In the following, let us incorporate the dependency on the auxiliary distribution into the rate region notations, by writing $\mathscr{Q}(\Bs;\uv)$ and $\mathscr{Q}(\Bs,M,\Tc;\uv)$ in lieu of $\mathscr{Q}(\Bs)$ and $\mathscr{Q}(\Bs,M,\Tc)$, respectively. By Theorem <ref>, which thus far we have proven to hold for finitely supported auxiliaries, we can assert that $\mathscr{Q}(\Bs;\langle \uv \rangle_\tau)$ is achievable. Therefore, the inner set limit $\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs;\langle \uv \rangle_\tau)$ is achievable. Upon writing out $\mathscr{Q}(\Bs;\langle \uv \rangle_\tau)$ according to its definition (<ref>) as

\begin{equation}
\mathscr{Q}(\Bs;\langle \uv \rangle_\tau)
= \bigcap_{M \in \mathscr{M}^\circ_{\mathbb{Z}}(L_\Bf)} \bigcup_{\Sc \in \mathscr{B}(M^*)} \bigcap_{\Tc \in \mathscr{B}([\Bs]_\Sc)} \mathscr{Q}(\Bs,M,\Tc;\langle \uv \rangle_\tau)
\end{equation}

and observing that this expression involves a finite number of set operations that are independent of $\tau$, we can argue that the inner set limit can be exchanged with the other set operations, so as to yield

\begin{equation}   \label{inner_set_limit_of_rate_region_with_clipping}
\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs;\langle \uv \rangle_\tau)
= \bigcap_{M \in \mathscr{M}^\circ_{\mathbb{Z}}(L_\Bf)} \bigcup_{\Sc \in \mathscr{B}(M^*)} \bigcap_{\Tc \in \mathscr{B}([\Bs]_\Sc)} \varliminf_{\tau \to \infty} \mathscr{Q}(\Bs,M,\Tc;\langle \uv \rangle_\tau).
\end{equation}

The set on the right-hand side of the last equation can in turn be expressed as

\begin{multline}
\mathscr{Q}(\Bs,M,\Tc;\langle \uv \rangle_\tau)
= \Bigl\{ (R_1,\dotsc,R_K) \in \mathbb{R}_+^K \colon \sum_{k\in\Tc} R_k < \\
H([\langle \uv \rangle_\tau]_\Tc) + \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} H(\Cs\Bs\langle \uv \rangle_\tau|Y_\tau) - H(\Bs\langle \uv \rangle_\tau|Y_\tau) \Bigr\}
\end{multline}

where $Y_\tau$ denotes the channel output induced by an auxiliary $\langle \uv \rangle_\tau$. Specifically, conditioned on any given $\uv = \us$ and given a modulation mapping $\xs(\us) = (x_1(u_1),\ldots, x_K(u_K))$, the variable $Y_\tau$ is distributed as $Y_\tau \sim P_{Y|\xv}(\cdot|\xs(\langle\us\rangle_\tau))$.

Given some real matrix $\Qs \in \mathbb{R}^{m \times n}$, an integer random vector $\uv \in \mathbb{Z}^n$ with finite entropy $H(\uv)$ and any modulation mapping $\xs(\us)$, we have that, as $\tau \to \infty$, the entropy $H(\Qs \langle \uv \rangle_\tau | Y_\tau)$ converges to $H(\Qs\uv | Y)$ uniformly in $\Qs$, in the sense that

\begin{equation}
\bigl| H(\Qs \langle \uv \rangle_\tau | Y_\tau) - H(\Qs\uv | Y) \bigr|
\leq \delta(\tau)
\end{equation}

for all $\Qs \in \mathbb{R}^{m \times n}$, where the function $\delta(\tau)$ only depends on the distribution of $(\uv,Y)$ (but not on $\Qs$) and satisfies $\lim_{\tau \to \infty} \delta(\tau) = 0$.

By Lemma <ref>, we have the limits

lim_τ→∞ H([⟨⟩_τ]_)
=    H([]_)

lim_τ→∞ H(⟨⟩_τ|Y_τ)
=    H(|Y)

lim_τ→∞ H(⟨⟩_τ|Y_τ)
=    H(|Y)

all of which are finite because $H(\uv)$, which upper bounds all three of them, is finite by assumption. Here, establishing the last two equalities (<ref>)–(<ref>) additionally requires the Dominated Convergence Theorem <cit.>, which is applicable due to $H(\uv|Y=y)$ being an upper bound both on $H(\Cs\Bs\langle \uv \rangle_\tau|Y=y)$ and on $H(\Bs\langle \uv \rangle_\tau|Y=y)$, and this upper bound is integrable with respect to the distribution of $Y$, since $H(\uv|Y) \leq H(\uv) < \infty$ by assumption.
As regards the infimum term, by Lemma <ref> we know that $H(\Cs\Bs\langle \uv \rangle_\tau|Y)$ converges to $H(\Cs\Bs\uv|Y)$ uniformly in $\Cs$, as $\tau \to \infty$. It follows that the order of limit-taking and infimization can be exchanged, yielding

\begin{equation}   \label{clipping_limit_2_bis}
\lim_{\tau \to \infty} \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} H(\Cs\Bs\langle \uv \rangle_\tau|Y_\tau)
= \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} H(\Cs\Bs\uv|Y).
\end{equation}

Furthermore, by Lemma <ref> we can revert all entropies back to the algebraic entropy notation:

=    ([]_)

=    _(|Y)

=    _(|Y).

Combining (<ref>), (<ref>) and (<ref>), and using algebraic entropy notation (<ref>), we conclude that

\begin{equation}
\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs,M,\Tc;\langle \uv \rangle_\tau)
= \mathscr{Q}(\Bs,M,\Tc;\uv)
\end{equation}

and therefore, substituting this back into (<ref>), we finally end up with

\begin{equation}   \label{quod_erat_demonstrandum_2}
\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs;\langle \uv \rangle_\tau)
= \mathscr{Q}(\Bs;\uv).
\end{equation}

This finalizes the proof of Theorem <ref>.[Note that actually, inner and outer set limits in $\tau \to \infty$ coincide, so we could as well strengthen the convergence claim (<ref>) to $\lim_{\tau \to \infty} \mathscr{Q}(\Bs;\langle \uv \rangle_\tau) = \mathscr{Q}(\Bs;\uv)$.]

§ PROOF OF THEOREM <REF>

We fix the coefficient matrix $\As \in \mathbb{Z}^{L \times K}$, the channel law $P_{Y|X_1,\dotsc,X_K} = P_{Y|\Xv}$ (satisfying the technical conditions in the statement of Theorem <ref>), modulation mappings $\xs(\us) := (x_1(u_1), \dotsc, x_K(u_K))$ and the distribution of an auxiliary vector $\uv = \bigl[ U_1, \dotsc, U_K \bigr]^\T \in \mathbb{R}^K$ with mutually independent entries and an absolutely continuous distribution (with respect to the Lebesgue measure).

In what follows, let $\nu > 0$ stand for a quantization resolution parameter. We will now append a subscript `$\nu$' to the notations of rate regions and quantities defined in (<ref>), (<ref>), (<ref>) and Definition <ref>, so as to signify that these are generated by an integer auxiliary $\floor{\nu\uv} \in \mathbb{Z}^K$ and modulation mapping $\us \mapsto \xs(\tfrac{1}{\nu}\us)$. Specifically, we will write these as $\mathscr{R}_\nu(\As)$, $\mathscr{Q}_\nu(\Bs)$, $\mathscr{Q}_\nu(\Bs,M,\Tc)$ and $J_\nu(\Bs,M)$, all of which being evaluated as per Theorem <ref>.

Each member of the sequence of rate regions $\left\{ \mathscr{R}_\nu(\As) \right\}_{\nu \in \mathbb{N}}$ is achievable as per Theorem <ref>. The proof of Theorem <ref> consists in showing that this sequence has an inner limit, as $\nu$ tends to infinity, that contains the rate region $\mathscr{R}(\As)$ (evaluated as per Theorem <ref>) with the continuous auxiliary $\uv \in \mathbb{R}^K$ and modulation mapping $\xs(\uv)$. That is, we will show that

\begin{equation}   \label{quod_est_demonstrandum_1}
\mathscr{R}(\As)
\subseteq \varliminf_{\nu \to \infty} \mathscr{R}_\nu(\As).
\end{equation}

The channel output variable induced by the continuous auxiliary $\uv$ and modulation mapping $\uv \mapsto \xs(\uv)$ shall be denoted as $Y$, in accordance with the system model from Section <ref>. By contrast, the channel output induced by the quantized auxiliary $\floor{\nu\uv}$ and modulation mapping $\us \mapsto \xs(\tfrac{1}{\nu}\us)$ shall be denoted as $Y_\nu$. This means that the joint distributions of $(\uv,Y)$ and $(\uv,Y_\nu)$ are specified as follows: conditioned on $\uv=\us$, we have that $Y$ and $Y_\nu$ are distributed respectively as

\begin{align}
Y &\sim P_{Y|\Xv}(\cdot|\xs(\us))
Y_\nu &\sim P_{Y|\Xv}\left(\cdot \middle| \xs(\tfrac{1}{\nu}\floor{\nu\us})\right)
\end{align}

Thus far, we have merely fixed the conditional distributions $P_{Y|\uv=\us}$ and $P_{Y_\nu|\uv=\us}$ and thus, together with the given distribution of $\uv \sim P_{\uv}$, we have determined the joint distributions of the pairs $(\uv,Y)$ and $(\uv,Y_\nu)$, respectively. But we have not specified a joint distribution on the triple $(\uv,Y,Y_\nu)$. Since all informational quantities (entropies and mutual information) involved in the present proof only depend on either $(\uv,Y)$ or $(\uv,Y_\nu)$, we remain free to choose an arbitrary coupling between $Y$ and $Y_\nu$ as long as the marginals are given by the distributions of $(\uv,Y)$ and $(\uv,Y_\nu)$ specified above. This degree of freedom will be exploited in the proof of Lemma <ref> stated further below.

The proof follows a similar two-stage outline as that of Theorem <ref>: we initially prove the theorem under the assumption of a compactly supported auxiliary, to then relax this assumption in a second and final step. To start off, let us state three auxiliary lemmata that will be needed at a later stage.

Since $\mathscr{R}_\nu(\As) = \bigcup_\Bs \mathscr{Q}_\nu(\Bs)$ where the union runs over $\Lambda_\mathbb{Z}(\Bs) \supseteq \Lambda_\mathbb{Z}(\As)$, and due to (<ref>), in order to prove (<ref>) it will suffice to prove

\begin{equation}   \label{quod_est_demonstrandum_2}
\mathscr{Q}(\Bs)
\subseteq \varliminf_{\nu \to \infty} \mathscr{Q}_\nu(\Bs).
\end{equation}

Recall that $\mathscr{Q}_\nu(\Bs)$ is given by [cf. (<ref>)]
=    ⋂_M ∈ℳ_ℤ^∘(L_)   ⋃_∈ℬ(M^*)   ⋂_∈ℬ([]_) 𝒬_ν(,M,).

Since, on the right-hand side of (<ref>), the triple $(M,\Sc,\Tc)$ is iterated over a finite set that does not depend on $\nu$, it follows that the order of set limit (as $\nu \to \infty$) and the set operations (unions and intersections) can be exchanged, leading to
_ν→∞ 𝒬_ν()
=    ⋂_M ∈ℳ_ℤ^∘(L_)   ⋃_∈ℬ(M^*)   ⋂_∈ℬ([]_) _ν→∞ 𝒬_ν(,M,)

where $\mathscr{Q}_\nu(\Bs,M,\Tc)$ is given by [cf. (<ref>)]

\begin{equation}
\mathscr{Q}_\nu(\Bs,M,\Tc)
= \Bigl\{ (R_1,\dotsc,R_K) \in \mathbb{R}_+^K \colon
\sum_{k\in\Tc} R_k < H([\floor{\nu\uv}]_\Tc) - H(\Bs\floor{\nu\uv}|Y_\nu) + J_\nu(\Bs,M) \Bigr\}
\end{equation}

and where [cf. (<ref>)]

\begin{equation}   \label{J_quantized}
= \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} H(\Cs\Bs\floor{\nu\uv}|Y_\nu).
\end{equation}

It will thus suffice to prove
_ν→∞ 𝒬_ν(,M,)

⊇    𝒬(,M,)

=    { (R_1,…,R_K) ∈ℝ_+^K ∑_k∈ R_k < ([]_) - _(|Y) + J(,M) }.
For this purpose we will prove the following three limits:

lim_ν→∞ { H([ν]_) - (L_-L_) log(ν) }
=    h([]_)

lim_ν→∞ { H(ν|Y_ν) - L_log(ν) }
=    h(|Y)

lim inf_ν→∞ { J_ν(,M) - L_log(ν) }
≥    J(,M).

