# Invertibility in Weak-Star Closed Algebras of Analytic Functions

Liming Yang1 1Department of Mathematics, Virginia Polytechnic and State
University, Blacksburg, VA 24061<EMAIL_ADDRESS>
###### Abstract.

For $K\subset\mathbb{C}$ a compact subset and $\mu$ a positive finite Borel
measure supported on $K,$ let $R^{\infty}(K,\mu)$ be the weak-star closure in
$L^{\infty}(\mu)$ of rational functions with poles off $K.$ We show that if
$R^{\infty}(K,\mu)$ has no non-trivial $L^{\infty}$ summands and $f\in
R^{\infty}(K,\mu),$ then $f$ is invertible in $R^{\infty}(K,\mu)$ if and only
if Chaumat’s map for $K$ and $\mu$ applied to $f$ is bounded away from zero on
the envelope with respect to $K$ and $\mu.$ The result proves the conjecture
$\diamond$ posed by J. Dudziak [6] in 1984.

###### Key words and phrases:

Analytic Capacity, Cauchy Transform, Subnormal Operator, and Spectral Mapping
Theorem

###### 2010 Mathematics Subject Classification:

Primary 47A15; Secondary 30C85, 31A15, 46E15, 47B38

##  1\. Introduction

For a Borel subset $B$ of the complex plane $\mathbb{C},$ let $M_{0}(B)$
denote the set of finite complex-valued Borel measures that are compactly
supported in $B$ and let $M_{0}^{+}(B)$ be the set of positive measures in
$M_{0}(B).$ The support of $\nu\in M_{0}(\mathbb{C}),$ $\text{spt}(\nu),$ is
the smallest closed set that has full $|\nu|$ measure. For a Borel set
$A\subset\mathbb{C},$ $\nu_{A}$ denotes $\nu$ restricted to $A.$

For a compact subset $K\subset\mathbb{C}$ and $\mu\in M_{0}^{+}(K),$ the
functions in $\mbox{Rat}(K):=\\{q:\mbox{$q$ is a rational function with poles
off $K$}\\}$ are members of $L^{\infty}(\mu)$. We let $R^{\infty}(K,\mu)$ be
the weak-star closure of $\mbox{Rat}(K)$ in $L^{\infty}(\mu).$ $C(K)$ denotes
the space of all continuous functions on $K$ while $R(K)$ denotes the uniform
closure in $C(K)$ of $\mbox{Rat}(K).$ Let $R(K)^{\perp}$ be the subset of
measures $\nu\in M_{0}(K)$ such that $\int fd\nu=0$ for each $f\in R(K).$ Then
$R(K)^{\perp}\cap L^{1}(\mu)$ is the space of weak-star continuous
annihilators of $R^{\infty}(K,\mu).$

There exists a Borel partition $\\{\Delta_{0},\Delta_{1}\\}$ of $K$ such that

$\displaystyle\ R^{\infty}(K,\mu)=L^{\infty}(\mu|_{\Delta_{0}})\oplus
R^{\infty}(K,\mu|_{\Delta_{1}})$ (1.1)

where $R^{\infty}(K,\mu|_{\Delta_{1}})$ contains no nontrivial $L^{\infty}$
summands (see [3, Proposition 1.16 on page 281] ). Call $R^{\infty}(K,\mu)$
pure if $\Delta_{0}=\emptyset$ in (1.1). Because of (1.1), we shall assume
that $R^{\infty}(K,\mu)$ is pure in this paper.

The envelope $E$ with respect to $K$ and $\mu\in M_{0}^{+}(K)$ is the set of
points $\lambda\in K$ such that there exists $\mu_{\lambda}\in M_{0}(K)$ that
is absolutely continuous with respect to $\mu$ such that
$\mu_{\lambda}(\\{\lambda\\})=0$ and $\int f(z)d\mu_{\lambda}(z)=f(\lambda)$
for each $f\in R(K).$ For $\lambda\in E$ and $f\in R^{\infty}(K,,\mu),$ set
$\rho(f)(\lambda)=\int fd\mu_{\lambda}.$ Clearly $\rho(f)(\lambda)$ is
independent of the particular $\mu_{\lambda}$ chosen. We thus have a map
$\rho$ called Chaumat’s map for $K$ and $\mu,$ which associates to each
function in $R^{\infty}(K,\mu)$ a point function on $E.$ Chaumat’s Theorem [2]
states the following: The map $\rho$ is an isometric isomorphism and a weak-
star homeomorphism from $R^{\infty}(K,\mu)$ onto
$R^{\infty}(\overline{E},{\mathfrak{m}}_{E}),$ where ${\mathfrak{m}}_{E}$ is
the area measure (Lebesgue measure on $\mathbb{C}$) ${\mathfrak{m}}$
restricted to $E$ (also see [3, Chaumat’s Theorem on page 288]) and
$\overline{B}$ denotes the closure of the set $B\subset\mathbb{C}$.

Our main theorem below proves the conjecture $\diamond$ posed by J. Dudziak in
[6, on page 386].

######  Theorem 1.1.

Let $K\subset\mathbb{C}$ be a compact subset and $\mu\in M_{0}^{+}(K).$
Suppose that $R^{\infty}(K,\mu)$ is pure, $E$ is the envelope for $K$ and
$\mu,$ and $\rho$ is the Chaumat’s map for $K$ and $\mu.$ If $f\in
R^{\infty}(K,\mu),$ then $f$ is invertible in $R^{\infty}(K,\mu)$ if and only
if there exists $\epsilon_{f}>0$ such that

$\ |\rho(f)(z)|\geq\epsilon_{f},~{}{\mathfrak{m}}_{E}-a.a..$

For $\mathcal{H}$ a complex separable Hilbert space,
$\mathcal{L}(\mathcal{H})$ denotes the space of all bounded linear operators
on $\mathcal{H}.$ The spectrum and essential spectrum of an operator
$A\in\mathcal{L}(\mathcal{H})$ are denoted $\sigma(A)$ and $\sigma_{e}(A)$,
respectively. An operator $S\in\mathcal{L}(\mathcal{H})$ is called subnormal
if there exists a complex separable Hilbert space $\mathcal{K}$ containing
$\mathcal{H}$ and a normal operator $N\in\mathcal{L}(\mathcal{K})$ such that
$N\mathcal{H}\subset\mathcal{H}$ and $S=N|_{\mathcal{H}}.$ Such an $N$ is
called a minimal normal extension (mne) of $S$ if the smallest closed subspace
of $\mathcal{K}$ containing $\mathcal{H}$ and reducing $N$ is $\mathcal{K}$
itself. Any two mnes of $S$ are unitarily equivalent in a manner that fixes
$S.$ The spectrum $\sigma(S)$ is the union of $\sigma(N)$ and some collection
of bounded components of $\mathbb{C}\setminus\sigma(N).$ The book [3] is a
good reference for basic information for subnormal operators.

For a subnormal operator $S\in\mathcal{L}(\mathcal{H})$ with $N=mne(S),$ let
$\mu$ be the scalar-valued spectral measure (svsm) for $N.$ Since
$\text{spt}(\mu)=\sigma(N)\subset\sigma(S),$ we see that
$\text{Rat}(\sigma(S))\subset L^{\infty}(\mu).$ Therefore, for $f\in
R^{\infty}(\sigma(S),\mu),$ the normal operator $f(N)$ is well defined and
$f(N)\mathcal{H}\subset\mathcal{H}.$ The operator $f(S):=f(N)|_{\mathcal{H}}$
defines a functional calculus for $S.$ We assume that
$R^{\infty}(\sigma(S),\mu)$ is pure. Let $E$ be the envelope for $\sigma(S)$
and $\mu$ and let $\rho$ be the Chaumat’s map from $R^{\infty}(\sigma(S),\mu)$
to $R^{\infty}(\sigma(S),{\mathfrak{m}}_{E}).$ Define $f(\sigma(S))$ the
closure of $\rho(f)(E).$ We define $cl(f,\sigma_{e}(S)),$ the cluster set of
$f$ on $\sigma_{e}(S),$ to be the subset of $\lambda\in\mathbb{C}$ such that
there exists a sequence $\\{z_{n}\\}\subset E$ and a $z_{0}\in\sigma_{e}(S)$
satisfying $z_{n}\rightarrow z_{0}$ and $\rho(f)(z_{n})\rightarrow\lambda.$

Combining Theorem 1.1 with [6, Corollary IV.6], we obtain the following
spectral mapping theorem for subnormal operators.

######  Corollary 1.2.

Let $S\in\mathcal{L}(\mathcal{H})$ be a subnormal operator and let $\mu$ be
the scalar-valued spectral measure for $N=mne(S).$ If
$R^{\infty}(\sigma(S),\mu)$ is pure and $f\in R^{\infty}(\sigma(S),\mu),$ then
$\sigma(f(S))\subset f(\sigma(S))$ and $\sigma_{e}(f(S))\subset
cl(f,\sigma_{e}(S)).$

In section 2, we review some results of analytic capacity and Cauchy transform
that are needed in our analysis. Section 3 reviews the modified Vitushkin
approximation scheme of Paramonov and proves some technical lemmas. In section
4, we introduce an algebra of “bounded and analytic functions” on a “nearly
open” subset. We obtain a criterion for functions in the algebra that will be
used in proving Theorem 1.1. In section 5, we introduce the concepts of non-
removable boundary and removable set that are more appropriate than that of
the envelope in studying $R^{\infty}(K,\mu).$ We obtain a criterion for
functions in $R^{\infty}(K,\mu)$ in section 6. We then prove Theorem 1.1 in
section 7.

