) the numerator is continuous in $\beta$ because $\hat{w}^{-1}_1(\epsilon)$ is continuous in $\beta$ (which we can see by inspection of (<ref>)) and by the fact that $f_A(\cdot)$ is continuous, and 3) the denominator is continuous also because $\hat{w}^{-1}_1(\epsilon)$ is continuous in $\beta$ and because $F_A$ has no mass points.

We now return to the Proof of Part <ref> of <Ref>. Let us temporarily write $\hat{w}_1(\epsilon; \beta)$ and $\hat{w}^{-1}_1(\epsilon; \beta)$ to be explicit about the dependence of these functions on $\beta$.

Consider a sequence of real numbers, with generic element $\Delta\beta$, that converges to $0$.
Since $\hat{w}^{-1}_1(\epsilon; \beta)$ is differentiable with respect to $\beta$,

\begin{align*}
\hat{w}^{-1}_1(\epsilon;\beta+\Delta\beta)=\hat{w}^{-1}_1(\epsilon; \beta)+\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta+o(\Delta\beta).
\end{align*}

Letting $\psi(\Delta\beta):=-\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta-o(\Delta\beta)$, we get $\hat{w}^{-1}_1(\epsilon; \beta)=\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta)+\psi(\Delta\beta)$. Hence,

\begin{align}
\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta+\Delta\beta)=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta)+\psi(\Delta\beta); \beta+\Delta\beta).
\label{eq1.lemma8}
\end{align}

Since $\hat{w}_1(\epsilon; \beta)$ is differentiable with respect to $\epsilon$,

\begin{align}
\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta)+\psi(\Delta\beta); \beta+\Delta\beta)&=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta) \tag*{} \\
&+\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot \psi(\Delta\beta)+o(\psi(\Delta\beta)).
\label{eq2.lemma8}
\end{align}

Because $\hat{w}_1$ and $\hat{w}^{-1}_1$ are inverse functions over the relevant values of $\epsilon$ and $\beta$ it is the case that   $\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)=\epsilon=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)$. Therefore, together with (<ref>) and (<ref>),

\begin{align*}
\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta+\Delta\beta)=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)+\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot \psi(\Delta\beta)+o(\psi(\Delta\beta)).
\end{align*}

By substituting the definition of $\psi(\Delta\beta)$,[In the last line of the subsequent display equation, we have $\psi(\Delta\beta)$ in the denominator. Here we choose a subsequence of $\psi(\Delta\beta)$ so that $\psi(\Delta\beta)\neq 0$ for all $\Delta\beta$. More precisely, denoting the originally chosen sequence of $\Delta\beta$s as $\{\Delta\beta_k\}_{k=1}^\infty$, there exists a sufficiently large index $k'$ such that $\psi(\Delta\beta_k)\neq 0$ for all $k\geq k'$. To see this, suppose that, for any $k'$, there exists $k''\geq k'$ such that $\psi(\Delta\beta_{k''})=0$. Then, there exists $\{\Delta\beta_\ell\}_{\ell=1}^\infty \subseteq \{\Delta\beta_k\}_{k=1}^\infty$ such that $\Delta\beta_\ell\rightarrow 0$ and $\psi(\Delta\beta_\ell)=0$ for all $\ell$. By the definition of $\psi(\cdot)$, we get $\cfrac{o(\Delta\beta_\ell)}{\Delta\beta_\ell}=-\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}$. We obtain a contradiction to $\cfrac{o(\Delta\beta_\ell)}{\Delta\beta_\ell}\rightarrow 0$ as $\ell\rightarrow \infty$.]

\begin{align}
&\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta+\Delta\beta) \tag*{} \\
&=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)+\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot \Bigl(-\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta-o(\Delta\beta)\Bigr)+o(\psi(\Delta\beta)) \tag*{} \\
&=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot \cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot o(\Delta\beta) \tag*{} \\
&+o(\psi(\Delta\beta)) \tag*{} \\
&=\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot \cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot o(\Delta\beta) \tag*{} \\
&+\cfrac{o(\psi(\Delta\beta))}{\psi(\Delta\beta)}\cdot \psi(\Delta\beta). \tag*{}
\end{align}

It follows that

\begin{align*}
&\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta+\Delta\beta)-\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta) \\
&=\underbrace{-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot \cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta}_{(A)}
\underbrace{-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}\cdot o(\Delta\beta)}_{(B)}
\underbrace{+\cfrac{o(\psi(\Delta\beta))}{\psi(\Delta\beta)}\cdot \psi(\Delta\beta)}_{(C)}.
\end{align*}

Now, divide both sides by $\Delta\beta$ and let $\Delta\beta\rightarrow 0$. We consider each term.