In summary, it suffices to prove (<ref>)–(<ref>) in order to prove Theorem <ref>.

The equality (<ref>) follows directly from Lemma <ref> applied to the right-invertible coefficient matrix $\Qs = [\Is]_\Tc \in \{0,1\}^{(L_\Bf-L_\Cf) \times K}$. As for (<ref>), the proof is more involved and will require several steps. We will first make use of the following lemma.

For any auxiliary variables $\uv \in \mathbb{R}^K$, modulation mappings $\xs(\uv)$ and a channel law $P_{Y|X_1,\dotsc,X_K}$ that all conform with the assumptions of Theorem <ref>, it holds that
lim_ν→∞ { H( ν | Y_ν) - H( ν | Y ) }
= 0

for any constant real-valued matrix $\Qs$. Furthermore, this convergence is uniform in $\Qs$, in that for any $\epsilon > 0$, there exists an $\nu_0(\epsilon) > 0$ such that for any $\nu > \nu_0(\epsilon)$ and any $\Qs$, we have

\begin{equation}
\bigl| H\bigl( \Qs \floor{\nu\uv} \big| Y_\nu \bigr) - H\bigl( \Qs \floor{\nu\uv} \big| Y \bigr) \bigr|
< \epsilon.
\end{equation}

See Appendix <ref>.

Lemma <ref> allows us to swap out $Y_\nu$ for $Y$ in the conditional entropy term on the left-hand side of (<ref>), yielding
lim_ν→∞ { H(ν|Y_ν) - L_log(ν) }
=    lim_ν→∞ { H(ν|Y) - L_log(ν) }

=    ∫_lim_ν→∞ { H( ν | Y=y ) - L_log(ν) } P_Y(y)

=    ∫_ℋ_(|Y=y) P_Y(y)

=    ℋ_(|Y).

The exchange of limit and integration carried out in the second equality of (<ref>) follows from the Dominated Convergence Theorem <cit.>, which is applicable because by Lemmata <ref> and <ref>, the integrand can be bounded from above by
H( ν | Y=y ) - L_log(ν)
≤    H( ν | Y=y ) + L_log( ‖‖_1/νL_ )

≤    H(  | Y=y ) + L_log( ‖‖_1/L_ )

and from below by
H( ν | Y=y ) - L_log(ν)
≥    H( ν | Y=y ) - L_log( ‖‖_1/νL_ )

≥    h(|Y=y) - L_log( ‖‖_1/L_ ).

Both the upper bound (<ref>) and lower bound (<ref>) are integrable in $y$ because $H(\floor{\Bs\uv}|Y)$ and $h(\Bs\uv|Y)$ are finite. The former is finite because $0 \leq H(\floor{\Bs\uv}|Y) \leq H(\floor{\Bs\uv})$ where $H(\floor{\Bs\uv})$ is finite by Lemma <ref>, whereas the latter equals $h(\Bs\uv|Y) = h(\Bs\uv) - I(\Bs\uv;Y)$, where $h(\Bs\uv)$ is finite by Lemma <ref> and $I(\Bs\uv;Y)$ is finite by the theorem's assumption of a finite sum mutual information $I(\uv;Y)$.

Now, only the inequality (<ref>) remains to be proven. Due to the uniform convergence stated in Lemma <ref>, we can swap out $Y_\nu$ for $Y$ in the limit on the left-hand side of (<ref>), i.e., if we define

\begin{equation}   \label{J_tilde_quantized}
\tilde{J}_\nu(\Bs,M)
\triangleq \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} H(\Cs\Bs\floor{\nu\uv}|Y)
\end{equation}

then $\lim_{\nu \to \infty} \bigl\{ J_\nu(\Bs,M) - \tilde{J}_\nu(\Bs,M) \bigr\} = 0$, hence

\begin{equation}   \label{liminf_J}
\liminf_{\nu \to \infty} \bigl\{ J_\nu(\Bs,M) - L_\Cf \log(\nu) \bigr\}
= \liminf_{\nu \to \infty} \bigl\{ \tilde{J}_\nu(\Bs,M) - L_\Cf \log(\nu) \bigr\}.
\end{equation}

Recall that the matrix $\Cs\Bs$ can be represented by Smith normal decomposition as [cf. Definition <ref>]

\begin{equation}   \label{SNF_of_CB}
\Cs\Bs = \Ss(\Cs\Bs) \Sigmas(\Cs\Bs) \Ts(\Cs\Bs)
\end{equation}

where $\Ss(\Cs\Bs) \in \mathbb{Z}^{L_\Cf \times L_\Cf}$ is unimodular, $\Sigmas(\Cs\Bs) \in \mathbb{Z}^{L_\Cf \times L_\Cf}$ is diagonal with positive diagonal entries $\sigma_1(\Cs\Bs),\dotsc,\sigma_{L_\Cf}(\Cs\Bs)$, and $\Ts(\Cs\Bs) \in \mathbb{Z}^{L_\Cf \times K}$ is right-invertible.

Let $\Omegas^\T \in \mathbb{Z}^{K \times L_\Cf}$ denote the matrix obtained from the Korkin–Zolotarev reduction of $\Ts(\Cs\Bs)^\T$. In other words, the columns of $\Omegas^\T$ are in Korkin–Zolotarev reduced form, and generate the same lattice as the columns of $\Ts(\Cs\Bs)^\T$. Consider partition cells

\begin{equation}
\Pc_\is
= \bigl\{ \us \in \mathbb{R}^K \colon \Omegas\floor{\us} = \is \bigr\}
\end{equation}

indexed by $\is \in \mathbb{Z}^{L_\Cf}$, which can be equivalently parametrized as
_   =    ⋃_∈ℤ^K-L_ { ∈ℝ^K  = ^♯+ _⊥}

=    ^♯+ Λ(_⊥) + [0,1)^K.

We have that the sets $\Pc_\is$ are mutually disjoint and $\{ \Pc_\is \colon \is \in \mathbb{Z}^{L_\Cf} \}$ constitutes a partition of $\mathbb{R}^K$, in that $\mathbb{R}^K = \bigcup_{\is \in \mathbb{Z}^{L_\Cf}} \Pc_\is$. Consider the entropy $H(\Cs\Bs\floor{\nu\uv}|Y)$, which can be written out as
=    H(ν|Y)

=    ∫_∑_∈ℤ^L_ Φ( { ν = | Y=y } ) P_Y(y)

=    ∫_∑_∈ℤ^L_ Φ( P_|Y=y(ν^-1_) ) P_Y(y)

where $\Phi(x) = -x\log(x)$. Let $\Qc_\nu$ denote the $\sigma$-algebra generated by the partition $\{ \nu^{-1}\Pc_\is \colon \is \in \mathbb{Z}^{L_\Cf} \}$. Then the operation $\us \mapsto \Omegas\floor{\nu\us}$ (i.e., scaling by $\nu$, followed by flooring and left-multiplication with $\Omegas$) can be interpreted as a quantization operation with respect to the atoms of $\Sigma_\nu$. Accordingly, we can write

\begin{equation}   \label{entropy_as_quantization}
= H(\Omegas\floor{\nu\uv}|Y)
= H(\lceil \uv \rfloor_{\Qc_\nu}|Y).
\end{equation}

Let us assume from now on that $\uv$ has compact support, with a support set contained in the cube $\tau \mathscr{S}_\infty^K$
and let $\tilde{\Qc}_\nu$ denote the $\sigma$-algebra generated by the quantization cells $\{ \nu^{-1}\Pc_\is \cap \tau \mathscr{S}_\infty^K \colon \is \in \mathbb{Z}^{L_\Cf} \}$. Then

\begin{equation}   \label{entropy_as_quantization_bis}
= H(\lceil \uv \rfloor_{\tilde{\Qc}_\nu}|Y).
\end{equation}

Next, we will upper-bound the volume of the $\is$-th quantization cell $\nu^{-1}\Pc_\is \cap \tau \mathscr{S}_\infty^K$.
For this purpose, we first outer-bound said quantization cell as follows:
=    ν^-1 ( ^♯+ Λ(_⊥) + [0,1)^K ) ∩τ𝒮_∞^K

⊆    ( ν^-1 ( ^♯+ Λ(_⊥) ) ∩(τ+ν^-1)𝒮_∞^K ) + [0,ν^-1)^K.

To see why this holds, bear in mind that a vector $\us$ belongs to the quantization cell $\nu^{-1}\Pc_\is \cap \tau\mathscr{S}_\infty^K$ if and only if there exists a pair of vectors $(\js,\ts) \in \mathbb{Z}^{K-L_\Cf} \times [0,1)^K$ such that $\us = \nu^{-1}(\Omegas^\sharp\is + \Omegas_\perp\js + \ts)$ and $\lVert \us \rVert_\infty \leq \tau$. The latter implies that $\nu^{-1} \lVert \Omegas^\sharp\is + \Omegas_\perp\js \rVert_\infty \leq \tau + \nu^{-1}$. Hence $\us = \us_1 + \us_2$ is the sum of a vector $\us_1 = \nu^{-1}(\Omegas^\sharp\is + \Omegas_\perp\js)$ satisfying $\us_1 \in \nu^{-1}(\Omegas^\sharp\is + \Lambda(\Omegas_\perp))$ as well as $\lVert \us_1 \rVert_\infty \leq \tau + \nu^{-1}$, and a vector $\us_2 = \nu^{-1}\ts$ satisfying $\us_2 \in [0,\nu^{-1})^K$. This means that $\us$ belongs to the set written out on the right-hand side of (<ref>). Hence with the help of the inclusion relation (<ref>), we can now upper-bound the volume of $\nu^{-1}\Pc_\is \cap \tau\mathscr{S}_\infty^K$ as follows:
=    ν^-K(_∩ντ𝒮_∞^K)

≤    ν^-K ( ( ( ^♯+ Λ(_⊥) ) ∩(ντ+1)𝒮_∞^K ) + [0,1)^K )

=    ν^-K | ( ^♯+ Λ(_⊥) ) ∩(ντ+1)𝒮_∞^K |

=    ν^-K | Λ(_⊥) ∩( (ντ+1)𝒮_∞^K - ^♯) |

≤    ν^-K | Λ(_⊥) ∩( 2ντ𝒮_∞^K - ^♯) |,

where the last inequality holds for $\nu\tau \geq 1$.
To further upper-bound the volume of quantization cells $\nu^{-1}\Pc_\is \cap \tau\mathscr{S}_\infty^K$, we will now elaborate an upper bound on the right-hand side of (<ref>) via the following lemma.

Let $\Fs = [ \fs_1 , \dotsc , \fs_L ] \in \mathbb{R}^{K \times L}$ denote a tall matrix with linearly independent columns, and let $\Gs = [ \gs_1 , \dotsc , \gs_L ] \in \mathbb{R}^{K \times L}$ denote a Gram–Schmidt orthogonalization of $\Fs$, with orthogonal columns $\gs_\ell, \ell=1,\dotsc,L$ constructed from successive orthogonal projections as follows:

=    _1

_ℓ   =    _ℓ- ∑_ℓ'=1^ℓ-1 _ℓ^_ℓ'/‖_ℓ' ‖_2^2 _ℓ',
$\ell = 2, \dotsc, L$.

The number of lattice points from $\Lambda(\Fs)$ that lie within a $K$-dimensional infinity-norm ball of radius $\rho$ and shifted by an arbitrary vector $\ts \in \mathbb{R}^K$ is upper-bounded as
| Λ() ∩(ρ𝒮_∞^K + ) |
≤    ∏_ℓ=1^L ⌈2√(K)ρ/‖_ℓ‖_2 ⌉.