##  2\. Preliminaries

If $B\subset\mathbb{C}$ is a compact subset, then we define the analytic
capacity of $B$ by

$\ \gamma(B)=\sup|f^{\prime}(\infty)|,$

where the supremum is taken over all those functions $f$ that are analytic in
$\mathbb{C}_{\infty}\setminus B,$ where
$\mathbb{C}_{\infty}=\mathbb{C}\cup\\{\infty\\},$ such that $|f(z)|\leq 1$ for
all $z\in\mathbb{C}_{\infty}\setminus B$; and
$f^{\prime}(\infty):=\lim_{z\rightarrow\infty}z[f(z)-f(\infty)].$ The analytic
capacity of a general subset $F$ of $\mathbb{C}$ is given by:

$\ \gamma(F)=\sup\\{\gamma(B):B\subset F\text{ compact}\\}.$

Let ${\mathfrak{m}}$ denote the Lebesgue measure on the complex plane (the
area measure). The following elementary property can be found in [8, Theorem
VIII.2.3],

$\displaystyle\ {\mathfrak{m}}(F)\leq 4\pi\gamma(F)^{2}.$ (2.1)

We will write $\gamma-a.a.$, for a property that holds everywhere, except
possibly on a set of analytic capacity zero. Good sources for basic
information about analytic capacity are Chapter VIII of [8], Chapter V of [3],
[7], and [12].

For $\nu\in M_{0}(\mathbb{C})$ and $\epsilon>0,$ $\mathcal{C}_{\epsilon}(\nu)$
is defined by

$\ \mathcal{C}_{\epsilon}(\nu)(z)=\int_{|w-z|>\epsilon}\dfrac{1}{w-z}d\nu(w).$

The (principal value) Cauchy transform of $\nu$ is defined by

$\displaystyle\ \mathcal{C}(\nu)(z)=\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(\nu)(z)\ $ (2.2)

for all $z\in\mathbb{C}$ for which the limit exists. If $\lambda\in\mathbb{C}$
and

$\displaystyle\ \tilde{\nu}(\lambda):=\int\frac{d|\nu|}{|z-\lambda|}<\infty,$
(2.3)

then $\lim_{\epsilon\rightarrow 0}\mathcal{C}_{\epsilon}(\nu)(\lambda)$
exists. Therefore, a standard application of Fubini’s Theorem shows that
$\mathcal{C}(\nu)\in L^{s}_{\mbox{loc}}(\mathbb{C})$, for $0<s<2$. In
particular, it is defined for ${\mathfrak{m}}-a.a.,$ and clearly
$\mathcal{C}(\nu)$ is analytic in
$\mathbb{C}_{\infty}\setminus\mbox{spt}(\nu).$ In fact, from Corollary 2.3
below, we see that (2.2) is defined for $\gamma-a.a..$ Throughout this paper,
the Cauchy transform of a measure always means the principal value of the
transform. In the sense of distributions,

$\displaystyle\ \bar{\partial}\mathcal{C}(\nu)=-\pi\nu.$ (2.4)

The maximal Cauchy transform is defined by

$\ \mathcal{C}_{*}(\nu)(z)=\sup_{\epsilon>0}|\mathcal{C}_{\epsilon}(\nu)(z)|.$

A related capacity, $\gamma_{+},$ is defined for subsets $F$ of $\mathbb{C}$
by:

$\ \gamma_{+}(F)=\sup\|\mu\|,$

where the supremum is taken over $\mu\in M_{0}^{+}(F)$ for which
$\|\mathcal{C}(\mu)\|_{L^{\infty}(\mathbb{C})}\leq 1.$ Since $\mathcal{C}\mu$
is analytic in $\mathbb{C}_{\infty}\setminus\mbox{spt}(\mu)$ and
$|(\mathcal{C}(\mu)^{\prime}(\infty)|=\|\mu\|$, we have:
$\gamma_{+}(F)\leq\gamma(F)$ for all subsets $F$ of $\mathbb{C}$.

X. Tolsa has established the following astounding results. See [11] (also
Theorem 6.1 and Corollary 6.3 in [12]) for (1) and (2). See [10, Proposition
2.1] (also [12, Proposition 4.16]) for (3).

######  Theorem 2.1.

(Tolsa 2003) (1) $\gamma_{+}$ and $\gamma$ are actually equivalent. That is,
there is an absolute constant $A_{T}$ such that, for all $F\subset\mathbb{C},$

$\ \gamma(F)\leq A_{T}\gamma_{+}(F).$

(2) Semiadditivity of analytic capacity:

$\ \gamma\left(\bigcup_{i=1}^{m}F_{i}\right)\leq
A_{T}\sum_{i=1}^{m}\gamma(F_{i})$

where $F_{1},F_{2},...,F_{m}\subset\mathbb{C}$ ($m$ may be $\infty$).

(3) There is an absolute constant $C_{T}>0$ such that, for $a>0$, we have:

$\ \gamma(\\{\mathcal{C}_{*}(\nu)\geq a\\})\leq\dfrac{C_{T}}{a}\|\nu\|.$

Given three distinct points $x,y,z\in\mathbb{C}$, let $R(x,y,z)$ be the radius
of the circle passing through $x,y,$ and $z$. The Menger curvature is defined
by $c(x,y,z)=\frac{1}{R(x,y,z)}.$ If two or three of the points coincide, we
define $c(x,y,z)=0.$

For $\eta\in M_{0}^{+}(\mathbb{C})$, define the curvature of $\eta$ as

$\ c^{2}(\eta)=\int\int\int c(x,y,z)^{2}d\eta(x)d\eta(y)d\eta(z)$

and

$\
c^{2}_{\epsilon}(\eta)=\int\int\int_{\underset{\underset{|x-y|>\epsilon}{|z-x|>\epsilon}}{|y-z|>\epsilon}}c(x,y,z)^{2}d\eta(x)d\eta(y)d\eta(z).$

Define

$\
N_{2}(\eta)=\sup_{\epsilon>0}\sup_{\|f\|_{L^{2}(\eta)}=1}\|\mathcal{C}_{\epsilon}(f\eta)\|_{L^{2}(\eta)}.$

A measure $\eta\in M_{0}^{+}(\mathbb{C})$ is $a$-linear growth if
$\eta(\mathbb{D}(\lambda,\delta))\leq a\delta$ for $\lambda\in\mathbb{C}\text{
and }\delta>0,$ where
$\mathbb{D}(\lambda,\delta):=\\{z:~{}|z-\lambda|<\delta\\}.$ If $\eta\in
M_{0}^{+}(\mathbb{C})$ is $a$-linear growth, then there is an absolute
constant $C>0$ so that

$\displaystyle\left|\|{\mathcal{C}}_{\epsilon}\eta\|_{L^{2}(\eta)}^{2}-\dfrac{1}{6}c^{2}_{\epsilon}(\eta)\right|\
\leq C\,a^{2}\|\eta\|$ (2.5)

(see [12, Proposition 3.3]). We use $C_{1},C_{2},...$ for absolute constants
that may change from one step to the next.

######  Proposition 2.2.

If $F\subset\mathbb{C}$ is a compact subset and $\eta\in M_{0}^{+}(F),$ then
the following properties are true.
(1) If $\|{\mathcal{C}}\eta\|_{L^{\infty}(\mathbb{C})}\leq 1,$ then $\eta$ is
$1$-linear growth and
$\|\mathcal{C}_{\epsilon}(\eta)\|_{L^{\infty}(\mathbb{C})}\leq C_{1}$ for all
$\epsilon>0.$
(2) If $\eta$ is $1$-linear growth and
$\|\mathcal{C}_{\epsilon}(\eta)\|_{L^{\infty}(\mathbb{C})}\leq 1$ for all
$\epsilon>0,$ then $c^{2}(\eta)\leq C_{2}\eta(F).$
(3) If $\eta$ is $1$-linear growth and $c^{2}(\eta)\leq\eta(F),$ then there
exists a subset $A\subset F$ such that $\eta(F)\leq 2\eta(A)$ and
$N_{2}(\eta|_{A})\leq C_{3}.$
(4) If $N_{2}(\eta)\leq 1,$ then there exists some function $w$ supported on
$F$, with $0\leq w\leq 1$ such that $\eta(F)\ \leq\ 2\int wd\eta$ and
$\|\mathcal{C}_{\epsilon}(w\eta)\|_{L^{\infty}(\mathbb{C})}\ \leq\ C_{4}$ for
all $\epsilon>0.$

See the proof of [12, Theorem 4.14] on page 113 and page 114 for (1) and (3).
(2) follows from (2.5). (4) follows from [12, Lemma 4.7].

Combining Theorem 2.1 (1), Proposition 2.2, and [11] (or [12, Theorem 8.1]),
we get the following corollary. The reader may also see [1, Corollary 3.1].

######  Corollary 2.3.

If $\nu\in M_{0}(\mathbb{C}),$ then there exists
$\mathcal{Q}\subset\mathbb{C}$ with $\gamma(\mathcal{Q})=0$ such that
$\lim_{\epsilon\rightarrow 0}\mathcal{C}_{\epsilon}(\nu)(z)$ exists for
$z\in\mathbb{C}\setminus\mathcal{Q}$.

######  Corollary 2.4.

Let $\eta\in M_{0}^{+}(\mathbb{C})$ such that $\|\mathcal{C}(\eta)\|\leq 1$.
If $F$ is a compact subset and $\gamma(F)=0$, then $\eta(F)=0$.

###### Proof.

Suppose $\eta(F)>0$. By Proposition 2.2 (1) & (2), $c^{2}(\eta)<\infty$. Since
$c^{2}(\eta|_{F})\leq c^{2}(\eta),$ we see that $\gamma(F)>0$ by Proposition
2.2 (3) & (4), and Theorem 2.1 (1). This is a contradiction. ∎

######  Lemma 2.5.

Let $\mu\in M_{0}^{+}(\mathbb{C}).$ Suppose $f_{n},f\in L^{1}(\mu)$ such that
$\|f_{n}-f\|_{L^{1}(\mu)}\rightarrow 0$ and $f_{n}\rightarrow f,~{}\mu-a.a..$
Then, for $\epsilon>0$, there exists a subset $A_{\epsilon}$ with
$\gamma(A_{\epsilon})<\epsilon$ and a subsequence $\\{f_{n_{k}}\\}$ so that
$\\{\mathcal{C}(f_{n_{k}}\mu)\\}$ uniformly converges to $\mathcal{C}(f\mu)$
on $\mathbb{C}\setminus A_{\epsilon}$.