Term (A):
Since $\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)}{\partial v}$ is continuous with respect to $\beta$ at almost all $(\hat{w}^{-1}_1(\epsilon; \beta); \beta)$ by <Ref>,

\begin{align*}
\lim_{\Delta\beta\rightarrow 0}\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}=\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)}{\partial v}.
\end{align*}

\begin{align*}
\lim_{\Delta\beta\rightarrow 0}\cfrac{(A)}{\Delta\beta}=-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)}{\partial v}\cdot \cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}.
\end{align*}

Term (B): As proven above, $\lim_{\Delta\beta\rightarrow 0}\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta+\Delta\beta); \beta+\Delta\beta)}{\partial v}=\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)}{\partial v}$. Together with $\lim_{\Delta\beta\rightarrow 0}\cfrac{o(\Delta\beta)}{\Delta\beta}=0$, it implies

\begin{align*}
\lim_{\Delta\beta\rightarrow 0}\cfrac{(B)}{\Delta\beta}=0.
\end{align*}

Term (C): Since $\psi(\Delta\beta)\rightarrow 0$ when $\Delta\beta\rightarrow 0$, we get

\begin{align}
\lim_{\Delta\beta\rightarrow 0}\cfrac{o(\psi(\Delta\beta))}{\psi(\Delta\beta)}=0.
\label{eq3}
\end{align}

\begin{align*}
\lim_{\Delta\beta\rightarrow 0}\cfrac{\psi(\Delta\beta)}{\Delta\beta}&=\lim_{\Delta\beta\rightarrow 0}\cfrac{-\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}\cdot \Delta\beta-o(\Delta\beta)}{\Delta\beta} \\
&=-\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}-\lim_{\Delta\beta\rightarrow 0}\cfrac{o(\Delta\beta)}{\Delta\beta} \\
&=-\cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}.
\end{align*}

Together with (<ref>), it implies

\begin{align*}
\lim_{\Delta\beta\rightarrow 0}\cfrac{(C)}{\Delta\beta}=0.
\end{align*}

Combining the conclusions regarding the three terms, we obtain that for almost all $\epsilon$ and almost all $\beta$

\begin{align*}
\lim_{\Delta\beta\rightarrow 0}\cfrac{\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta+\Delta\beta)-\hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta) }{\Delta\beta}=-\cfrac{\partial \hat{w}_1(\hat{w}^{-1}_1(\epsilon; \beta); \beta)}{\partial v}\cdot \cfrac{\partial \hat{w}^{-1}_1(\epsilon; \beta)}{\partial \beta}.
\end{align*}

This concludes the Proof of Part <ref>.

Next, we will show that $\hat w_1$ is Lipschitz continuous in $v \in [0,1)$. To do so, first note that $\hat w_1(v)$ is constant and equal to $0$ for $v \in [0,\phi(0)].$ So, consider $v \in [\phi(0),1).$ In that case, $\hat w_1(\cdot)$ is the inverse function of $\phi(\cdot)$ defined over $[0,E]$. We will show that there exists a constant $c>0$ such that $\phi(\epsilon_1)-\phi(\epsilon_2) \ge c (\epsilon_1-\epsilon_2)$ for every $\epsilon_1,\epsilon_2 \in [0,E]$ with $\epsilon_1>\epsilon_2$.

From the argument in the proof of <Ref>, it follows that for any $\beta<\bar \beta$ and any $\underline{\epsilon}>0$, $\phi(\epsilon)-\epsilon>\delta_\beta>\delta_{\bar \beta}>0$ for all $\epsilon\in[\underline\epsilon,1].$ By a similar argument and the assumption that
$w_2$ is continuous and is not the identity function, for any $\beta<\bar\beta$, there exists $c''>0$ such that $\phi(0)-0>c''.$ Letting $\underline \epsilon=c''/2$, the non-decreasing property of $\phi(\cdot)$ implies that $\phi(\epsilon)-\epsilon>c''/2$ for all $\epsilon\in[0,\underline \epsilon)$. Therefore, letting $c'=\min\{c''/2,\delta_{\bar\beta}\}$, $\phi(\epsilon)-\epsilon>c'$ for all $ \epsilon \in [0,1]$.