Let us define the $L$-dimensional cube

\begin{equation}
\tilde{\mathscr{S}}_\infty^L
= \left\{ \sum_{\ell=1}^L \alpha_\ell \frac{\gs_\ell}{\lVert \gs_\ell \rVert_2} \colon (\alpha_1, \dotsc, \alpha_L)^\T \in [-1,1]^L \right\}
\subset \mathbb{R}^K
\end{equation}

which results from rotating the unit cube $\mathscr{S}_\infty^L = [-1,1]^L$ (embedded in $\mathbb{R}^K$ by zero-padding) about the origin to have its vertices aligned with the orthogonal basis $\Gs$. Alternatively, to represent $\tilde{\mathscr{S}}_\infty^L$ we may use the sumset notation

\begin{equation}
\tilde{\mathscr{S}}_\infty^L
= \sum_{\ell=1}^L [-1,1] \frac{\gs_\ell}{\lVert \gs_\ell \rVert_2}.
\end{equation}

We have the inclusion relations
Λ() ∩(ρ𝒮_∞^K + )
(a)⊂    () ∩(ρ𝒮_∞^K + )

(b)⊂    () ∩(ρ√(K)𝒮_2^K + )

(c)⊂    () ∩(ρ√(K)𝒮_2^K + (^)^-1^)

(d)⊂    () ∩(ρ√(K)𝒮̃_∞^L + (^)^-1^).

Here, Step $(a)$ is simply due to $\Lambda(\Fs) \subset \Span(\Fs) = \Span(\Gs)$; Step $(b)$ follows from $\mathscr{S}_\infty^K \subset \sqrt{K}\mathscr{S}_2^K$; Step $(c)$ holds because, if we decompose $\ts = \ts_\perp + \ts_{\|}$ into a vector $\ts_\perp = (\Is - \Gs(\Gs^\T\Gs)^{-1}\Gs^\T)\ts$ that is orthogonal to $\Span(\Gs)$ and a vector $\ts_{\|} = \Gs(\Gs^\T\Gs)^{-1}\Gs^\T\ts$ that lies in $\Span(\Gs)$, then the intersection of the $L$-dimensional plane $\Span(\Gs)$ and the $K$-dimensional sphere $\mathscr{S}_2^K$ shifted by $\ts$, is maximal when the orthogonal component $\ts_\perp$ is zero; Step $(d)$ follows because $\tilde{\mathscr{S}}_\infty^L \subset \Span(\Gs)$ includes all elements of $\Gs$ with Euclidean norm smaller or equal to unity. Intersecting both sides of (<ref>) with $\Lambda(\Fs)$ we end up with

\begin{equation}   \label{truncated_lattice_outer_bounding}
\Lambda(\Fs) \cap (\rho\mathscr{S}_\infty^K + \ts)
\subset \Lambda(\Fs) \cap (\rho\sqrt{K}\tilde{\mathscr{S}}_\infty^L + \ts_{\|})
\end{equation}

where $\ts_{\|}$ is an abbreviation for the projection of $\ts$ onto the column space of $\Gs$, as defined above. Hence, from (<ref>) we infer that

\begin{equation}   \label{truncated_lattice_cardinality_upper_bound}
\bigl| \Lambda(\Fs) \cap (\rho\mathscr{S}_\infty^K + \ts) \bigr|
\leq \bigl| \Lambda(\Fs) \cap (\rho\sqrt{K}\tilde{\mathscr{S}}_\infty^L + \ts_{\|}) \bigr|.
\end{equation}

Since the lattice $\Lambda(\Fs)$ can be expressed as an iterated union

\begin{equation}
\Lambda(\Fs)
= \bigcup_{\alpha_L \in \mathbb{Z}} \dotso \bigcup_{\alpha_1 \in \mathbb{Z}} \left\{ \sum_{\ell=1}^L \alpha_\ell \fs_\ell \right\}
\end{equation}

we can evaluate the cardinality $\bigl| \Lambda(\Fs) \cap (\rho\sqrt{K}\tilde{\mathscr{S}}_\infty^L + \ts_{\|}) \bigr|$ by an iterated sum as
| Λ() ∩(ρ√(K)𝒮̃_∞^L + _) |
=    ∑_α_L ∈ℤ …∑_α_1 ∈ℤ 1{ ∑_ℓ=1^L α_ℓ_ℓ∈(ρ√(K)𝒮̃_∞^L + _) }

where the indicator function $\mathds{1}\{\cdot\}$ equals one if the statement in braces is true, and zero otherwise. The iterated sum (<ref>) can be decomposed into partial summations
=    ∑_α_1 ∈ℤ 1{ ∑_ℓ=1^L α_ℓ_ℓ∈( ρ√(K)𝒮̃_∞^L + _ ) }

=    ∑_α_ℓ+1 ∈ℤ S_ℓ(α_ℓ+1,…,α_L),
for $\ell = 1,\dotsc,L-1$

such that the last sum $S_L = \sum_{\alpha_L \in \mathbb{Z}} S_{L-1}(\alpha_L)$ equals the cardinality $\bigl| \Lambda(\Fs) \cap (\rho\sqrt{K}\tilde{\mathscr{S}}_\infty^L + \ts_{\|}) \bigr|$, which we seek to upper-bound. We will now prove by induction that the following upper bound holds:

\begin{equation}   \label{upper_bound_for_induction}
\leq \prod_{k=1}^\ell \left\lceil \frac{2\rho\sqrt{K}}{\lVert \gs_k \rVert_2} \right\rceil \times \mathds{1}{\left\{ \sum_{\ell'=\ell+1}^L \alpha_{\ell'} \sum_{\ell''=\ell+1}^L \frac{\fs_{\ell'}^\T\gs_{\ell''}}{\lVert \gs_{\ell''} \rVert_2^2} \gs_{\ell''} \in \bigl( \rho\sqrt{K}\tilde{\mathscr{S}}_\infty^L + \ts_{\|} \bigr) \right\}}.
\end{equation}

For the first partial sum, we can prove the bound (<ref>) as follows:
(a)=    ∑_α_1 ∈ℤ 1{ ( α_1 + ∑_ℓ=2^L α_ℓ_ℓ^_1/‖_1 ‖_2^2 ) _1 + ∑_ℓ=2^L α_ℓ∑_ℓ'=2^L _ℓ^_ℓ'/‖_ℓ' ‖_2^2 _ℓ' ∈( ρ√(K)𝒮̃_∞^L + _ ) }

(b)=    ∑_α_1 ∈ℤ 1{ ( α_1 + ∑_ℓ=2^L α_ℓ_ℓ^_1/‖_1 ‖_2^2 ) _1 - _1^/‖_1 ‖_2^2 _1 ∈[-1,1] ρ√(K)_1/‖_1 ‖_2 } ×

×1{ ∑_ℓ=2^L α_ℓ∑_ℓ'=2^L _ℓ^_ℓ'/‖_ℓ' ‖_2^2 _ℓ' - ∑_ℓ'=2^L _ℓ'^/‖_ℓ' ‖_2^2 _ℓ' ∈∑_ℓ'=2^L [-1,1] ρ√(K)_ℓ'/‖_ℓ' ‖_2 }

(c)≤    ⌈2ρ√(K)/‖_1 ‖_2 ⌉1{ ∑_ℓ=2^L α_ℓ∑_ℓ'=2^L _ℓ^_ℓ'/‖_ℓ' ‖_2^2 _ℓ' ∈( ρ√(K)𝒮̃_∞^L + _ ) }.
Here, Step $(a)$ consists in expanding $\sum_{\ell=1}^L \alpha_\ell \fs_\ell$ along the basis $\Gs$; in Step $(b)$ we have exploited the orthogonality of $\Gs$ and the geometry of $\tilde{\Bs}_\infty^L$ to factorize the indicator function; in Step $(c)$ we have upper-bounded the first indicator function by the ceiling term.
By comparison with (<ref>), we confirm that the upper bound holds indeed for $\ell = 1$.
If the bound (<ref>) holds for some $\ell \geq 1$, we prove that it must also hold for $\ell+1$, as follows:

=    ∑_α_ℓ+1 ∈ℤ S_ℓ(α_ℓ+1,…,α_L)

≤    ∏_k=1^ℓ⌈2ρ√(K)/‖_k ‖_2 ⌉∑_α_ℓ+1 ∈ℤ 1{ ∑_ℓ'=ℓ+1^L α_ℓ' ∑_ℓ”=ℓ+1^L _ℓ'^_ℓ”/‖_ℓ” ‖_2^2 _ℓ” ∈( ρ√(K)𝒮̃_∞^L + _ ) }

(a)=    ∏_k=1^ℓ⌈2ρ√(K)/‖_k ‖_2 ⌉∑_α_ℓ+1 ∈ℤ 1{ α_ℓ+1 ( _ℓ+1 + ∑_ℓ”=ℓ+2^L _ℓ+1^_ℓ”/‖_ℓ” ‖_2^2 _ℓ” ) + ∑_ℓ'=ℓ+2^L α_ℓ' ∑_ℓ”=ℓ+1^L _ℓ'^_ℓ”/‖_ℓ” ‖_2^2 _ℓ”

∈( ρ√(K)𝒮̃_∞^L + _ ) }

(b)=    ∏_k=1^ℓ⌈2ρ√(K)/‖_k ‖_2 ⌉∑_α_ℓ+1 ∈ℤ 1{ α_ℓ+1 _ℓ+1 + ∑_ℓ'=ℓ+2^L α_ℓ' _ℓ'^_ℓ+1/‖_ℓ+1 ‖_2^2 _ℓ+1 - _ℓ+1^/‖_ℓ+1 ‖_2^2 _ℓ+1

∈[-1,1] ρ√(K)_ℓ+1/‖_ℓ+1 ‖_2 } ×

×1{ α_ℓ+1 ∑_ℓ”=ℓ+2^L _ℓ+1^_ℓ”/‖_ℓ” ‖_2^2 _ℓ” + ∑_ℓ'=ℓ+2^L α_ℓ' ∑_ℓ”=ℓ+2^L _ℓ'^_ℓ”/‖_ℓ” ‖_2^2 _ℓ” - ∑_ℓ”=ℓ+2^L _ℓ”^/‖_ℓ” ‖_2^2 _ℓ”

∈∑_ℓ”=ℓ+2^L [-1,1] ρ√(K)_ℓ”/‖_ℓ” ‖_2 }

(c)≤    ∏_k=1^ℓ+1 ⌈2ρ√(K)/‖_k ‖_2 ⌉1{ ∑_ℓ'=ℓ+1^L α_ℓ' ∑_ℓ”=ℓ+2^L _ℓ'^_ℓ”/‖_ℓ” ‖_2^2 _ℓ” - ∑_ℓ”=ℓ+2^L _ℓ”^/‖_ℓ” ‖_2^2 _ℓ”

∈∑_ℓ”=ℓ+2^L [-1,1] ρ√(K)_ℓ”/‖_ℓ” ‖_2 }

=    ∏_k=1^ℓ+1 ⌈2ρ√(K)/‖_k ‖_2 ⌉1{ ∑_ℓ'=ℓ+1^L α_ℓ' ∑_ℓ”=ℓ+2^L _ℓ'^_ℓ”/‖_ℓ” ‖_2^2 _ℓ” ∈( ρ√(K)𝒮̃_∞^L + _ ) }.
The steps are similar as before: in Step $(a)$ we expand along the basis $\Gs$; in Step $(b)$ we factorize the indicator function; in Step $(c)$ we upper-bound the first indicator function. This concludes the argument by induction. Therefore, the inequality (<ref>) holds for $\ell = L-1$, hence

\begin{equation}
= \bigl| \Lambda(\Fs) \cap (\rho\sqrt{K}\tilde{\mathscr{S}}_\infty^L + \ts_{\|}) \bigr|
\leq \prod_{k=1}^L \left\lceil \frac{2\rho\sqrt{K}}{\lVert \gs_k \rVert_2} \right\rceil
\end{equation}

which in combination with (<ref>) concludes the proof of Lemma <ref>.