###### Proof.

From Corollary 2.3, we let $\mathcal{Q}_{1}\subset\mathbb{C}$ with
$\gamma(\mathcal{Q}_{1})=0$ such that the principal values of
$\mathcal{C}(f_{n}\mu)(z)$ for $n\geq 1$ and $\mathcal{C}(f\mu)(z)$ exist for
$z\in\mathbb{C}\setminus\mathcal{Q}_{1}$. Define

$\
A_{nm}=\left\\{z\in\mathbb{C}\setminus\mathcal{Q}_{1}:~{}|\mathcal{C}(f_{n}\mu)(z)-\mathcal{C}(f\mu)(z)|\geq\dfrac{1}{m}\right\\}.$

Since

$\displaystyle\ |\mathcal{C}(f_{n}\mu)(z)-\mathcal{C}(f\mu)(z)|$
$\displaystyle=\lim_{\epsilon\rightarrow
0}|\mathcal{C}_{\epsilon}(f_{n}\mu)(z)-\mathcal{C}_{\epsilon}(f\mu)(z)|$
$\displaystyle\leq\mathcal{C}_{*}((f_{n}-f)\mu)(z),$

applying Theorem 2.1 (3), we get

$\
\gamma(A_{nm})\leq\gamma\left\\{\mathcal{C}_{*}((f_{n}-f)\mu)(z)\geq\dfrac{1}{m}\right\\}\leq
C_{T}m\|f_{n}-f\|_{L^{1}(\mu)}.$

Choose $n_{m}$ so that $\|f_{n_{m}}-f\|_{L^{1}(\mu)}\leq\frac{1}{m2^{m}}$ and
we have $\gamma(A_{n_{m}m})\leq\frac{C_{T}}{2^{m}}$.

Set $B_{k}=\cup_{m=k}^{\infty}A_{n_{m}m}$. Applying Theorem 2.1 (2), there
exists $k_{0}$ so that $\displaystyle\frac{A_{T}C_{T}}{2^{k_{0}-1}}<\epsilon$
and

$\ \gamma(B_{k_{0}}\cup\mathcal{Q}_{1})\leq
A_{T}\sum_{m=k}^{\infty}\gamma(A_{n_{m}m})\leq
A_{T}C_{T}\sum_{m=k_{0}}^{\infty}\dfrac{1}{2^{m}}<\epsilon.$

Then on $(B_{k_{0}}\cup\mathcal{Q}_{1})^{c}$, $\mathcal{C}(f_{n_{m}}\mu)(z)$
converges to $\mathcal{C}(f\mu)(z)$ uniformly. ∎

######  Lemma 2.6.

Let $\\{\nu_{j}\\}\subset M_{0}(\mathbb{C}).$ Then for $\epsilon>0$, there
exists a Borel subset $F$ such that $\gamma(F^{c})<\epsilon$ and
$\mathcal{C}_{*}(\nu_{j})(z)\leq M_{j}<\infty$ for $z\in F$.

###### Proof.

Let $A_{j}=\\{\mathcal{C}_{*}(\nu_{j})(z)\leq M_{j}\\}$. By Theorem 2.1 (3),
we can select $M_{j}>0$ so that
$\gamma(A_{j}^{c})<\frac{\epsilon}{2^{j+1}A_{T}}.$ Set
$F=\cap_{j=1}^{\infty}A_{j}.$ Then applying Theorem 2.1 (2), we get

$\ \gamma(F^{c})\leq A_{T}\sum_{j=1}^{\infty}\gamma(A_{j}^{c})<\epsilon.$

∎

######  Lemma 2.7.

Suppose that $\\{g_{n}\\}\subset L^{1}(\mu)$ and $F$ is a bounded subset with
$\gamma(F)>0$. Then there exists $\eta\in M_{0}^{+}(F)$ satisfying:

(1) $\eta$ is $1$-linear growth,
$\|\mathcal{C}_{\epsilon}(\eta)\|_{L^{\infty}(\mathbb{C})}\leq 1$ for all
$\epsilon>0,$ and $\gamma(F)\leq C_{5}\|\eta\|$;

(2) $\mathcal{C}_{*}(g_{n}\mu)\in L^{\infty}(\eta)$;

(3) there exists a subsequence $f_{k}(z)=\mathcal{C}_{\epsilon_{k}}(\eta)(z)$
such that $f_{k}$ converges to $f\in L^{\infty}(\mu)$ in weak-star topology,
and $f_{k}(\lambda)$ converges to $f(\lambda)=\mathcal{C}(\eta)(\lambda)$
uniformly on any compact subset of $F^{c}$ as $\epsilon_{k}\rightarrow 0$.
Moreover, for $n\geq 1,$

$\displaystyle\ \int
f(z)g_{n}(z)d\mu(z)=-\int\mathcal{C}(g_{n}\mu)(z)d\eta(z).$ (2.6)

###### Proof.

From Lemma 2.6, we find a compact subset $F_{1}\subset F$ such that
$\gamma(F\setminus F_{1})<\frac{\gamma(F)}{2A_{T}}$ and
$\mathcal{C}_{*}(g_{n}\mu)(z)\leq M_{n}<\infty$ for $z\in F_{1}$. Using
Theorem 2.1 (2), we get
$\gamma(F_{1})\geq\frac{1}{A_{T}}\gamma(F)-\gamma(F\setminus
F_{1})\geq\frac{1}{2A_{T}}\gamma(F).$ Using Theorem 2.1 (1) and Proposition
2.2 (1), there exists $\eta\in M_{0}^{+}(F_{1})$ satisfying (1). So (2) holds.
Clearly,

$\displaystyle\
\int\mathcal{C}_{\epsilon}(\eta)(z)g_{n}d\mu=-\int\mathcal{C}_{\epsilon}(g_{n}\mu)(z)d\eta$
(2.7)

for $n\geq 1$. We can choose a sequence
$f_{k}(\lambda)=\mathcal{C}_{\epsilon_{k}}(\eta)(\lambda)$ that converges to
$f$ in $L^{\infty}(\mu)$ weak-star topology and $f_{k}(\lambda)$ uniformly
tends to $f(\lambda)$ on any compact subset of $F^{c}$. On the other hand, by
Corollary 2.3 and Corollary 2.4,
$|\mathcal{C}_{\epsilon_{k}}(g_{n}\mu)(z)|\leq M_{n},~{}\eta-a.a.$ and
$\lim_{k\rightarrow\infty}\mathcal{C}_{\epsilon_{k}}(g_{n}\mu)(z)=\mathcal{C}(g_{n}\mu)(z),~{}\eta-a.a.$.
Applying the Lebesgue dominated convergence theorem to (2.7), we get (2.6). ∎

For $\nu\in M_{0}(\mathbb{C}),$ define
$\Theta_{\nu}(\lambda):=\lim_{\delta\rightarrow
0}\frac{|\nu|(\mathbb{D}(\lambda,\delta))}{\delta}$ if the limit exists. The
following lemma follows from [12, Lemma 8.12].

######  Lemma 2.8.

Let $B\subset\mathbb{C}$ be a bounded measurable subset and $g\in
L^{1}({\mathfrak{m}}_{B}).$ Then

$\ \Theta_{g{\mathfrak{m}}_{B}}(z)=0,~{}\gamma-a.a..$

######  Definition 2.9.

Let $\mathcal{Q}$ be a set with $\gamma(\mathcal{Q})=0$. Let $f(z)$ be a
function defined on $\mathbb{D}(\lambda,\delta_{0})\setminus\mathcal{Q}$ for
some $\delta_{0}>0.$ The function $f$ has a $\gamma$-limit $a$ at $\lambda$ if

$\ \lim_{\delta\rightarrow
0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\cap\\{|f(z)-a|>\epsilon\\})}{\delta}=0$

for all $\epsilon>0$. If in addition, $f(\lambda)$ is well defined and
$a=f(\lambda)$, then $f$ is $\gamma$-continuous at $\lambda$.

The following lemma is straightforward (see [4, Corollary 2.5]).

######  Lemma 2.10.

If $f(z)$ and $g(z)$ are $\gamma$-continuous at $\lambda,$ then $f(z)+g(z)$
and $f(z)g(z)$ are $\gamma$-continuous at $\lambda.$ If in addition
$g(\lambda)\neq 0,$ then $\frac{f(z)}{g(z)}$ is $\gamma$-continuous at
$\lambda.$

The following lemma is from [1, Lemma 3.2].

######  Lemma 2.11.

Let $\nu\in M_{0}(\mathbb{C})$ and assume that for some $\lambda$ in
$\mathbb{C}$ we have:

  * (a)

$\underset{\delta\rightarrow
0}{\overline{\lim}}\dfrac{|\nu|(\mathbb{D}(\lambda,\delta))}{\delta}=0$ and

  * (b)

$\mathcal{C}(\nu)(\lambda)=\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(\nu)(\lambda)$ exists.

Then the Cauchy transform $\mathcal{C}(\nu)(z)$ is $\gamma$-continuous at
$\lambda$.

##  3\. Modified Vitushkin scheme of Paramonov and some lemmas

If a compact subset $F$ is contained in $\mathbb{D}(a,\delta)$ and $f$ is
bounded and analytic on $\mathbb{C}_{\infty}\setminus F$ with $f(\infty)=0,$
we consider the Laurent expansion of $f$ for
$z\in\mathbb{C}\setminus\mathbb{D}(a,\delta),$

$\ f(z)=\sum_{n=1}^{\infty}\dfrac{c_{n}(f,a)}{(z-a)^{n}}.$

As $c_{1}(f,a)$ does not depend on the choice of $a$, we define
$c_{1}(f)=c_{1}(f,a).$ The coefficient $c_{2}(f,a)$ does depend on $a$.
However, if $c_{1}(f)=0$, then $c_{2}(f,a)$ does not depend on $a$, and in
this case, we define $c_{2}(f)=c_{2}(f,a).$

Let $\varphi$ be a smooth function with compact support. Vitushkin’s
localization operator $T_{\varphi}$ is defined by

$\
(T_{\varphi}f)(\lambda)=\dfrac{1}{\pi}\int\dfrac{f(z)-f(\lambda)}{z-\lambda}\bar{\partial}\varphi(z)d{\mathfrak{m}}(z),$

where $f\in L^{1}_{loc}(\mathbb{C})$. Clearly,
$(T_{\varphi}f)(z)=-\frac{1}{\pi}\mathcal{C}(\varphi\bar{\partial}f{\mathfrak{m}})(z).$
Consequently, in the sense of distributions,

$\ \bar{\partial}(T_{\varphi}f)(z)=\varphi(z)\bar{\partial}f(z).$

Therefore, $T_{\varphi}f$ is analytic outside of
$\text{supp}(\bar{\partial}f)\cap\text{supp}(\varphi).$ If
$\text{supp}(\varphi)\subset\mathbb{D}(a,\delta),$ then

$\displaystyle(T_{\varphi}f)(\infty)=0,~{}c_{1}(T_{\varphi}f)=-\dfrac{1}{\pi}\int
f(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}(z),$ $\displaystyle
c_{2}(T_{\varphi}f,a)=\dfrac{1}{\pi}\int(z-a)f(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}(z).$

The following is the estimate of the norm of $T_{\varphi}f.$

$\displaystyle\ \|T_{\varphi}f\|_{\infty}\leq
4\|f\|_{\infty}\delta\|\bar{\partial}\varphi\|.$ (3.1)

See [8, VIII.7.1] for the details of $T_{\varphi}.$

Let $\delta>0$. We say that $\\{\varphi_{ij},S_{ij},\delta\\}$ is a smooth
partition of unity subordinated to $\\{2S_{ij}\\},$ if the following
assumptions hold:

(i) $S_{ij}$ is a square with vertices
$(i\delta,j\delta),~{}((i+1)\delta,j\delta),~{}(i\delta,(j+1)\delta),$ and
$((i+1)\delta,(j+1)\delta);$

(ii) $c_{ij}$ is the center of $S_{ij}$ and $\varphi_{ij}$ is a $C^{\infty}$
smooth function supported in $\mathbb{D}(c_{ij},\delta)\subset 2S_{ij}$ and
with values in $[0,1]$;

(iii)

$\
\|\bar{\partial}\varphi_{ij}\|\leq\frac{C_{13}}{\delta},~{}\sum\varphi_{ij}=1.$

(See [8, VIII.7] for details).

Let $f\in C(\mathbb{C}_{\infty})$ (or $f\in L^{\infty}(\mathbb{C})$) with
compact support. Define $f_{ij}=T_{\varphi_{ij}}f,$ then $f_{ij}\neq 0$ for
only finite many $(i,j).$ Clearly,

$\ f(z)=\sum_{f_{ij}\neq 0}f_{ij}(z).$

The standard Vitushkin approximation scheme requires us to construct functions
$a_{ij}$ such that $f_{ij}-a_{ij}$ has triple zeros at $\infty$, which
requires us to estimate both $c_{1}(a_{ij})$ and $c_{2}(a_{ij},c_{ij})$ (see
[8, section 7 on page 209]).

The main idea of P. V. Paramonov [9] is that one does not actually need to
estimate each coefficient $c_{2}(a_{ij},c_{ij})$. It suffices to estimate the
sum of coefficients $\sum_{j\in J_{il}}c_{2}(a_{ij},c_{ij})$ for a special
non-intersecting partition $\\{J_{il}\\}$ of $J_{i}=\\{j_{min}\leq j\leq
j_{max}\\},$ where $j_{min}=\min\\{j:~{}f_{ij}\neq 0\\}$ and
$j_{max}=\max\\{j:~{}f_{ij}\neq 0\\}.$

We now state the modified Vitushkin approximation scheme by P. V. Paramonov.
We fix a bounded Borel subset $F\subset\mathbb{C}.$ Let

$\displaystyle\ \alpha_{ij}=\gamma(\mathbb{D}(c_{ij},k_{1}\delta)\cap F),$
(3.2)

where $k_{1}\geq 3$ is a fixed integer. Let
$m_{1}=\min\\{i,j,~{}\mathbb{D}(c_{ij},k_{1}\delta)\cap F\neq\emptyset\\}$ and
$m_{2}=\max\\{i,j,~{}\mathbb{D}(c_{ij},k_{1}\delta)\cap F\neq\emptyset\\}.$
Note both $m_{1}$ and $m_{2}$ are finite. Set
$min_{i}=\min\\{j:~{}\alpha_{ij}\neq 0\\}$ and
$max_{i}=\max\\{j:~{}\alpha_{ij}\neq 0\\}$. Let $J_{i}=\\{j:~{}min_{i}\leq
j\leq max_{i}\\}$. We will also fix a constant $C_{0}>1.$

######  Definition 3.1.

For $j_{1}+1\in J_{i},$ let

$\displaystyle\ J=\\{j:~{}j_{1}+1\leq j\leq j_{1}+s_{1}+s_{2}+s_{3}\\}.$ (3.3)

Choose positive integers $s_{1},$ $s_{2}$, and $s_{3}$ so that

$\displaystyle\
\delta\leq\sum_{j=j_{1}+1}^{j_{1}+s_{1}}\alpha_{ij}<\delta+k_{1}\delta,$ (3.4)
$\ 6k_{1}+4C_{0}<s_{2}\leq 6k_{1}+4C_{0}+1,$

and

$\displaystyle\
\delta\leq\sum_{j=j_{1}+s_{1}+s_{2}+1}^{j_{1}+s_{1}+s_{2}+s_{3}}\alpha_{ij}<\delta+k_{1}\delta$
(3.5)

hold. With this choice, we say that $J$ is a complete group.

We now present a detailed description of the procedure of partitioning $J_{i}$
into groups. We split each $J_{i}$ into (finitely many) non-intersecting
groups $J_{il}$, $l=1,...,l_{i}$, as follows. Starting from the lowest index
$min_{i}$ in $J_{i}$ we include in $J_{i1}$ (going upwards and without jumps
in $J_{i}$) all indices until we have collected a minimal (with respect to the
number of elements) complete group $J_{i1}$. Then we repeat this procedure for
$J_{i}\setminus J_{i1}$, and so on. After we have constructed all the complete
groups $J_{i1},...,J_{il_{i}-1}$ in this way (there may be none), then what
remains is the last portion $J_{il_{i}}=J_{i}\setminus(J_{i1}\cup...\cup
J_{il_{i}-1})$ of indices in $J_{i}$, which includes no complete groups. We
call this portion $J_{il_{i}}$ an incomplete group of indices (clearly, there
is at most one incomplete group for each $i$).

######  Definition 3.2.

Let $\delta>0,$ $\alpha_{ij},$ and $F$ be defined as in (3.2). Define

$\ P_{1}(\delta,\alpha_{ij},F,k_{1},C_{0})=\\{J_{il}:~{}1\leq l\leq
l_{i}-1,~{}m_{1}\leq i\leq m_{2}\\}$

to be the collection of all complete groups and

$\ P_{2}(\delta,\alpha_{ij},F,k_{1},C_{0})=\\{J_{il_{i}}:~{}m_{1}\leq i\leq
m_{2}\\}$

to be the set of all incomplete groups as $i$ ranges from $m_{1}$ to $m_{2}.$
The modified Vitushkin scheme of Paramonov is defined to be

$\
P(\delta,\alpha_{ij},F,k_{1},C_{0})=P_{1}(\delta,\alpha_{ij},F,k_{1},C_{0})\cup
P_{2}(\delta,\alpha_{ij},F,k_{1},C_{0}).$

For a group $J$ (complete or incomplete) with row index $i,$ let

$\ J^{\prime}(z)=\\{j\in J:~{}\alpha_{ij}>0\text{ and
}|z-c_{ij}|>3k_{1}\delta\\},$

and

$\ L_{J}^{\prime}(z)=\sum_{j\in
J^{\prime}(z)}\left(\dfrac{\delta\alpha_{ij}}{|z-c_{ij}|^{2}}+\dfrac{\delta^{3}}{|z-c_{ij}|^{3}}\right).$

Define $L_{J}(z)=L_{J}^{\prime}(z)$ if $\\{j\in
J:~{}\alpha_{ij}>0\\}=J^{\prime}(z)$, otherwise,
$L_{J}(z)=1+L_{J}^{\prime}(z)$. For $g_{ij}$ that is bounded and analytic on
$\mathbb{C}\setminus F_{ij},$ where $F_{ij}$ is a compact subset of
$\mathbb{D}(c_{ij},k_{1}\delta),$ define

$\displaystyle\ g_{J}=\sum_{j\in J}g_{ij},~{}c_{1}(g_{J})=\sum_{j\in
J}c_{1}(g_{ij}),~{}c_{2}(g_{J})=\sum_{j\in J}c_{2}(g_{ij},c_{ij}).$ (3.6)

######  Definition 3.3.

A sequence of functions $\\{g_{ij}\\}$ is applicable to
$P(\delta,\alpha_{ij},F,k_{1},C_{0})$ and $\mu\in M_{0}^{+}(\mathbb{C})$ if
the following assumptions hold, for some absolute constant $C_{7}$:

(1) $g_{ij}$ is bounded and analytic on
$\mathbb{C}_{\infty}\setminus\overline{\mathbb{D}(c_{ij},k_{1}\delta)};$

(2) $g_{ij}(\infty)=c_{1}(g_{ij})=0;$

(3) $\|g_{ij}\|_{L^{\infty}(\mu)}\leq C_{7}$;

(4) $|c_{n}(g_{ij},c_{ij})|\leq C_{7}(k_{1}\delta)^{n-1}\alpha_{ij}$ for
$n\geq 1.$

The following lemma is straightforward.