Let $\Gs = [ \gs_1, \dotsc, \gs_{K-L_\Cf} ] \in \mathbb{R}^{K \times (K-L_\Cf)}$ denote the Gram–Schmidt orthogonalization of

\begin{equation}
\Omegas_\perp = \bigl[ \omegas_1^\perp, \dotsc, \omegas_{K-L_\Cf}^\perp \bigr] \in \mathbb{Z}^{K \times (K-L_\Cf)}.
\end{equation}

That is, $\gs_\ell$ is constructed from the vectors $(\omegas_1^\perp, \dotsc, \omegas_\ell^\perp)$ by successive projections as described in (<ref>)–(<ref>).
Applying Lemma <ref> on the lattice $\Lambda(\Omegas_\perp)$ intersected with a ball of radius $\rho$ and arbitrary shift $-\ts$, we get
| Λ(_⊥) ∩(ρ𝒮_∞^K - ) |
≤    ∏_ℓ=1^K-L_ ⌈2√(K) ρ/‖_ℓ‖_2 ⌉

=    (2√(K) ρ)^K-L_/√((^))
(for $2\sqrt{K} \rho \geq \max_{\ell \in [K-L_\Cf]} \lVert \gs_\ell \rVert_2$)

because $\prod_{\ell=1}^{K-L_\Cf} \lVert \gs_\ell \rVert_2^2 = \det(\Gs^\T\Gs) = \det(\Omegas_\perp^\T\Omegas_\perp) = \det(\Omegas\Omegas^\T)$.
Note that
max_ℓ∈[K-L_] ‖_ℓ‖_2
(a)≤    max_ℓ∈[K-L_] ‖_ℓ^⊥‖_2

(b)≤     max_ℓ∈[K-L_] √(ℓ+3/4) λ_2,ℓ(_⊥)

≤    √(K-L_+3/4) λ_2,K-L_(_⊥)

(c)≤    √(K-L_+3/4) ∏_ℓ=1^K-L_ λ_2,ℓ(_⊥)

(d)≤    √(K-L_+3/4) 2^K-L_ (_⊥^_⊥)

(e)≤    √(K) 2^K (^).

Here, Step $(a)$ holds because $\gs_\ell$ results from a (contractive) projection of $\omegas_\ell^\perp$, hence $\lVert \gs \rVert_2 \leq \lVert \omegas_\ell^\perp \rVert_2$ for all $\ell$; Step $(b)$ involves the fact that $\Omegas_\perp$ is in reduced Korkin–Zolotarev form and the inequality (<ref>); Step $(c)$ makes use of the fact that the successive minima of $\Omegas_\perp$ are lower-bounded by unity because $\Omegas_\perp$ is an integer lattice; Step $(d)$ involves Lemma <ref>; Step $(e)$ uses $\det(\Omegas_\perp^\T\Omegas_\perp) = \det(\Omegas\Omegas^\T)$ and $L_\Cf \geq 0$ as well as $3 \leq 3K$.
From this inequality we can deduce a sufficient condition for the inequality (<ref>) to hold, leading to a weaker version of (<ref>), namely
| Λ(_⊥) ∩(ρ𝒮_∞^K - ) |
≤    (2√(K) ρ)^K-L_/√((^))
(for $\rho \geq 2^{K-1} \det(\Omegas\Omegas^\T)$).

Evaluating (<ref>) for $\rho = 2\nu\tau$ and noticing that the condition $2\nu\tau \geq 2^{K-1} \det(\Omegas\Omegas^\T)$ in (<ref>) already implies the condition $\nu\tau \geq 1$ from (<ref>) (due to $K \geq 2$ and $\det(\Omegas\Omegas^\T) \geq 1$ since $\Omegas$ is an integer lattice), we can combine (<ref>) and (<ref>) to obtain
≤    (4√(K)τ)^K-L_/√((^)) ν^-L_
(for $\nu\tau \geq 2^{K-2} \det(\Omegas\Omegas^\T)$).

Based on (<ref>), Lemma <ref> and the bound (<ref>), we can state the following lower bound on the entropy $H(\Cs\Bs\floor{\nu\uv}|Y)$:

=    H(⌈⌋__ν|Y)

≥    h(|Y) - max_∈ℤ^L_ log((ν^-1_ ∩τ𝒮_∞^K))

≥    h(|Y) - log( (4√(K)τ)^K-L_/√((^)) ) + L_log(ν)
(for $\nu\tau \geq 2^{K-2} \det(\Omegas\Omegas^\T)$).

On the other hand, we can derive an upper bound on $H(\Cs\Bs\floor{\nu\uv}|Y)$ by means of Lemmata <ref> [Inequality (<ref>)] and <ref>, as follows:
H(ν|Y) - L_log(ν)
≤    H(ν) - L_log(ν) + L_log( ‖‖_1/L_ )

≤    H() + L_log( ‖‖_1/L_ ).

Let us define the quantity

\begin{equation}
= \min_{\Cs \in \mathscr{C}(M)} \left\{ H(\floor{\Cs\Bs\uv}) + L_\Cf \log{\left( \frac{\lVert \Cs\Bs \rVert_1}{L_\Cf} \right)} \right\}
\end{equation}

which is an upper bound on $\tilde{J}_\nu(\Bs,M) - L_\Cf\log(\nu)$.
In the optimization problem that defines the quantity $\tilde{J}_\nu(\Bs,M)$ [cf. (<ref>)], a matrix $\Cs \in \mathscr{C}(M)$ is strictly suboptimal, in the sense that $H(\Cs\Bs\floor{\nu\uv}|Y) > \tilde{J}_\nu(\Bs,M)$, if the following holds:

\begin{equation}   \label{sufficient_condition_for_suboptimality}
H(\Cs\Bs\floor{\nu\uv}|Y) - L_\Cf\log(\nu)
> H_0(\Bs,M).
\end{equation}

We will now derive a sufficient condition on $\Cs$ for being strictly suboptimal, so as to eventually reduce the domain of optimization in (<ref>) to a finite set. For this purpose, consider that (<ref>) certainly holds if [cf. (<ref>)] the following two conditions on $\nu$ and $\Omegas$ are satisfied:

h(|Y) - log( (4√(K)τ)^K-L_/√((^)) )
>    H_0(,M)

ντ   ≥    2^K-2 (^).

(Recall that $\Omegas$ is a function of $\Bs$ and $\Cs$).
As these two conditions (<ref>)–(<ref>) involve the determinant $\det(\Omegas\Omegas^\T)$, we now wish to translate these into sufficient conditions that involve $\det(\Cs\Cs^\T)$.
Consider the following inequality, which relates $\det(\Omegas\Omegas^\T)$ and $\det(\Cs\Cs^\T)$:
λ_min(^)^L_ (^)
(a)≤    (^^)

(b)=    (())^2 (()()^)

=    (())^2 (^)

(c)≤    (())^2 (^).

The inequality $(a)$ is obtained by upper-bounding the positive semidefinite matrix $\Bs\Bs^\T$ by the scaled identity matrix $\lambda_\mathrm{max}(\Bs\Bs^\T) \Is$, where $\lambda_\mathrm{max}(\Bs\Bs^\T) = \max_{\xs \in \mathbb{R}^{L_\Bf}} \xs^\T\Bs\Bs^\T \xs / \lVert \xs \rVert_2^2$ denotes the maximum eigenvalue of $\Bs\Bs^\T$; the equality $(b)$ results from applying the Smith normal decomposition (<ref>) and recalling that $\Ss(\Cs\Bs)$ is unimodular; Step $(c)$, in turn, is an application of the following auxiliary result.

For any integer matrix $\Bs \in \mathbb{Z}^{\ell \times k}$ with $\ell \leq k$ and full column rank, and for any right-invertible matrix $\Cs \in \mathbb{Z}^{r \times n}$ ($r \leq n$) we have $\det(\Sigmas(\Cs\Bs)) \leq \det(\Sigmas(\Bs))$.

Consider the Smith normal decomposition $\Cs = \tilde{\Ss}(\Cs) \tilde{\Sigmas}(\Cs) \tilde{\Ts}(\Cs)$ and the reduced Smith normal decomposition $\Bs = \Ss(\Bs) \Sigmas(\Bs) \Ts(\Bs)$. Since $\Cs$ is right-invertible by assumption, we have $\tilde{\Sigmas}(\Cs) = [\Is_{r \times r} \ \mynull_{r \times (n-r)}]$.
Since $\tilde{\Ts}(\Cs)$ and $\tilde{\Ss}(\Cs)$ are unimodular, on the one hand, $\det(\Sigmas(\Bs)) = \det(\tilde{\Ts}(\Cs)\Bs)$ and on the other hand, $\det(\Sigmas(\Cs\Bs)) = \det(\tilde{\Sigmas}(\Cs)\tilde{\Ts}(\Cs)\Bs) = \det([\tilde{\Ts}(\Cs)\Bs]_{[r]})$.
The product $\tilde{\Ts}(\Cs)\Bs \in \mathbb{Z}^{\ell \times k}$ has full column rank, much like $\Bs$.
Therefore, in the statement of Lemma <ref> we can assume without loss of generality that $\Cs$ is of the form $[\Is_{r \times r} \ \mynull_{r \times (n-r)}]$ and thus restate the lemma as the inequality

\begin{equation}
\det(\Sigmas([\Bs]_{[r]})) \leq \det(\Sigmas(\Bs)).
\end{equation}

To prove the latter, in turn, it will suffice to prove that, if $\Bs' \in \mathbb{Z}^{(\ell-1) \times k}$ denotes the matrix $\Bs$ with one row removed, we have

\begin{equation}
\det(\Sigmas(\Bs')) \leq \det(\Sigmas(\Bs)).
\end{equation}

Without loss of generality, let $\Bs' = [\Bs]_{[2{:}\ell]}$, that is, $\Bs'$ is obtained from removing the first row of $\Bs$. Let us denote the $\ell \times \ell$ submatrices of $\Bs$ as $\Bs_i \in \mathbb{Z}^{\ell \times \ell}$, of which there are $\binom{k}{\ell}$. As a direct consequence of the definition of elementary divisors (<ref>), the determinant $\det(\Sigmas(\Bs))$ is equal to the (positive) greatest common divisor of all $\ell \times \ell$ minors of $\Bs$, i.e.,

\begin{equation}
\det(\Sigmas(\Bs))
= \gcd\left( \det(\Bs_1), \det(\Bs_2), \dotsc \right).
\end{equation}

Let us now develop each of these determinants along the elements of the first row of $\Bs_i$, that is,

\begin{equation}
\det(\Bs_i)
= \sum_{j=1}^{\ell} (-1)^{j+1} [\Bs_i]_{1,j} \det([\Bs_i]_{\setminus 1, \setminus j})
\end{equation}

where $[\Bs_i]_{\setminus 1, \setminus j}$ denotes the matrix $\Bs_i$ with the first line and $j$-th column removed. Note that these submatrices $[\Bs_i]_{\setminus 1, \setminus j}$ are in fact $(\ell-1) \times (\ell-1)$ submatrices of $\Bs'$. Clearly, by iterating over all $(i,j) \in [\binom{\ell}{k}] \times [\ell-1]$, the submatrices $[\Bs_i]_{\setminus 1, \setminus j}$ will reach all submatrices of $\Bs'$ (each submatrix multiple times). Consequently, the greatest common divisor of the minors of $\Bs'$, which is

\begin{multline}
\det(\Sigmas(\Bs'))
= \gcd\Bigl( \det([\Bs_1]_{\setminus 1, \setminus 1}), \dotsc, \det([\Bs_1]_{\setminus 1, \setminus (\ell-1)}), \det([\Bs_2]_{\setminus 1, \setminus 1}), \dotsc \\
\dotsc, \det([\Bs_2]_{\setminus 1, \setminus (\ell-1)}), \dotsc \Bigr)
\end{multline}

divides every $\det(\Bs_i)$, and is therefore a divisor of $\det(\Sigmas(\Bs))$. Since $\Bs$ has full row rank, we have $\det(\Sigmas(\Bs)) \neq 0$ and therefore the claim $\det(\Sigmas(\Bs')) \leq \det(\Sigmas(\Bs))$ follows, thereby concluding the proof of Lemma <ref>.