######  Lemma 3.4.

If $f(z)$ is bounded and analytic on
$\mathbb{C}_{\infty}\setminus\overline{\mathbb{D}(a,\delta)}$ with
$f(\infty)=0,$ $\|f\|\leq 1,$ $F\subset\overline{\mathbb{D}(a,\delta)},$ and
for $n\geq 1,$

$\ |c_{n}(f,a)|\leq C_{8}n\delta^{n-1}\gamma(F).$

Then for $|z-a|>2\delta,$

$\
\left|f(z)-\dfrac{c_{1}(f)}{z-a}-\dfrac{c_{2}(f,a)}{(z-a)^{2}}\right|\leq\dfrac{C_{9}\gamma(F)\delta^{2}}{|z-a|^{3}}.$

Therefore, if $\\{g_{ij}\\}$ is applicable to
$P(\delta,\alpha_{ij},F,k_{1},C_{0}),$ then for $|z-c_{ij}|>3k_{1}\delta,$

$\displaystyle\
|g_{ij}(z)|\leq\dfrac{C_{10}\delta\alpha_{ij}}{|z-c_{ij}|^{2}}+\dfrac{C_{10}\delta^{3}}{|z-c_{ij}|^{3}}.$
(3.7)

The following lemma follows easily from Definitions 3.2 and (3.7).

######  Lemma 3.5.

Let $\\{g_{ij}\\}$ be applicable to $P(\delta,\alpha_{ij},F,k_{1},C_{0})$ and
$\mu\in M_{0}^{+}(\mathbb{C})$. If $J$ is a group (complete or incomplete)
with row index $i,$ then

$\displaystyle|g_{J}(z)|\leq
C_{11}L_{J}(z),~{}\|g_{J}\|(\|g_{J}\|_{L^{\infty}(\mu)})\leq C_{11},$
$\displaystyle c_{1}(g_{J})=0,~{}|c_{2}(g_{J})|\leq C_{11}\delta^{2}.$

The following key lemma is due to [9, Lemma 2.7].

######  Lemma 3.6.

Let $\\{g_{ij}\\}$ be applicable to $P(\delta,\alpha_{ij},F,k_{1},C_{0})$ and
$\mu\in M_{0}^{+}(\mathbb{C}).$ Suppose

(H1) there exists $h_{ij}\in L^{\infty}(\mu)$ that is bounded and analytic on
$\mathbb{C}_{\infty}\setminus\overline{\mathbb{D}(c_{ij},k_{1}\delta)}$ and
satisfies

$\displaystyle\begin{aligned} \ &h_{ij}(\infty)=0,~{}\|h_{ij}\|\leq
C_{0},~{}c_{1}(h_{ij})=\alpha_{ij},\\\ \ &|c_{n}(h_{ij},c_{ij})|\leq
C_{0}(k_{1}\delta)^{n-1}\alpha_{ij}\text{ for }n\geq 1.\end{aligned}$ (3.8)

Then for each complete group $J_{il}\in
P_{1}(\delta,\alpha_{ij},F,k_{1},C_{0})$, there exists a function $h_{J_{il}}$
that has the following form

$\displaystyle\ \begin{aligned} \
&H^{i}_{jk}:=\dfrac{\delta}{|c_{ik}-c_{ij}|}(\lambda_{ik}^{j}h_{ik}-\lambda_{ij}^{k}h_{ij}),\\\
\ &h_{J_{il}}^{0}=\sum_{j\in J_{il}^{d}}\sum_{k\in J_{il}^{u}}H^{i}_{jk},\\\ \
&h_{J_{il}}=\dfrac{c_{2}(g_{J_{il}})}{c_{2}(h_{J_{il}}^{0})}h_{J_{il}}^{0},\
\end{aligned}$ (3.9)

where $J_{il}^{d}=(j_{1}+1,...,j_{1}+s_{1})$ and
$J_{il}^{u}=(j_{1}+s_{1}+s_{2}+1,...,j_{1}+s_{1}+s_{2}+s_{3})$, and there
exist $\lambda_{ij}^{k},\lambda_{ik}^{j}\geq 0$ for $j\in J_{il}^{d}$ and
$k\in J_{il}^{u}$ such that (H2)-(H5) below are satisfied:

(H2)

$\ \sum_{j\in J_{il}^{d}}\lambda_{ik}^{j}\leq 1,~{}\sum_{k\in
J_{il}^{u}}\lambda_{ij}^{k}\leq 1;$

(H3)

$\ \sum_{j\in J_{il}^{d}}\sum_{k\in
J_{il}^{u}}\lambda_{ij}^{k}\alpha_{ij}=\delta,~{}\sum_{j\in
J_{il}^{d}}\sum_{k\in
J_{il}^{u}}\lambda_{ik}^{j}\alpha_{ik}=\delta,~{}\lambda_{ik}^{j}\alpha_{ik}=\lambda_{ij}^{k}\alpha_{ij};$

(H4) $c_{1}(H^{i}_{jk})=0,$ $s_{2}>6k_{1}+4C_{0},$ and (see (2.28) in [9] for
details)

$\
|c_{2}(H^{i}_{jk})-\lambda_{ij}^{k}\alpha_{ij}\sqrt{-1}\delta|\leq\dfrac{\lambda_{ij}^{k}\alpha_{ij}\delta}{2};$

(H5) if $|z-c_{ij}|>3k_{1}\delta$ and $|z-c_{ik}|>3k_{1}\delta$, then

$\ |H^{i}_{jk}(z)|\leq
C_{13}\left(\dfrac{\lambda_{ij}^{k}\alpha_{ij}\delta}{|z-c_{ij}|^{2}}+\dfrac{\lambda_{ik}^{j}\alpha_{ik}\delta}{|z-c_{ik}|^{2}}\right)$

and for all $z\in\mathbb{C},$

$\displaystyle|h_{J_{il}}(z)|\leq C_{14}L_{J_{il}}(z),~{}\|h_{J_{il}}\|\leq
C_{14},$ $\displaystyle
c_{1}(h_{J_{il}})=0,~{}c_{2}(h_{J_{il}})=c_{2}(g_{J_{il}}).$

For $P_{0}\subset P(\delta,\alpha_{ij},F,k_{1},C_{0}),$ define

$\ S(P_{0})=\bigcup_{J_{il}\in P_{0}.~{}j\in
J_{il},~{}\alpha_{ij}>0}2k_{1}S_{ij}.$

With above lemmas and slight modifications of the proof of (2.35) in [9], we
have the following lemma (also see [5, Lemma 6.7]).

######  Lemma 3.7.

Let $\\{g_{ij}\\}$ be applicable to $P(\delta,\alpha_{ij},F,k_{1},C_{0})$ and
$\mu\in M_{0}^{+}(\mathbb{C})$. Let $\\{h_{ij}\\}$ satisfy the assumption in
Lemma 3.6. Set $P_{1}=P_{1}(\delta,\alpha_{ij},F,k_{1},C_{0})$ and
$P_{2}=P_{2}(\delta,\alpha_{ij},F,k_{1},C_{0}).$ Let $P_{1}^{\prime}\subset
P_{1}$ and $P_{2}^{\prime}\subset P_{2}.$ Suppose that for each complete group
$J_{il},$ $h_{J_{il}}$ is constructed as in Lemma 3.6. Set
$\Psi_{J_{il}}(z)=g_{J_{il}}(z)-h_{J_{il}}(z).$ Then

$\displaystyle\ \sum_{J_{il}\in
P_{1}^{\prime}}|\Psi_{J_{il}}(z)|+\sum_{J_{il_{i}}\in
P_{2}^{\prime}}|g_{J_{il_{i}}}(z)|\leq
C_{15}\min\left(1,\dfrac{\delta}{dist(z,S(P_{1}^{\prime}\cup
P_{2}^{\prime}))}\right)^{\frac{1}{5}}.$ (3.10)

######  Lemma 3.8.

Let $\eta_{ij}\in M_{0}^{+}(\mathbb{D}(c_{ij},k_{1}\delta)\cap F)$ be such
that

$\ \|{\mathcal{C}}\eta_{ij}\|\leq C_{0},~{}\|\eta_{ij}\|=\alpha_{ij}.$

Set $h_{ij}=-{\mathcal{C}}\eta_{ij}.$ Let $N\geq 12k_{1}$ be an integer. Set
$S=S_{i_{0},j_{0}}$ and $J_{j_{0}}^{N}=\\{j_{0}-N,...,j_{0},...,j_{0}+N\\}$
for given $i_{0}$ and $j_{0}.$ For a given $i_{0}-N\leq i\leq i_{0}+N,$ let
$h_{J_{il}}$ be defined as in Lemma 3.6 (H1) for $J_{il}\in P_{1}.$ Then

$\displaystyle\ \int_{S}|h_{J_{il}}(z)|d{\mathfrak{m}}(z)\leq
C_{17}{\mathfrak{m}}(S)\left(\dfrac{1}{N^{2}}+\sum_{j\in J_{il}\cap
J_{j_{0}}^{N}}\dfrac{\alpha_{ij}}{\delta}\right).$ (3.11)

###### Proof.

Clearly, $h_{ij}$ satisfies (3.8). Let $J_{il}^{d},$ $J_{il}^{u},$ and
$H^{i}_{jk}$ be defined as in (H1) of Lemma 3.6. For $z\in S,$ we have the
following estimation cases for $H^{i}_{jk}.$

Case I: If $j\in J_{il}^{d}\setminus J_{j_{0}}^{N}$ and $k\in
J_{il}^{u}\setminus J_{j_{0}}^{N}$, then $|z-c_{ij}|\geq
N\delta>12k_{1}\delta,$ $|z-c_{ik}|>12k_{1}\delta,$ and by (H5) of Lemma 3.6,
we have

$\ |H^{i}_{jk}(z)|\leq
C_{18}\dfrac{\lambda_{ij}^{k}\alpha_{ij}+\lambda_{ik}^{j}\alpha_{ik}}{N^{2}\delta}.$

From (H3) of Lemma 3.6, we infer that

$\ \begin{aligned} \ H_{il}(z):=&\sum_{j\in J_{il}^{d}\setminus
J_{j_{0}}^{N}}\sum_{k\in J_{il}^{u}\setminus J_{j_{0}}^{N}}|H^{i}_{jk}(z)|\\\
\ \leq&\dfrac{C_{17}}{N^{2}\delta}\left(\sum_{j\in J_{il}^{d}}\sum_{k\in
J_{il}^{u}}\lambda_{ij}^{k}\alpha_{ij}+\sum_{j\in J_{il}^{d}}\sum_{k\in
J_{il}^{u}}\lambda_{ik}^{j}\alpha_{ik}\right)\\\ \
\leq&\dfrac{2C_{18}}{N^{2}}\ \end{aligned}$