To complement (<ref>), one can obtain inequality in reverse direction as follows:
λ_max(^)^L_ (^)
≥    (^^)

=    (())^2 (^)

≥    (^)

where the last inequality follows from $\Sigmas(\Cs\Bs)$ being an integer non-singular matrix.
Combining (<ref>)–(<ref>) with (<ref>) and (<ref>), and upon rearranging the terms in (<ref>), we obtain a new pair of sufficient conditions (on $\Cs$ and $\nu$) that together imply (<ref>)–(<ref>), namely

λ_min(^)^L_/2 √((^))/(())
>    (4√(K)τ)^K-L_ e^ H_0(,M) - h(|Y)

ντ   ≥    2^K-2 λ_max(^)^L_ (^).

In summary, we have shown that if $\Cs$ is such that $\det(\Cs\Cs^\T)$ is so large as to satisfy (<ref>), then for all $\nu$ large enough to satisfy (<ref>), the matrix $\Cs$ is not a solution to the minimization problem. In other words, matrices $\Cs$ satisfying (<ref>) can be excluded from the minimization domain, to such effect that the minimization domain can be reduced to a finite set and thus the order of limit-taking and minimization can be exchanged. Specifically, if we denote the constant

\begin{equation}   \label{def:C_0}
\triangleq \frac{(4\sqrt{K}\tau)^{K-L_\Cf} \det(\Sigmas(\Bs))}{\lambda_\mathrm{min}(\Bs\Bs^\T)^{L_\Cf/2}} e^{ H_0(\Bs,M) - h(\uv|Y) }
\end{equation}

then we have [cf. (<ref>)–(<ref>)]
lim inf_ν→∞ { J_ν(,M) - L_log(ν) }
=    lim inf_ν→∞ { J̃_ν(,M) - L_log(ν) }

=    lim inf_ν→∞ { inf_∈𝒞_ℤ(M) H(ν|Y) - L_log(ν) }

=    min_∈𝒞_ℤ(M) (^) ≤Θ lim inf_ν→∞ { H(ν|Y) - L_log(ν) }

=    min_∈𝒞_ℤ(M) (^) ≤Θ _(|Y)
for any $\Theta \geq C_0(\Bs,M)$. Taking the limit as $\Theta \to \infty$, we finally end up with
lim inf_ν→∞ { J_ν(,M) - L_log(ν) }
=    inf_∈𝒞_ℤ(M) _(|Y).

This completes the proof of (<ref>) and thus of Theorem <ref> for absolutely continuous auxiliaries of bounded support. To remove this limitation, we will now use a “truncation” argument, similar to the “clipping” argument used in the proof of Theorem <ref>. Based on an absolutely continuous auxiliary $\uv \in \mathbb{R}^K$ with distribution $P_{\uv}$, let us define the truncated auxiliary $\uv_\tau \in \mathbb{R}^K$ (for any $\tau$ sufficiently large to satisfy $P_{\uv}(\tau\mathscr{S}_\infty^K) > 0$) as

\begin{equation}   \label{truncated_auxiliary}
P_{\uv_\tau}(V) = \frac{P_{\uv}(V \cap \tau\mathscr{S}_\infty^K)}{P_{\uv}(\tau\mathscr{S}_\infty^K)}
\quad \text{(for all measurable $V$).}
\end{equation}

In other words, if $f_{\uv}$ and $f_{\uv_\tau}$ denote the probability density functions of $\uv$ and $\uv_\tau$, respectively, then $f_{\uv_\tau}$ is given by

\begin{equation}
f_{\uv_\tau}(\us) = \frac{f_{\uv}(\us)}{\int_{\tau\mathscr{S}_\infty^K} f_{\uv}(\us) \intd\us} \mathds{1}\bigl\{ \us \in \tau\mathscr{S}_\infty^K \bigr\}.
\end{equation}

As already done in the proof of Theorem <ref>, let us incorporate the dependency on the auxiliary distribution into the rate region notations, by writing $\mathscr{Q}(\Bs;\uv)$ and $\mathscr{Q}(\Bs,M,\Tc;\uv)$ in lieu of $\mathscr{Q}(\Bs)$ and $\mathscr{Q}(\Bs,M,\Tc)$, respectively. By Theorem <ref>, which thus far we have proven to hold for auxiliaries with bounded support, we can assert that $\mathscr{Q}(\Bs;\uv_\tau)$ is achievable. Therefore, the inner set limit $\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs;\uv_\tau)$ is achievable. Upon writing out $\mathscr{Q}(\Bs;\uv_\tau)$ according to its definition (<ref>) as

\begin{equation}
\mathscr{Q}(\Bs;\uv_\tau)
= \bigcap_{M \in \mathscr{M}^\circ_{\mathbb{Z}}(L_\Bf)} \bigcup_{\Sc \in \mathscr{B}(M^*)} \bigcap_{\Tc \in \mathscr{B}([\Bs]_\Sc)} \mathscr{Q}(\Bs,M,\Tc;\uv_\tau)
\end{equation}

and observing that this expression involves a finite number of set operations that are independent of $\tau$, we can argue that the inner set limit can be exchanged with the other set operations, so as to yield

\begin{equation}   \label{inner_set_limit_of_rate_region_with_truncation}
\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs;\uv_\tau)
= \bigcap_{M \in \mathscr{M}^\circ_{\mathbb{Z}}(L_\Bf)} \bigcup_{\Sc \in \mathscr{B}(M^*)} \bigcap_{\Tc \in \mathscr{B}([\Bs]_\Sc)} \varliminf_{\tau \to \infty} \mathscr{Q}(\Bs,M,\Tc;\uv_\tau).
\end{equation}

The set on the right-hand side of the last equation can in turn be expressed as [cf. (<ref>)]

\begin{multline}
\mathscr{Q}(\Bs,M,\Tc;\uv_\tau)
= \biggl\{ (R_1,\dotsc,R_K) \in \mathbb{R}_+^K \colon \sum_{k\in\Tc} R_k < \\
h([\uv_\tau]_\Tc) + \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} h(\Ts(\Cs\Bs)\uv_\tau|Y_\tau) - h(\Ts(\Bs)\uv_\tau|Y_\tau) \biggr\}
\end{multline}

where $Y_\tau$ denotes the channel output induced by the auxiliary $\uv_\tau$. Specifically, for any value $\us$ belonging to the support of $\uv_\tau$, conditioned on $\uv_\tau = \us$ and given a modulation mapping $\xs(\us)$, the variable $Y_\tau$ is distributed as $Y_\tau \sim P_{Y|\xv}(\cdot|\xs(\us))$.

Let $(\uv,Y) \in \mathbb{R}^K \times \Yc$ and $(\uv_\tau,Y_\tau) \in \mathbb{R}^K \times \Yc$ denote pairs of random variables as defined above, where $\uv \sim P_{\uv}$ is absolutely continuous with finite differential entropy $h(\uv)$ and finite entropy $H(\floor{\uv})$, and where $\uv_\tau \sim P_{\uv_\tau}$ denotes its truncated counterpart as defined in (<ref>).
Given some real matrix $\Qs \in \mathbb{R}^{M \times K}$ with full row rank, we have that, as $\tau \to \infty$, the entropy $h(\Qs\uv_\tau|Y_\tau)$ converges to $h(\Qs\uv|Y)$ uniformly in $\Qs$, in the sense that there exists an upper bound on the absolute value of the difference of differential entropies

\begin{equation}
\bigl| h(\Qs\uv_\tau|Y_\tau) - h(\Qs\uv|Y) \bigr|
\leq \delta(\tau)
\end{equation}

that does not depend on $\Qs$, and satisfies $\lim_{\tau \to \infty} \delta(\tau) = 0$.

By Lemma <ref>, we have the limits

lim_τ→∞ h([_τ]_)
=    h([]_)

lim_τ→∞ h(() _τ|Y_τ)
=    h(()|Y)

lim_τ→∞ h(() _τ|Y_τ)
=    h(()|Y).

Regarding (<ref>), since Lemma <ref> further asserts that the convergence is uniform, we can exchange the order of limit-taking and infimization can be exchanged, yielding

\begin{equation}   \label{truncation_limit_2_bis}
\lim_{\tau \to \infty} \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} h(\Ts(\Cs\Bs)\uv_\tau|Y_\tau)
= \inf_{\Cs \in \mathscr{C}_\mathbb{Z}(M)} h(\Ts(\Cs\Bs)\uv|Y).
\end{equation}

Furthermore, by Lemma <ref> we can revert all entropies back to the algebraic entropy notation:

=    ([]_)

=    _(|Y)

=    _(|Y).

Combining (<ref>), (<ref>) and (<ref>), and using algebraic entropy notation (<ref>), we conclude that

\begin{equation}
\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs,M,\Tc;\uv_\tau)
= \mathscr{Q}(\Bs,M,\Tc;\uv)
\end{equation}

and therefore, substituting back into (<ref>), we finally end up with

\begin{equation}   \label{quod_erat_demonstrandum_truncation}
\varliminf_{\tau \to \infty} \mathscr{Q}(\Bs;\uv_\tau)
= \mathscr{Q}(\Bs;\uv).
\end{equation}

This finalizes the proof of Theorem <ref>.

§ PROOF OF LEMMA <REF>

By <cit.>, we have

\begin{align*}
I(\floor{\nu\uv} ; \nu\uv-\floor{\nu\uv})
= \lim_{m\to\infty} I\bigl(\floor{\nu\uv}; \bigfloor{m(\nu\uv-\floor{\nu\uv})}\bigr).
\end{align*}

The mutual information $I\bigl(\floor{\nu\uv}; \bigfloor{m(\nu\uv-\floor{\nu\uv})}\bigr)$ can be upper-bounded as follows:
I(ν; m(ν-ν))

=    H(ν) + H(mν- mν) - H(mν- mν, ν)

(a)=    H(ν) + H(mν- mν) - H(mν)

(b)≤    H(ν) + nlog(m+1) - H(mν)

where step $(a)$ follows because there is a one-to-one correspondence between $\bigl(\bigfloor{m\nu\uv - m\floor{\nu\uv}}, \floor{\nu\uv}\bigr)$ and $\floor{m\nu\uv}$ and step $(b)$ follows because each of the $n$ coordinates of $\bigfloor{m\nu\uv - m\floor{\nu\uv}}$ takes at most $m+1$ values. Taking the limit as $\nu \to \infty$, we thus get an upper bound
I(ν ; ν- ν)
≤    H(ν) + lim_m→∞ { nlog(m+1) - H(mν) }

=    H(ν) - nlog(ν) - h().

Finally, taking the limit as $\nu \to \infty$ we obtain the desired result
lim_ν→∞ I( ν ; ν- ν )
=    0.

This concludes the proof.

§ PROOF OF LEMMA <REF>

The statement of Theorem <ref> distinguishes two mutually exclusive assumptions:

* Case 1: Staircase mappings $x_k(u_k)$ with finite images;

* Case 2: Linear mappings $x_k(u_k) = \beta_k u_k$ and a linear additive noise system equation.

The proof of this lemma requires distinct treatment for each of these assumptions, and will thus be structured in two parts.

§.§ Case 1: Staircase mappings with finite images

Let the $|\Xc_k|$ locations of jump discontinuities of the staircase function $x_k(u_k)$ be denoted as $\upsilon_1^{(k)} < \upsilon_2^{(k)} < \dotsc$ and let

\begin{equation}
\Upsilon_\nu^{(k)}
= \left[ \upsilon_1^{(k)} - \tfrac{1}{\nu} ; \upsilon_1^{(k)} + \tfrac{1}{\nu} \right] \cup \left[ \upsilon_2^{(k)} - \tfrac{1}{\nu} ; \upsilon_2^{(k)} + \tfrac{1}{\nu} \right] \cup \dotso
\end{equation}

denote the closed $1/\nu$-radius neighborhood of the set of discontinuity locations. Furthermore, let $\Upsilon_\nu = \Upsilon_\nu^{(1)} \times \dotso \times \Upsilon_\nu^{(K)}$ be the Cartesian product of these neighborhoods. Let us define the binary random variable

\begin{equation}
\begin{cases}
0   &   \text{if $\uv \notin \Upsilon_\nu$} \\
1   &   \text{if $\uv \in \Upsilon_\nu$}.
\end{cases}
\end{equation}

Note that $\lim_{\nu \to \infty} \P\{E_\nu = 0\} = 1$ as a consequence of $\uv$ being absolutely continuous and $\Upsilon_\nu$ having vanishing Lebesgue measure as $\nu \to \infty$. Moreover, note that the event $E_\nu = 0$ implies that $x(\floor{\uv}_\nu) = x(\uv)$. As a consequence, we can choose to couple the channel output variables $Y$ and $Y_\nu$ in a way that, conditioned on $E_\nu = 0$, we have $Y = Y_\nu$. Since mutual informations $I(E_\nu ; Y_\nu)$, $I(E_\nu ; Y_\nu | \Qs\floor{\uv}_\nu)$, $I(E_\nu ; Y)$ and $I(E_\nu ; Y | \Qs\floor{\uv}_\nu)$ are all upper-bounded by $H(E_\nu)$,
we infer that

| I(E_ν; Y_ν) - I(E_ν; Y_ν| _ν) |    ≤    H(E_ν),

| I(E_ν; Y) - I(E_ν; Y | _ν) |    ≤    H(E_ν)

whose left-hand sides tend to zero as $\nu \to \infty$ because of $\lim_{\nu \to \infty} H(E_\nu) = 0$. By the chain rule for mutual information, we have

I(E_ν; Y_ν) - I(E_ν; Y_ν| _ν)
=    I(_ν; Y_ν) - I(_ν; Y_ν| E_ν),

I(E_ν; Y) - I(E_ν; Y | _ν)
=    I(_ν; Y) - I(_ν; Y | E_ν).