Case II: If $j\in J_{il}^{d}\cap J_{j_{0}}^{N}$ and $k\in J_{il}^{u}\setminus
J_{j_{0}}^{N}$, then $|z-c_{ik}|>12k_{1}\delta$ and by (H1) and (H4) of Lemma
3.6, we have

$\ |H^{i}_{jk}(z)|\leq
C_{19}\left(\dfrac{\lambda_{ik}^{j}\alpha_{ik}}{N\delta}+\lambda_{ij}^{k}|\mathcal{C}(\eta_{ij})(z)|\right)=C_{19}\lambda_{ij}^{k}\left(\dfrac{\alpha_{ij}}{N\delta}+|\mathcal{C}(\eta_{ij})(z)|\right);$

Case III: If $j\in J_{il}^{d}\setminus J_{j_{0}}^{N}$ and $k\in J_{il}^{u}\cap
J_{j_{0}}^{N}$, then $|z-c_{ij}|>12k_{1}\delta$ and by (H1) and (H4) of Lemma
3.6, we have

$\ |H^{i}_{jk}(z)|\leq
C_{20}\lambda_{ik}^{j}\left(\dfrac{\alpha_{ik}}{N\delta}+|\mathcal{C}(\eta_{ik})(z)|\right);$

Case IV: If $j\in J_{il}^{d}\cap J_{j_{0}}^{N}$ and $k\in J_{il}^{u}\cap
J_{j_{0}}^{N}$, then by (H1) of Lemma 3.6, we have

$\ |H^{i}_{jk}(z)|\leq
C_{21}(\lambda_{ik}^{j}|\mathcal{C}(\eta_{ik})(z)|+\lambda_{ij}^{k}|\mathcal{C}(\eta_{ij})(z)|).$

Combining Cases I-IV, we estimate $h_{J_{il_{d}}}$ as the following:

$\ \begin{aligned} \ |h_{J_{il}}(z)|\leq&H_{il}(z)+C_{19}\sum_{j\in
J_{il}^{d}\cap J_{j_{0}}^{N}}\sum_{k\in J_{il}^{u}\setminus
J_{j_{0}}^{N}}\lambda_{ij}^{k}\left(\dfrac{\alpha_{ij}}{N\delta}+|\mathcal{C}(\eta_{ij})(z)|\right)\\\
\ &+C_{20}\sum_{j\in J_{il}^{d}\setminus J_{j_{0}}^{N}}\sum_{k\in
J_{il}^{u}\cap
J_{j_{0}}^{N}}\lambda_{ik}^{j}\left(\dfrac{\alpha_{ik}}{N\delta}+|\mathcal{C}(\eta_{ik})(z)|\right)\\\
\ &+C_{21}\sum_{j\in J_{il}^{d}\cap J_{j_{0}}^{N}}\sum_{k\in J_{il}^{u}\cap
J_{j_{0}}^{N}}(\lambda_{ik}^{j}|\mathcal{C}(\eta_{ik})(z)|+\lambda_{ij}^{k}|\mathcal{C}(\eta_{ij})(z)|).\end{aligned}$

Using (H2), we get

$\ \begin{aligned} \
|h_{J_{il}}(z)|\leq&\dfrac{C_{21}}{N^{2}}+C_{23}\sum_{j\in J_{il}^{d}\cap
J_{j_{0}}^{N}}\left(\dfrac{\alpha_{ij}}{N\delta}+|\mathcal{C}(\eta_{ij})(z)|\right)\\\
\ &+C_{23}\sum_{k\in J_{il}^{u}\cap
J_{j_{0}}^{N}}\left(\dfrac{\alpha_{ik}}{N\delta}+|\mathcal{C}(\eta_{ik})(z)|\right).\end{aligned}$

Hence,

$\ \begin{aligned} \ &\int_{S}|h_{J_{il}}(z)|d{\mathfrak{m}}(z)\\\ \
\leq&\dfrac{C_{23}}{N^{2}}{\mathfrak{m}}(S)+C_{22}\sum_{j\in J_{il}\cap
J_{j_{0}}^{N}}\left(\dfrac{\alpha_{ij}}{N\delta}{\mathfrak{m}}(S)+\|\eta_{ij}\|({\mathfrak{m}}(S))^{\frac{1}{2}}\right)\\\
\
\leq&\dfrac{C_{23}}{N^{2}}{\mathfrak{m}}(S)+C_{24}{\mathfrak{m}}(S)(\frac{1}{N}+1)\sum_{j\in
J_{il}\cap J_{j_{0}}^{N}}\dfrac{\alpha_{ij}}{\delta}.\end{aligned}$

This completes the proof. ∎

##  4\. The algebra $H^{\infty}(\mathcal{D})$

For a bounded measurable subset $\mathcal{D}\subset\mathbb{C},$ define

$\ L^{\infty}({\mathfrak{m}}_{\mathcal{D}})=\\{f\in
L^{\infty}(\mathbb{C}):~{}f(z)=0,~{}z\in\mathbb{C}\setminus\mathcal{D}\\}.$

######  Definition 4.1.

The subset $\mathcal{D}\subset\mathbb{C}$ is $\gamma$-open if there exists a
subset $\mathcal{Q}$ with ${\mathfrak{m}}(\mathcal{Q})=0$ such that for
$\lambda\in\mathcal{D}\setminus\mathcal{Q}$,

$\displaystyle\ \lim_{\delta\rightarrow
0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\setminus\mathcal{D})}{\delta}=0.$
(4.1)

If $\gamma(\mathcal{Q})=0,$ then $\mathcal{D}$ is called strong $\gamma$-open.

Clearly, by (2.1), if $\mathcal{D}$ is strong $\gamma$-open, then
$\mathcal{D}$ is $\gamma$-open.

######  Definition 4.2.

Let $\mathcal{D}\subset\mathbb{C}$ be a bounded measurable subset. Assume that
$\mathcal{D}$ is $\gamma$-open. Let $H_{\mathbb{C}\setminus\mathcal{D}}$ be
the set of functions $f(z)$ such that $f(z)$ is bounded and analytic on
$\mathbb{C}\setminus E_{f}$ for some compact subset
$E_{f}\subset\mathbb{C}\setminus\mathcal{D}.$ Define $H^{\infty}(\mathcal{D})$
to be the weak-star closed subalgebra of
$L^{\infty}({\mathfrak{m}}_{\mathcal{D}})$ generated by functions in
$H_{\mathbb{C}\setminus\mathcal{D}}.$

If $\mathcal{D}$ is a bounded open subset, then $H^{\infty}(\mathcal{D})$ is
the algebra of bounded and analytic functions on $\mathcal{D}.$ [4, Main
Theorem] gives an interesting example that $H^{\infty}(\mathcal{D})$ is
different from the algebra of bounded and analytic functions on an open
subset. The aim of this section is to prove the following theorem.

######  Theorem 4.3.

Let $\mathcal{D}$ be a $\gamma$-open bounded measurable subset. Let $f\in
L^{\infty}({\mathfrak{m}}_{\mathcal{D}})$ be given with
$\|f\|_{L^{\infty}({\mathfrak{m}}_{\mathcal{D}})}\leq 1$. If there exists
$C_{f}>0$ (depending on $f$) such that for $\lambda\in\mathbb{C}$, $\delta>0$,
a smooth non-negative function $\varphi$ with support in
$\mathbb{D}(\lambda,\delta),$ $\varphi(z)\leq 1,$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{25}}{\delta},$
we have

$\displaystyle\
\left|\int(z-\lambda)^{n}f(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{\mathcal{D}}(z)\right|\leq
C_{f}\delta^{n}\gamma(\mathbb{D}(\lambda,k_{1}\delta)\setminus\mathcal{D}),$
(4.2)

where $k_{1}\geq 2$ is a given integer and $n\geq 0,$ then there exists a
sequence of functions $\\{f_{n}\\},$ where $f_{n}$ is analytic off a compact
subset $F_{n}\subset\mathbb{C}\setminus\mathcal{D}$ and
$\|f_{n}\|_{\mathbb{C}}\leq C_{26}C_{f},$ such that $f_{n}(z)\rightarrow
f(z),~{}{\mathfrak{m}}_{\mathcal{D}}-a.a..$

We fix $f\in L^{\infty}({\mathfrak{m}}_{\mathcal{D}})$ with
$\|f\|_{L^{\infty}({\mathfrak{m}}_{\mathcal{D}})}\leq 1$ and
$f(z)=0,~{}z\in\mathbb{C}\setminus\mathcal{D}.$ Replacing $f$ by
$\frac{f}{C_{f}},$ we assume $C_{f}=1.$ Let $\\{\varphi_{ij},S_{ij},\delta\\}$
be a smooth partition of unity as in last section. Set
$\alpha_{ij}=\gamma(\mathbb{D}(\lambda,k_{1}\delta)\setminus\mathcal{D}).$ We
write $f$ as the following:

$\displaystyle\
f=\sum_{ij}f_{ij}=\sum_{2S_{ij}\cap\mathcal{D}\neq\emptyset}f_{ij},$ (4.3)

where $f_{ij}=T_{\varphi_{ij}}f.$ From the assumption (4.2), we see that, for
$n\geq 1,$

$\displaystyle\ \begin{aligned} \
|c_{n}(f_{ij},c_{ij})|=&\left|\int(z-\lambda)^{n-1}f(z)\dfrac{\partial\varphi_{ij}(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|\\\
\ \leq&C_{27}\delta^{n-1}\alpha_{ij}.\ \end{aligned}$ (4.4)

We use the notation $P,P_{1},P_{2},$ for $F=\mathbb{C}\setminus\mathcal{D}$ in
the last section.