Combining (<ref>) and (<ref>), we infer that

| I(_ν; Y_ν) - I(_ν; Y_ν| E_ν) |
≤    H(E_ν),

| I(_ν; Y) - I(_ν; Y | E_ν) |
≤    H(E_ν).

Let us now focus on the two conditional mutual information terms appearing in (<ref>), which can be written out respectively as

I(_ν; Y_ν| E_ν)
I(_ν; Y_ν| E_ν= 0) {E_ν= 0}

+ I(_ν; Y_ν| E_ν= 1) {E_ν= 1},

I(_ν; Y | E_ν)
I(_ν; Y | E_ν= 0) {E_ν= 0}

+ I(_ν; Y | E_ν= 1) {E_ν= 1}.

Due to the coupling between $Y$ and $Y_\nu$ that we have chosen, which imposes that $Y=Y_\nu$ whenever $E_\nu = 0$, it holds that $I(\Qs\floor{\uv}_\nu ; Y_\nu | E_\nu = 0) = I(\Qs\floor{\uv}_\nu ; Y | E_\nu = 0)$. Subtracting (<ref>) from (<ref>) and taking the absolute value, we obtain the equality

\begin{multline}
\Bigl| I(\Qs\floor{\uv}_\nu ; Y | E_\nu) - I(\Qs\floor{\uv}_\nu ; Y_\nu | E_\nu) \Bigr| \\
= \Bigl| I(\Qs\floor{\uv}_\nu ; Y | E_\nu = 1) - I(\Qs\floor{\uv}_\nu ; Y_\nu | E_\nu = 1) \Bigr| \cdot \P\{E_\nu = 1\}.   \label{MI_squeeze_final}
\end{multline}

As to the mutual information terms appearing on the right-hand side of (<ref>), they can be bounded as follows:

I(_ν; Y | E_ν= 1)
=    I(_ν; Y | ∈Υ_ν)

≤    I(; Y | ∈Υ_ν)

≤    sup_P_U I(;Y)

=    sup_P_X I(X;Y)

≤    log||

I(_ν; Y_ν| E_ν= 1)
=    I(_ν; Y_ν| ∈Υ_ν)

≤    I(_ν; Y_ν| ∈Υ_ν)

≤    sup_P_U I(;Y)

=    sup_P_X I(X;Y)

≤    log||

where $|\Xc| = \prod_{k=1}^K |\Xc_k|$. It follows that the left-hand side of (<ref>) is upper-bounded by $\log|\Xc| \cdot \P\{E_\nu = 1\}$, i.e.,

\begin{equation}   \label{MI_vanishing_limits_2}
\bigl| I(\Qs\floor{\uv}_\nu ; Y | E_\nu) - I(\Qs\floor{\uv}_\nu ; Y_\nu | E_\nu) \bigr|
\leq \log|\Xc| \cdot \P\{E_\nu = 1\}.
\end{equation}

Combining (<ref>) and (<ref>), we get
| H(_ν| Y) - H(_ν| Y_ν) |
=    | I(_ν; Y) - I(_ν; Y_ν) |

≤    log|| ·{E_ν= 1} + 2 H(E_ν).

Since this upper bound does not depend on $\Qs$ and vanishes as $\nu \to \infty$, this concludes the proof for Case 1.

§.§ Case 2: Linear mappings and linear additive noise channel

Now assume that we fix a linear mapping $x_k(u_k) = \beta_k u_k$. Let us define, conditioned on values $(U_1,\dotsc,U_K) = (u_1,\dotsc,u_K)$ (in short $\uv=u$) of the auxiliary variables, the outputs $Y$ and $Y_\nu$ as being

=    ∑_k=1^K h_k β_k u_k + Z

Y_ν   =    ∑_k=1^K h_k β_k u_k_ν+ Z

respectively, with real-valued weights $\beta_k$, where $Z$ is an independent (not necessarily Gaussian) additive noise. We choose the coupling of $Y$ and $Y_\nu$ to be such that they are generated by the same realization of $Z$. Specifically, this means that, conditioned on $\uv = u$, their difference is given by

\begin{equation}   \label{coupling}
Y - Y_\nu
= \sum_{k=1}^K h_k \beta_k (u_k - \floor{u_k}_\nu)
\end{equation}

and is thus a linear function of the quantization error vector

\begin{equation}
\dv_\nu
= \uv - \floor{\uv}_\nu.
\end{equation}

We first bound the entropy difference from above as follows:
H( _ν| Y_ν) - H( _ν| Y )
=    I( _ν; Y ) - I( _ν; Y_ν)

≤    I( _ν; Y , Y_ν, _ν) - I( _ν; Y_ν)

=    I( _ν; _ν| Y_ν)

≤    I( _ν; _ν| Y_ν)

≤    I( _ν; _ν).

The second equality is due to $Y$ being a deterministic function of $(Y_\nu,\dv_\nu)$; the second inequality is the data processing inequality; the third and last inequality holds because $\dv_\nu \rightarrow \uv \rightarrow Y_\nu$ forms a Markov chain.

Next, we bound the same entropy difference from below:
H( _ν| Y_ν) - H( _ν| Y )
=    I( _ν; Y ) - I( _ν; Y_ν)

≥    I( _ν; Y ) - I( _ν; _ν, Y )

=    -I( _ν; _ν| Y )

≥    -I( _ν; _ν| Y ).

Here, the first inequality is due to $Y_\nu$ being a deterministic function of $(\dv_\nu,Y)$; the second inequality is the data processing inequality.

The upper bound (<ref>) tends to zero as $\nu \to \infty$ as a consequence of Lemma <ref>. As for the lower bound (<ref>), we have

\begin{equation}
\lim_{\nu \to \infty} I\bigl( \floor{\uv}_\nu ; \dv_\nu \bigl| Y \bigr)
= \int_\Yc \lim_{\nu \to \infty} I\bigl( \floor{\uv}_\nu ; \dv_\nu \bigl| Y = y \bigr) P_Y(\mathrm{d}y)
= 0
\end{equation}

by the Dominated Convergence Theorem <cit.>. The latter can be applied because $I\bigl( \floor{\uv}_\nu ; \dv_\nu \bigl| Y = y \bigr)$ has an integrable upper bound. To derive such a bound, reproducing the steps in the proof of Lemma <ref> up to (<ref>) first leads to
I( _ν; _ν| Y=y )
≤    H( _ν| Y=y ) - K log(ν) - h(|Y=y).

Then, the discrete entropy on the right-hand side of (<ref>) can be upper-bounded as follows: let $k$ and $k'$ denote $K$-dimensional integer vectors and define

\begin{equation}
= \P\bigl\{ k + \tfrac{k'}{\nu} \leq \uv < k + \tfrac{k'+1}{\nu} \big| Y=y \bigr\}
\end{equation}

where inequalities on vectors are to be interpreted entrywise. Note that

\begin{equation}
\sum_{k' \in [0:\nu-1]^K} p_{k,k'}^\nu
= \P\bigl\{ k \leq \uv < k + 1 \big| Y=y \bigr\}.
\end{equation}

By the log-sum inequality,
H( _ν| Y=y )
=    ∑_k ∈ℤ^K ∑_k' ∈[0:ν-1]^K p_k,k'^νlog1/p_k,k'^ν

≤    ∑_k ∈ℤ^K ( ∑_k' ∈[0:ν-1]^K p_k,k'^ν) log( ν^K/∑_k' ∈[0:ν-1]^K p_k,k'^ν )

=    Klog(ν) + H(  | Y=y ).
Combining the latter with (<ref>), we obtain

\begin{equation}
I\bigl( \floor{\uv}_\nu ; \dv_\nu \big| Y=y \bigr)
\leq H\bigl( \floor{\uv} \big| Y=y \bigr) - h(\uv|Y=y).
\end{equation}

This bound is integrable due to $I(\uv;Y)$, $H(\floor{\uv})$ and $h(\uv)$ being finite.

In summary, the upper and lower bound (<ref>) and (<ref>) on the entropy difference $H\bigl( \Qs \floor{\uv}_\nu \big| Y_\nu \bigr) - H\bigl( \Qs \floor{\uv}_\nu \big| Y \bigr)$ both tend to zero. Since these bounds do not depend on $\Qs$, this convergence is uniform. This concludes the proof for Case 2.

§ PROOF OF THEOREM <REF>

This proof follows a similar logic as the proof of Theorem <ref> (although the absence of the minimum-entropy term $J(\Bs,M)$ drastically simplifies the proof).

We set $\mathbb{U} = \mathbb{Z}$ and fix the pmf $p_{\uv}(\us) = \prod_{k=1}^K p_{U_k}(u_k)$, the modulation mappings $\xs(\us)$ as well as the matrix $\Bs_{[j]} \in \mathbb{Z}^{j \times K}$. Let $Y_\q$ denote the channel output induced by the (finite-field) auxiliary $\phi_\q(\uv)$, in the context of Theorem <ref>. We prove Theorem <ref> in two steps: first, assuming that $\uv$ is finitely supported, and then, we relax this finiteness assumption.

If the support of $\uv$ is contained in the finite cube $\llbracket 2\tau + 1 \rrbracket^K$ for some $\tau > 0$, then for $\q > 2\tau + 1$ we have $H(\phi_\q([\uv]_\Tc))) = H([\uv]_\Tc)$. Furthermore, for $\q \geq (2\tau+1) \lVert \Bs_{[j]} \rVert_\infty$, there is a one-to-one correspondence between $\phi_\q(\Bs_{[j]}\uv)$ and $\Bs_{[j]}\uv$ such that all instances of $Y_\q$ can be replaced by $Y$, hence $H(\phi_\q(\Bs_{[j]}\uv)|Y_\q) = H(\Bs_{[j]}\uv|Y) = \Hc_{\Bs_{[j]}}(\uv|Y)$.
Consequently, the right-hand side of (<ref>), as evaluated in the context of Theorem <ref>, becomes equal to

\begin{multline}
H(\phi_\q(U_k)) + H(\phi_\q(\Bs_{[j-1]}\uv)|Y_\q) - H(\phi_\q(\Bs_{[j]}\uv)|Y_\q) \\
= H(U_k) + H(\Bs_{[j-1]}\uv|Y) - H(\Bs_{[j]}\uv|Y)
\quad \text{(for $\q \geq (2\tau+1) \lVert \Bs \rVert_\infty$).}
\end{multline}

This proves Theorem <ref> for finitely supported auxiliaries. To lift this restriction, now consider an auxiliary $\uv \in \mathbb{Z}^K$ with arbitrary support but finite entropy $H(\uv)$. Using the clipping operation $\langle \cdot \rangle_\tau$ as defined in (<ref>) and denoting as $Y_\tau$ the channel output induced by the clipped auxiliary $\langle \uv \rangle_\tau$, we have the limits

lim_τ→∞ H(⟨U_k ⟩_τ)
=    H(U_k)
= (U_k)

lim_τ→∞ H(_[j-1]⟨⟩_τ|Y_τ)
=    H(_[j-1]|Y)
= __[j-1](|Y)

lim_τ→∞ H(_[j]⟨⟩_τ|Y_τ)
=    H(_[j]|Y)
= __[j](|Y).

Here, the last two limits follow from Lemma <ref>. This concludes the proof of Theorem <ref>.

§ PROOF OF THEOREM <REF>

We set $(\mathbb{U},\mathbb{A}) = (\mathbb{R},\mathbb{Z})$ and fix the pdf $f_{\uv}(\us) = \prod_{k=1}^K f_{U_k}(u_k)$, the modulation mappings $\xs(\us)$ as well as the matrix $\Bs_{[j]} \in \mathbb{Z}^{j \times K}$. Let $Y_\nu$ denote the channel output induced by the integer-valued auxiliary $\lfloor \nu\uv \rfloor$, in the context of Theorem <ref>.

By Lemma <ref>, we have

lim_ν→∞ { H(_[j-1]⌊ν⌋|Y_ν) - H(_[j-1]⌊ν⌋|Y) }
=    0

lim_ν→∞ { H(_[j]⌊ν⌋|Y_ν) - H(_[j]⌊ν⌋|Y) }
=    0.

Hence, $Y_\nu$ can be swapped out for $Y$ in these limits, so we obtain

lim_ν→∞ H(⌊νU_k ⌋)
=    h(U_k)
= (U_k)

lim_ν→∞ H(_[j-1]⌊ν⌋| Y_τ)
=    h((_[j-1])|Y)
= __[j-1](|Y)

lim_ν→∞ H(_[j]⌊ν⌋| Y_τ)
=    h((_[j])|Y)
= __[j](|Y).

This concludes the proof of Theorem <ref>.

§ EVALUATION OF $\MATHSCR{R}(\AS)$ FOR $(K,L)=(2,2)$

We evaluate $\mathscr{R}(\As) = \bigcup_{\Bs} \mathscr{Q}(\Bs)$ for the special case $K = L = 2$.
Let us denote $\Bs = \begin{bsmallmatrix} b_{2,1} & b_{1,2} \\ b_{2,1} & b_{2,2} \end{bsmallmatrix} = \begin{bsmallmatrix} \bs_1^\T \\ \bs_2^\T \end{bsmallmatrix}$. We distinguish two cases based on the rank of the matroid $M$.

The rank-$0$ matroid $M = (\{1,2\},\emptyset)$ In this case, the index sets $\Sc$ and $\Tc$ are equal to $\Sc = \Tc = \{1,2\}$, hence we obtain the sum-rate bound
R_1 + R_2
<    () - _(|Y)

=    () - (|Y)

=    I(;Y)

=    I(;Y).

The rank-$1$ matroids

Since $\Bs$ has full rank, either the diagonal entries $(b_{1,1},b_{2,2})$ or the off-diagonal entries $(b_{1,2},b_{2,1})$ are both non-zero, for otherwise $\Bs$ would have an all-zero row or column. Since we can freely permute the rows of $\Bs$ without affecting the value of $\mathscr{Q}(\Bs)$, we shall assume without loss of generality that the diagonal entries $(b_{1,1},b_{2,2})$ are non-zero. In the following, we make case distinctions based on the off-diagonals:

* The off-diagonals are $b_{1,2} = b_{2,1} = 0$:

* Case $M = (\{1,2\},\{1\})$: the index sets are equal to $\Sc = \Tc = \{2\}$, hence
<    (U_2) - (|Y) + (U_1|Y)

=    I(U_2;Y,U_1).

* Case $M = (\{1,2\},\{2\})$: the index sets are equal to $\Sc = \Tc = \{1\}$, hence
<    (U_1) - (|Y) + (U_2|Y)

=    I(U_1;Y,U_2).

* Case $M = (\{1,2\},\{1\},\{2\})$: the index set $\Sc$ is iterated over $\Sc \in \{\{1\},\{2\}\}$:

* For $\Sc = \{1\}$, the index set $\Tc$ is equal to $\Tc = \{1\}$, hence
<    (U_1) - (|Y) + inf_c_1 ≠0,  c_2 ≠0 _[ c_1 b_1,1  c_2 b_2,2 ](|Y).

* For $\Sc = \{2\}$, the index set $\Tc$ is equal to $\Tc = \{2\}$, hence
<    (U_2) - (|Y) + inf_c_1 ≠0,  c_2 ≠0 _[ c_1 b_1,1  c_2 b_2,2 ](|Y).

Since algebraic entropy is invariant against scaling of the coefficient matrix, in the infimum term in (<ref>)–(<ref>), the coefficients $(b_{1,1},b_{2,2})$ can be set to $(1,1)$ without loss of generality. In other words, we can set $\Bs$ to the identity matrix.
The inequalities (<ref>)–(<ref>) are to be combined by a logical `or' (due to union-taking over $\Sc$), and the result is to be combined with (<ref>) and (<ref>) by a logical `and' operation (due to the intersection over matroids), hence we obtain, all in all,

<    I(U_1;Y,U_2)

<    I(U_2;Y,U_1)

min{ R_1 - (U_1) , R_2 - (U_2) }
<    - (|Y) + inf_c_1 ≠0,  c_2 ≠0 _[ c_1  c_2 ](|Y).

* The off-diagonals are $b_{1,2} \neq 0$ and $b_{2,1} = 0$:

* Case $M = (\{1,2\},\{1\})$: the index sets are equal to $\Sc = \Tc = \{2\}$, hence we obtain the rate bound
R_2    <    (U_2) - (|Y) + __1^(|Y).

* Case $M = (\{1,2\},\{2\})$: The index set $\Sc$ is equal to $\Sc = \{1\}$, and $\Tc$ is iterated over $\Tc \in \{\{1\},\{2\}\}$, hence we obtain the pair of rate bounds

<    (U_1) - (|Y) + (U_2|Y)
= I(U_1;Y,U_2)

<    (U_2) - (|Y) + (U_2|Y).

where the equality in (<ref>) follows from Lemmata <ref> and <ref>.

* Case $M = (\{1,2\},\{1\},\{2\})$: the index set $\Sc$ is iterated over $\Sc \in \{\{1\},\{2\}\}$. We make further case distinctions:

* For $\Sc = \{1\}$, the index set $\Tc$ is iterated over $\Tc \in \{\{1\},\{2\}\}$, hence we obtain the pair of inequalities

<    (U_1) - (|Y) + inf_c_1 ≠0,  c_2 ≠0 _[c_1 b_1,1  c_1 b_1,2 + c_2 b_2,2](|Y)

<    (U_2) - (|Y) + inf_c_1 ≠0,  c_2 ≠0 _[c_1 b_1,1  c_1 b_1,2 + c_2 b_2,2](|Y).

* For $\Sc = \{2\}$, the index set $\Tc$ is equal to $\Tc = \{2\}$, hence we obtain once more the inequality (<ref>). The pair of inequalities (<ref>)–(<ref>) is thus combined by a logical `or' with (<ref>), hence what remains from this combination is only (<ref>).

The above three case distinctions on the matroid $M$ are to be combined by a logical `and' (due to intersections over matroids), thus leading to a set of four rate inequalities (<ref>), (<ref>)–(<ref>) and (<ref>) (since (<ref>) was removed by union-taking over $\Sc$) that need to be simultaneously satisfied. Upon combining these four inequalities, we can reduce them to just two:

<    I(U_1;Y,U_2)

<    (U_2) - (|Y) + inf_(c_1,c_2) ≠(0,0) _[c_1  c_2](|Y).

To see why (<ref>), (<ref>) and (<ref>) can be combined to (<ref>), observe that the infimum term in (<ref>) can be simply expressed as $\inf_{(c_1,c_2) \colon c_1 \neq 0} \Hc_{[c_1 \ c_2]}(\uv|Y)$.

* The off-diagonals are $b_{1,2} = 0$ and $b_{2,1} \neq 0$:

The derivations in this case are analogous to the previous case of $b_{1,2} \neq 0$ and $b_{2,1} = 0$, resulting in rate inequalities like (<ref>)–(<ref>), but with user indices swapped. That is,

<    (U_1) - (|Y) + inf_(c_1,c_2) ≠(0,0) _[c_1  c_2](|Y)

<    I(U_2;Y,U_1).

* The off-diagonals are $b_{1,2} \neq 0$ and $b_{2,1} \neq 0$:

* Case $M = (\{1,2\},\{1\})$: the index set $\Sc$ is equal to $\Sc = \{2\}$, and $\Tc$ is iterated over $\Tc \in \{\{1\},\{2\}\}$, hence we obtain the pair of rate bounds

<    (U_1) - (|Y) + __1^(|Y)

<    (U_2) - (|Y) + __1^(|Y).

* Case $M = (\{1,2\},\{2\})$: the index set $\Sc$ is equal to $\Sc = \{1\}$, and $\Tc$ is iterated over $\Tc \in \{\{1\},\{2\}\}$, hence we obtain the pair of rate bounds

<    (U_1) - (|Y) + __2^(|Y)

<    (U_2) - (|Y) + __2^(|Y).

* Case $M = (\{1,2\},\{1\},\{2\})$: the index set $\Sc$ is iterated over $\Sc \in \{\{1\},\{2\}\}$:

* For $\Sc = \{1\}$, the index set $\Tc$ is iterated over $\Tc \in \{\{1\},\{2\}\}$, hence we obtain the pair of inequalities

<    (U_1) - (|Y) + inf_c_1 ≠0, c_2 ≠0 _(|Y)

<    (U_2) - (|Y) + inf_c_1 ≠0, c_2 ≠0 _(|Y).

* For $\Sc = \{2\}$, the index set $\Tc$ is also iterated over $\Tc \in \{\{1\},\{2\}\}$ and we obtain the same inequalities as (<ref>)–(<ref>).

The above three case distinctions on the matroid $M$ are to be combined by a logical `and' (due to the intersection over matroids), leading to

<    (U_1) - (|Y) + inf_(c_1,c_2) ≠(0,0) _[c_1  c_2](|Y)

<    (U_2) - (|Y) + inf_(c_1,c_2) ≠(0,0) _[c_1  c_2](|Y).

Due to the union-taking over matrices $\Bs$, the above four case distinctions on the off-diagonals of $\Bs$ have to be combined by a logical `or'. Namely, the four groups of inequalities (<ref>)–(<ref>), (<ref>)–(<ref>), (<ref>)–(<ref>) and (<ref>)–(<ref>) have to be combined by logical `or'. As we shall see, the group of inequalities (<ref>)–(<ref>) (corresponding to a diagonal matrix $\Bs$) are implied by either of the latter three, that is, by either (<ref>)–(<ref>), (<ref>)–(<ref>) or (<ref>)–(<ref>). To confirm this, we first argue that the two side bounds (<ref>) and (<ref>) are implied by any of the aforementioned inequality pairs. This is because, for any $(c_1,c_2)$, we have

(U_1) - (|Y) + inf_(c_1,c_2) ≠(0,0) _[c_1  c_2](|Y)
≤    I(U_1;Y,U_2)

(U_2) - (|Y) + inf_(c_1,c_2) ≠(0,0) _[c_1  c_2](|Y)
≤    I(U_2;Y,U_1).

These two bounds are obtained when upper-bounding the infimum terms on the left-hand sides of (<ref>) and (<ref>) by setting $c_1=0$ and $c_2 \neq 0$ in the former, and $c_1 \neq 0$ and $c_2=0$ in the latter, respectively. As regards the third and last inequality (<ref>), clearly it is implied by either of the inequalities (<ref>), (<ref>), (<ref>) or (<ref>), because $\inf_{(c_1,c_2) \neq (0,0)} \Hc_{[c_1 \ c_2]}(\uv|Y) \leq \inf_{c_1 \neq 0,\, c_2 \neq 0} \Hc_{[c_1 \ c_2]}(\uv|Y)$. We conclude that in the union over $\Bs$, it suffices to consider diagonal matrices $\Bs$, which in turn can be reduced to the identity $\Bs=\Is$. Combining the rate inequalities (<ref>)–(<ref>) with the sum rate bound (<ref>), we finally obtain the rate region

R_1 + R_2
<    I(U_1,U_2;Y)

<    I(U_1;Y,U_2)

<    I(U_2;Y,U_1)

min{ R_1 - (U_1) , R_2 - (U_2) }
<    - (|Y) + inf_c_1 ≠0,  c_2 ≠0 _[ c_1  c_2 ](|Y).