######  Lemma 4.4.

Suppose that $\eta_{ij}\in
M_{0}^{+}(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{D})$ satisfying

$\displaystyle\ \|{\mathcal{C}}\eta_{ij}\|\leq C_{0}\text{ and
}\|\eta_{ij}\|=\alpha_{ij}.$ (4.5)

Then the following statements are true.
(1) Let $g_{ij}^{0}=-\frac{c_{1}(f_{ij})}{\alpha_{ij}}{\mathcal{C}}\eta_{ij},$
$g_{ij}=f_{ij}-g_{ij}^{0},$ and $h_{ij}=-{\mathcal{C}}\eta_{ij}.$ Then
$\\{g_{ij}\\}$ is applicable to $P$ and ${\mathfrak{m}}_{\mathcal{D}}$ (see
Definition 3.3), $\\{h_{ij}\\}$ satisfies the assumptions in Lemma 3.6,
$g_{J_{il}}$ defined in (3.6) satisfies the properties in Lemma 3.5, and
$h_{J_{il}}$ defined in (3.9) satisfies the properties in Lemma 3.6.
(2) Rewrite (4.3) as the following

$\ f=\sum_{J_{il}\in P_{1}}(g_{J_{il}}-h_{J_{il}})+\sum_{J_{il_{i}}\in
P_{2}}g_{J_{il_{i}}}+f_{\delta}$

where

$\displaystyle\ f_{\delta}=\sum_{J_{il}\in P}\sum_{j\in
J_{il}}g_{ij}^{0}+\sum_{J_{il}\in P_{1}}h_{J_{il}}.$ (4.6)

Then $f_{\delta}$ is bounded and analytic off a compact subset of
$\mathbb{C}\setminus\mathcal{D}$ and

$\displaystyle\ \|f_{\delta}\|_{L^{\infty}({\mathfrak{m}}_{\mathcal{D}})}\leq
C_{29}.$ (4.7)

(3) There exists $\delta_{n}\rightarrow 0$ as $n\rightarrow\infty$ such that

$\
\lim_{n\rightarrow\infty}\|f_{\delta_{n}}-f\|_{L^{1}({\mathfrak{m}}_{\mathcal{D}})}=0.$

###### Proof.

(Lemma 4.4 (1) and (2)): It is clear that

$\ |c_{n}(g_{ij}^{0},c_{ij})|\leq
C_{30}(k_{1}\delta)^{n-1}\alpha_{ij},~{}|c_{n}(h_{ij},c_{ij})|\leq(k_{1}\delta)^{n-1}\alpha_{ij}.$

(1) follows from Lemma 3.4 and (4.4).

(4.7) in (2) follows from (3.10). ∎

To prove Lemma 4.4 (3), we need several lemmas. Therefore, for the lemmas
below, we assume that there are $\eta_{ij}$ satisfying (4.5). The functions
$g_{ij}^{0},$ $g_{ij},$ $h_{ij},$ $g_{J_{il}},$ $h_{J_{il}},$ and $f_{\delta}$
are defined in Lemma 4.4. We fix a positive integer $N$ satisfying
$N>200(2k_{1}+3)^{2}$.

######  Lemma 4.5.

There exists $\mathcal{Q}_{f}$ with ${\mathfrak{m}}(\mathcal{Q}_{f})=0$ such
that for all $\epsilon>0,$

$\displaystyle\ \lim_{\delta\rightarrow
0}\dfrac{{\mathfrak{m}}(\mathbb{D}(\lambda,\delta))\cap\\{|f(z)-f(\lambda)|>\epsilon\\})}{\delta^{2}}=0$
(4.8)

at each point $\lambda\in\mathcal{D}\setminus\mathcal{Q}_{f}.$

###### Proof.

Let $\mathcal{D}_{f}\subset\mathcal{D}$ be the set of Lebesgue points for $f,$
that is, for $\lambda\in\mathcal{D}_{f},$

$\ \lim_{\delta\rightarrow
0}\dfrac{1}{{\mathfrak{m}}(\mathbb{D}(\lambda,\delta))}\int_{\mathbb{D}(\lambda,\delta)}|f(z)-f(\lambda)|d{\mathfrak{m}}(z)=0.$

Since

$\
\dfrac{{\mathfrak{m}}(\mathbb{D}(\lambda,\delta))\cap\\{|f(z)-f(\lambda)|>\epsilon\\})}{{\mathfrak{m}}(\mathbb{D}(\lambda,\delta))}\leq\dfrac{1}{\epsilon{\mathfrak{m}}(\mathbb{D}(\lambda,\delta))}\int_{\mathbb{D}(\lambda,\delta)}|f(z)-f(\lambda)|d{\mathfrak{m}}(z),$

(4.8) holds at each point of $\mathcal{D}_{f}.$ Set
$\mathcal{Q}_{f}=\mathcal{D}\setminus\mathcal{D}_{f},$ then
${\mathfrak{m}}(\mathcal{Q}_{f})=0$ by the Lebesgue differentiation theorem. ∎

Let $\lambda\in\mathcal{D}$ satisfy (4.1) and be such that $f$ satisfies (4.8)
at $\lambda.$ Define

$\displaystyle\
F(f,N,\lambda)=\left\\{z:~{}|f(z)-f(\lambda)|\geq\frac{1}{N^{3}}\right\\}.$
(4.9)

Let $\mathcal{D}(f,m,N)$ be the set of $\lambda\in\mathcal{D}$ such that

$\displaystyle\
{\mathfrak{m}}(F(f,N,\lambda)\cap\mathbb{D}(\lambda,\delta_{1}))<\dfrac{1}{N^{8}}\delta_{1}^{2}$
(4.10)

and

$\displaystyle\
\gamma(\mathbb{D}(\lambda,\delta_{1})\setminus\mathcal{D})<\dfrac{1}{N^{2}}\delta_{1}$
(4.11)

for $\delta_{1}\leq\frac{1}{m}$. From Definition 4.1 and Lemma 4.5, it is
straightforward to verify

$\displaystyle\
{\mathfrak{m}}\left(\mathcal{D}\setminus\bigcup_{m=1}^{\infty}\mathcal{D}(f,m,N)\right)=0.$
(4.12)

Set $S=S_{i_{0},j_{0}}$ for given $i_{0},j_{0}.$ and
$J_{k}^{N}=\\{k-N,...,k,...,k+N\\}.$

######  Lemma 4.6.

If $\delta<\frac{1}{2mN}$ and $\mathcal{D}(f,m,N)\cap S\neq\emptyset,$ then

$\displaystyle\ \sum_{i\in J_{i_{0}}^{N},j\in
J_{j_{0}}^{N}}\int_{\mathcal{D}(f,m,N)\cap
S}|f_{ij}(z)|d{\mathfrak{m}}(z)\leq\dfrac{C_{31}}{N}{\mathfrak{m}}(S).$ (4.13)

###### Proof.

Let $\lambda\in\mathcal{D}(f,m,N)\cap S.$ For $z,w\in F(f,N,\lambda)^{c}$, by
(4.9), we have $|f(z)-f(w)|<\frac{2}{N^{3}}$ and for $i\in J_{i_{0}}^{N},j\in
J_{j_{0}}^{N},$

$\ \begin{aligned} \
|f_{ij}(w)|\leq&\dfrac{1}{\pi}\int_{F(f,N,\lambda)}\dfrac{2\|f\|}{|z-w|}|\bar{\partial}\varphi_{ij}|d{\mathfrak{m}}(z)\\\
\
&+\dfrac{1}{N^{3}\pi}\int_{F(f,N,\lambda)^{c}}\dfrac{2}{|z-w|}|\bar{\partial}\varphi_{ij}|d{\mathfrak{m}}(z)\\\
\ \leq&C_{32}\sqrt{{\mathfrak{m}}(\mathbb{D}(\lambda,2N\delta)\cap
F(f,N,\lambda))}\|\bar{\partial}\varphi_{ij}\|+C_{32}\dfrac{1}{N^{3}}\\\ \
\leq&\dfrac{C_{33}}{N^{3}},\end{aligned}$

where (4.10) is used for the last step. Therefore,

$\displaystyle\ \begin{aligned} \ &\sum_{i\in J_{i_{0}}^{N},j\in
J_{j_{0}}^{N}}\int_{S\cap\mathcal{D}(f,m,N)}|f_{ij}(w)|d{\mathfrak{m}}(w)\\\ \
\leq&C_{34}\sum_{i\in J_{i_{0}}^{N},j\in
J_{j_{0}}^{N}}\left(\dfrac{{\mathfrak{m}}(S)}{N^{3}}+\int_{S\cap
F(f,N,\lambda)}|f_{ij}(w)|d{\mathfrak{m}}(w)\right)\\\
\leq&C_{35}(2N+1)^{2}\left(\dfrac{{\mathfrak{m}}(S)}{N^{3}}+{\mathfrak{m}}(\mathbb{D}(\lambda,2\delta)\cap
F(f,N,\lambda))\right)\\\ \ \leq&\dfrac{C_{36}}{N}{\mathfrak{m}}(S),\
\end{aligned}$ (4.14)

∎

where (4.10) is used again for the last step.

######  Lemma 4.7.

If $\delta<\frac{1}{2mN}$ and $\lambda\in\mathcal{D}(f,m,N)\cap
S\neq\emptyset,$ then

$\displaystyle\ \sum_{i\in J_{i_{0}}^{N},j\in J_{j_{0}}^{N}}\alpha_{ij}\leq
100(2k_{1}+3)^{2}\gamma(\mathbb{D}(\lambda,2N\delta)\setminus\mathcal{D})\leq
200(2k_{1}+3)^{2}\dfrac{\delta}{N}.$ (4.15)

###### Proof.

Let $F=\mathbb{D}(\lambda,2N\delta)\setminus\mathcal{D}.$ Then, by (4.11),
$\gamma(F)\leq\frac{2N\delta}{N^{2}}=\frac{2\delta}{N}.$ Therefore, for each
$\lambda\in\mathbb{C},$ the disk $\mathbb{D}(\lambda,\gamma(F))$ meets at most
$(2k_{1}+3)^{2}$ of sets
$\\{\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{D}\\}_{i\in
J_{i_{0}}^{N},j\in J_{j_{0}}^{N}}.$ Now (4.15) follows from [8, Theorem
VIII.2.7]. ∎

######  Lemma 4.8.

If $\delta<\frac{1}{2mN}$ and $\mathcal{D}(f,m,N)\cap S\neq\emptyset,$ then

$\displaystyle\ \sum_{i\in J_{i_{0}}^{N},j\in
J_{j_{0}}^{N}}\int_{S}|g_{ij}^{0}(z)|d{\mathfrak{m}}(z)\leq\dfrac{C_{37}}{N}{\mathfrak{m}}(S).$
(4.16)

###### Proof.

We have the following calculation.