§ EVALUATION OF $\MATHSCR{R}(\AS)$ FOR $(K,L)=(2,1)$

We evaluate $\mathscr{R}(\As) = \bigcup_{\Bs} \mathscr{Q}(\Bs)$ for the special case $K=2$, $L=1$. The coefficient matrix $\As = \as^\T = [a_1 \ a_2] \in \mathbb{A}^{1 \times 2}$ with $a_1 \neq 0$ and $a_2 \neq 0$ is a one-by-two row vector with non-zero entries. Since in the union operation in (<ref>), the matrix $\Bs$ is iterated over all matrices satisfying $\Lambda_\mathbb{A}(\Bs) \supset \Lambda_\mathbb{A}(\As)$, we infer that $\Bs$ must run over all full-rank matrices $\Bs \in \mathbb{A}^{2 \times 2}$ as well as over all those $\Bs \in \mathbb{A}^{1 \times 2}$ that are scalar multiples of $\as^\T$ (for which case it suffices to consider $\Bs=\as^\T$).
We make a case distinction based on the rank of $\Bs$.

§.§.§ The matrix $\Bs$ has rank $1$

In this case, $\Bs$ is a non-zero multiple of the coefficient vector $\as^\T$. Due to invariance of $\mathscr{Q}(\Bs)$ against scalar multiplication of $\Bs$, if will suffice to set $\Bs = \as^\T$. By virtue of the constraints on $(M, \Sc, \Tc)$ laid out under Equation (<ref>), we have that

* the matroid $M$ is the empty matroid $(\{1,2\},\emptyset)$;

* the index set $\Sc$ is the singleton set $\{1\}$;

* the index set $\Tc$ is iterated over $\{\{1\},\{2\}\}$.

The resulting rate region is thus
≜{ (R_1, R_2) ∈ℝ_+^2 R_1    <    (U_1) - _^(|Y)

R_2    <    (U_2) - _^(|Y) }.

§.§.§ The matrix $\Bs$ has rank $2$

For this case, the union $\bigcup_{\Bs} \mathscr{Q}(\Bs)$ was already computed in Appendix <ref> and shown to be equal to $\mathscr{Q}(\Is)$, which is described by inequalities (<ref>)–(<ref>).

§ PROOF OF LEMMA <REF>

As a first step, we show that if the inequality (<ref>) holds, then $\as^\T$ is also the only vector (with two non-zero entries) satisfying said inequality. As a consequence, $\as^\T$ is then necessarily the unique minimizer that achieves the infimum in (<ref>)–(<ref>).

To prove this, let $\hat{\as}^\T = [ \hat{a}_1 \ \hat{a}_2 ] \in \mathbb{A}^2 $ denote some other coefficient vector with non-zero entries, that is linearly independent of $\as^\T = [ a_1 \ a_2 ]$. Then we have
=    lim_ν→∞ H(⌊ν⌋|Y)/log(ν)

=    lim_ν→∞ { H(a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋|Y)/log(ν) + H(â_1 ⌊νU_1⌋+ â_2 ⌊νU_2⌋| Y, a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋)/log(ν) }

=    d_^(|Y) + d_^(|Y) - lim_ν→∞ I(â_1 ⌊νU_1⌋+ â_2 ⌊νU_2⌋; a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋| Y)/log(ν)

=    d_^(|Y) + d_^(|Y).
To prove the last equality, we argue that the mutual information $I(\hat{a}_1 \lfloor\nu U_1\rfloor + \hat{a}_2 \lfloor\nu U_2\rfloor ; a_1 \lfloor\nu U_1\rfloor + a_2 \lfloor\nu U_2\rfloor | Y)$ has a finite upper bound that is independent of $\nu$. In fact, by Lemma <ref> and the data-processing inequality,
I(â_1 ⌊νU_1⌋+ â_2 ⌊νU_2⌋; a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋| Y)

≤    2log(‖ ‖_1) + I(⌊â_1 νU_1 + â_2 νU_2 ⌋; ⌊a_1 νU_1 + a_2 νU_2 ⌋| Y)

≤    2log(‖ ‖_1) + I(â_1 U_1 + â_2 U_2 ; a_1 U_1 + a_2 U_2 | Y).

It now suffices to prove the finiteness of the mutual information term on the right-hand side of (<ref>).
We make a case distinction depending on the setting:

Theorems <ref> and <ref>
In these cases, $(U_1,U_2)$ is discrete and the mutual information term can be upper-bounded by discrete entropies as
I(â_1 U_1 + â_2 U_2 ; a_1 U_1 + a_2 U_2 | Y)
≤    H(â_1 U_1 + â_2 U_2 | Y)

≤    H(U_1) + H(U_2)

which is finite by assumption.

Theorem <ref>
In this case, $(U_1,U_2) \in \mathbb{R}^2$ is absolutely continuous and $I(U_1,U_2;Y)$ is assumed finite. The mutual information term can be expressed as
I(â_1 U_1 + â_2 U_2 ; a_1 U_1 + a_2 U_2 | Y)

=    h(â_1 U_1 + â_2 U_2 | Y) + h(a_1 U_1 + a_2 U_2 | Y) - h(â_1 U_1 + â_2 U_2 , a_1 U_1 + a_2 U_2 | Y)

=    h(â_1 U_1 + â_2 U_2 | Y) + h(a_1 U_1 + a_2 U_2 | Y) - h(U_1,U_2|Y) - log| ( [   ] ) |.
Given that $H(\lfloor \uv \rfloor |Y)$, $h(U_1|Y)$, $h(U_2|Y)$ and $h(U_1,U_2|Y)$ are finite, by Lemma <ref>, the first two differential entropy terms $h(\hat{a}_1 U_1 + \hat{a}_2 U_2 | Y)$ and $h(a_1 U_1 + a_2 U_2 | Y)$ are finite too.
This closes the previous argument. We conclude that in all cases,

\begin{equation*}
= d_{\as}(\uv|Y) + d_{\hat{\as}}(\uv|Y).
\end{equation*}

Hence, by the data-processing inequality,
=    lim inf_ν→∞ { H(⌊ν⌋|Y) - d(|Y) log(ν) }

=    lim inf_ν→∞ { H(a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋|Y) + H(â_1 ⌊νU_1⌋+ â_2 ⌊νU_2⌋|Y, a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋)

- d_^(|Y)log(ν) - d_^(|Y)log(ν) }

≤    lim inf_ν→∞ { H(a_1 ⌊νU_1⌋+ a_2 ⌊νU_2⌋|Y) + H(â_1 ⌊νU_1⌋+ â_2 ⌊νU_2⌋|Y)

- d_^(|Y)log(ν) - d_^(|Y)log(ν) }

=    _^(|Y) + _^(|Y).

As a consequence,

(U_1) - _^(|Y)
≤    (U_1) - (|Y) + _^(|Y)

(U_2) - _^(|Y)
≤    (U_2) - (|Y) + _^(|Y).

The sum of the right-hand sides of this pair of inequalities equals

\begin{equation*}
I(\uv;Y) + 2\Hc_{\as^\T}(\uv|Y) - \Hc(\uv|Y)
\end{equation*}

which due to (<ref>) is upper-bounded by the sum rate $I(\uv;Y)$.
Hence $\mathscr{Q}(\hat{\as}^\T)$, which is delimited by the left-hand side quantities of (<ref>)–(<ref>), is contained in $\mathscr{R}_\MAC$.
Therefore, $\as^\T$ is the only coefficient vector such that $\mathscr{Q}(\as^\T)$ is not contained in $\mathscr{R}_\MAC$, while for any other (linearly independent) vector $\hat{\as}^\T$, we have $\mathscr{Q}(\hat{\as}^\T) \subseteq \mathscr{R}_\MAC$. Consequently, if the inequality (<ref>) holds, then $\as^\T$ is the unique maximizer of the set $\mathscr{Q}(\as^\T)$, i.e., the unique minimizer of $\cs^\T \mapsto \Hc_{\cs^\T}(\uv|Y)$ over vectors $\cs^\T$ with non-zero entries [cf. (<ref>)–(<ref>)]. That is,

\begin{equation}   \label{unique_minimizer_of_entropy}
\as^\T
= \argmin_{c_1 \neq 0,\, c_2 \neq 0} \Hc_{[ c_1 \ c_2 ]}(\uv|Y).
\end{equation}

Since, as argued in (<ref>), (<ref>)–(<ref>) and Lemma <ref>, the condition (<ref>) entails a non-empty set $\mathscr{R}_\MAC \cap \overline{\mathscr{R}'}$, i.e., the set described by the inequalities (<ref>)–(<ref>). By substracting (<ref>) from (<ref>), and (<ref>) from (<ref>), it becomes clear that any element from $\mathscr{R}_\MAC \cap \overline{\mathscr{R}'}$ will satisfy

<    (U_1) - _^(|Y)

<    (U_2) - _^(|Y)

and thus, we have $\mathscr{R}_\MAC \cap \overline{\mathscr{R}'} \subseteq \mathscr{Q}(\as^\T)$. As a consequence [cf. (<ref>), (<ref>)],
=    ℛ_∪𝒬(^)

=    ℛ_∖( ℛ_∩ℛ' ) ∪𝒬(^)

=    ℛ_∪𝒬(^).
This proves (<ref>) and thus the first part of Lemma <ref>.

By contrast, if $\as^\T$ satisfies the reverse of the inequality (<ref>), then we distinguish two cases:

Case $\mathscr{R}_\LMAC \subsetneq \mathscr{R}_\MAC$
By Lemma <ref>, there exists a vector $\cs^\T$  such that $2 \Hc_{\cs^\T}(\uv|Y) < \Hc(\uv|Y)$. As argued before [cf. (<ref>)], this coefficient vector is the unique minimizer of $\tilde{\cs}^\T \mapsto \Hc_{\tilde{\cs}^\T}(\uv|Y)$, hence

\begin{equation}   \label{unique_entropy_minimum}
\Hc_{\cs^\T}(\uv|Y)
= \min_{\tilde{c}_1 \neq 0,\, \tilde{c}_2 \neq 0} \Hc_{[ \tilde{c}_1 \ \tilde{c}_2 ]}(\uv|Y).
\end{equation}

Yet since we have assumed, on the other hand, that $2 \Hc_{\as^\T}(\uv|Y) \geq \Hc(\uv|Y)$, it follows that $\as$ and $\cs$ are distinct, linearly independent vectors. Re-using the bounding steps performed in (<ref>) for a pair of linearly independent coefficient vectors, we get
≤    _^(|Y) + _^(|Y).

Similarly as in (<ref>)–(<ref>),
the latter allows us to write down the following two inequalities:

(U_1) - _^(|Y)
≤    (U_1) - (|Y) + _^(|Y)

(U_2) - _^(|Y)
≤    (U_2) - (|Y) + _^(|Y).

in which $\Hc_{\cs^\T}(\uv|Y)$ may be written as the minimum (<ref>).
Hence the set $\mathscr{Q}(\as^\T)$, which is delimited by the quantities on the left-hand sides of (<ref>)–(<ref>), is contained in $\mathscr{R}'$ [cf. (<ref>)]. Given that $2 \Hc_{\as^\T}(\uv|Y) \geq \Hc(\uv|Y)$, we also have that $\mathscr{Q}(\as^\T) \subset \mathscr{R}_\MAC$. Taken together, we have $\mathscr{Q}(\as^\T) \subseteq \mathscr{R}_\MAC \cap \mathscr{R}' = \mathscr{R}_\LMAC$. Therefore,
=    ℛ_∪𝒬(^)

=    ℛ_.

Case $\mathscr{R}_\LMAC = \mathscr{R}_\MAC$
By Lemma <ref>, there exists no vector $\cs^\T$ such that $2 \Hc_{\cs^\T}(\uv|Y) < \Hc(\uv|Y)$. Therefore, all sets $\mathscr{Q}(\cs^\T)$ (for all $\cs^\T$ with non-zero entries) are subsets of $\mathscr{R}_\LMAC = \mathscr{R}_\MAC$. As in the previous case, it follows that $\mathscr{R}(\as^\T) = \mathscr{R}_\LMAC$.

This proves (<ref>) and thus concludes the proof of Lemma <ref>.