$\ \begin{aligned} \
\int_{S}|g^{0}_{ij}(w)|d{\mathfrak{m}}(w)\leq&C_{38}\int\int_{S}\dfrac{1}{|z-w|}d{\mathfrak{m}}(w)d\eta_{ij}(z)\\\
\ \leq&C_{39}\sqrt{{\mathfrak{m}}(S)}\alpha_{ij}.\ \end{aligned}$

The proof now follows from Lemma 4.7. ∎

######  Lemma 4.9.

If $\delta<\frac{1}{2mN}$ and $S\cap\mathcal{D}(f,m,N)\neq\emptyset,$ then

$\displaystyle\
\int_{S\cap\mathcal{D}(f,m,N)}|f(z)-f_{\delta}(z)|d{\mathfrak{m}}(z)\leq\dfrac{C_{40}}{N^{\frac{1}{5}}}{\mathfrak{m}}(S).$
(4.17)

###### Proof.

Set $\Psi_{J_{il}}=g_{J_{il}}-h_{J_{il}}$ Define

$\ R_{1}=\\{J_{il}\in P_{1}:~{}i\in J_{i_{0}}^{N}\text{ and }J_{il}\cap
J_{j_{0}}^{N}\neq\emptyset\\}$

and

$\ R_{2}=\\{J_{il_{i}}\in P_{2}:~{}i\in J_{i_{0}}^{N}\text{ and
}J_{il_{i}}\cap J_{j_{0}}^{N}\neq\emptyset\\}.$

By definition of a complete group (see (3.3), (3.4), and (3.5)) and Lemma 4.7,
for each $i\in J_{i_{0}}^{N},$ there are at most two $J_{il}\in R_{1}.$ So we
rewrite

$\ R_{1}=\\{J_{i}^{d},J_{i}^{u}:~{}i\in J_{i_{0}}^{N}\\},$

where $j_{1}<j_{2}$ for $j_{1}\in J_{i}^{d}$ and $j_{2}\in J_{i}^{u}.$ For
$z\in S,$ from Lemma 3.7, we see that

$\displaystyle\ \sum_{J_{il}\in P_{1}\setminus
R_{1}}|\Psi_{J_{il}}(z)|+\sum_{J_{il_{i}}\in P_{2}\setminus
R_{2}}|g_{J_{il_{i}}}(z)|\leq\dfrac{C_{41}}{N^{\frac{1}{5}}}.$ (4.18)

For $i\in J_{i_{0}}^{N},$ $J_{i}^{d},J_{i}^{u}\in R_{1},$ and $J_{il_{i}}\in
R_{2},$ The index groups $J_{i}^{d}\setminus J_{j_{0}}^{N},$
$J_{i}^{u}\setminus J_{j_{0}}^{N},$ and $J_{il_{i}}\setminus J_{j_{0}}^{N}$
can be viewed as incomplete groups. Hence, using Lemma 3.7, we get

$\ \sum_{i\in J_{i_{0}}^{N}}(|g_{J_{i}^{d}\setminus
J_{j_{0}}^{N}}(z)|+|g_{J_{i}^{u}\setminus
J_{j_{0}}^{N}}(z)|+|g_{J_{il_{i}}\setminus
J_{j_{0}}^{N}}(z)|)\leq\dfrac{C_{42}}{N^{\frac{1}{5}}}$

Combining with Lemma 4.6 and Lemma 4.8, we have

$\displaystyle\ \int_{S\cap\mathcal{D}(f,m,N)}\left(\sum_{J_{il}\in
R_{1}}|g_{J_{il}}(z)|+\sum_{J_{il_{i}}\in
R_{2}}|g_{J_{il_{i}}}(z)|\right)d{\mathfrak{m}}(z)\leq\dfrac{C_{43}}{N^{\frac{1}{5}}}{\mathfrak{m}}(S).$
(4.19)

Using Lemma 3.8 and Lemma 4.7, we conclude that

$\displaystyle\ \int_{S\cap\mathcal{D}(f,m,N)}\sum_{i\in
J_{i_{0}}^{N}}(|h_{J_{i}^{d}}(z)|+|h_{J_{i}^{u}}(z)|)d{\mathfrak{m}}(z)\leq\dfrac{C_{44}}{N}{\mathfrak{m}}(S).$
(4.20)

Combining (4.18), (4.19), and (4.20), we prove the lemma. ∎

Since ${\mathfrak{m}}(\mathcal{D})<\infty,$ we obtain the following corollary.

######  Corollary 4.10.

If $\delta<\frac{1}{2mN},$ then

$\
\int_{\mathcal{D}(f,m,N)}|f(z)-f_{\delta}(z)|d{\mathfrak{m}}(z)\leq\dfrac{C_{45}}{N^{\frac{1}{5}}}.$

###### Proof.

(Lemma 4.4 (3)): There exists $m_{N}$ such that, from (4.12),

$\ {\mathfrak{m}}(\mathcal{D}\setminus\mathcal{D}(f,m_{N},N))<\dfrac{1}{N}.$

Set $\delta_{N}=\frac{1}{4m_{N}N}$. Clearly, by Corollary 4.10 and (4.7),

$\ \|f_{\delta_{N}}-f\|_{L^{1}({\mathfrak{m}}_{\mathcal{D}})}\rightarrow 0.$

∎

###### Proof.

(Theorem 4.3): Applying Theorem 2.1 (1), there exists $\eta_{ij}^{0}\in
M_{0}^{+}(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{D})$ such that
$\|\eta_{ij}^{0}\|\geq\frac{1}{2A_{T}}\alpha_{ij}$ and
$\|{\mathcal{C}}(\eta_{ij}^{0})\|_{\mathbb{C}}\leq 1.$ Set
$\eta_{ij}=\frac{\eta_{ij}^{0}}{\|\eta_{ij}^{0})\|}\alpha_{ij}.$ Then
$\eta_{ij}$ satisfies (4.5).

From Lemma 4.4 (3), we see that $f_{N}:=f_{\delta_{N}}\rightarrow
f,~{}{\mathfrak{m}}_{\mathcal{D}}-a.a.$ (by passing to a subsequence). Thus,
$f_{N}\rightarrow f$ in $L^{\infty}({\mathfrak{m}}_{\mathcal{D}})$ weak-star
topology. This implies $f\in H^{\infty}(\mathcal{D})$ as $f_{N}\in
H^{\infty}(\mathcal{D})$. This proves the theorem. ∎

##  5\. Non-removable boundary and removable set

For a compact subset $K\subset\mathbb{C}$ and $\mu\in M_{0}^{+}(K),$ we assume
$R^{\infty}(K,\mu)$ is pure.

######  Definition 5.1.

The envelope $E$ for $K$ and $\mu$ is the set of points $\lambda\in K$ such
that there exists $\mu_{\lambda}\in M_{0}(K)$ that is absolutely continuous
with respect to $\mu$ such that $\mu_{\lambda}(\\{\lambda\\})=0$ and $\int
f(z)d\mu_{\lambda}(z)=f(\lambda)$ for each $f\in R(K).$

The elementary properties of $E$ are listed below.

######  Proposition 5.2.

(a) $E$ is the set of weak-star continuous homomorphisms on
$R^{\infty}(K,\mu)$ (see [3, Proposition VI.2.5]).
(b) $E$ is a nonempty Borel set with area density one at each of its points
(see [3, Proposition VI.2.8]).
(c) $\text{int}(E)=\text{int}(\overline{E})$ (see [3, Proposition VI.3.9]).
(d) $\overline{E}$ is the union of $\text{spt}\mu$ and some collection of
bounded components $\mathbb{C}\setminus\text{spt}\mu$ (see [3, Proposition
VI.3.11]).

For $\lambda\in E$ and $f\in R^{\infty}(K,,\mu),$ set $\rho(f)(\lambda)=\int
fd\mu_{\lambda}.$ Clearly $\rho(f)(\lambda)$ is independent of the particular
$\mu_{\lambda}$ chosen. We thus have a map $\rho$ called Chaumat’s map for $K$
and $\mu,$ which associates to each function in $R^{\infty}(K,\mu)$ a point
function on $E.$ Chaumat’s Theorem [2] states as the following: The map $\rho$
is an isometric isomorphism and a weak-star homeomorphism from
$R^{\infty}(K,\mu)$ onto $R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$ (also
see [3, Chaumat’s Theorem on page 288]). Therefore, we make the following
assumptions:
(a) $R^{\infty}(K,\mu)$ is pure (has no non-trivial $L^{\infty}$ summands);
(b) $K=\overline{E},$ where $E$ is the envelope for $R^{\infty}(K,\mu);$
(c) $\mu={\mathfrak{m}}_{E}.$

######  Definition 5.3.

Let $\Lambda=\\{g_{n}\\}\subset R(\overline{E})^{\perp}\cap
L^{1}({\mathfrak{m}}_{E})$ be a $L^{1}({\mathfrak{m}}_{E})$ norm dense subset.
The non-removable boundary for $\Lambda$ is defined by

$\
\mathcal{F}(\Lambda)=\bigcap_{n=1}^{\infty}\\{z:~{}\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(g_{n}{\mathfrak{m}}_{E})(z)\text{ exists,
}\mathcal{C}(g_{n}{\mathfrak{m}}_{E})(z)=0\\}$

and the removable set for $\Lambda$ is defined by

$\
\mathcal{R}(\Lambda)=\bigcup_{n=1}^{\infty}\\{z:~{}\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(g_{n}{\mathfrak{m}}_{E})(z)\text{ exists,
}\mathcal{C}(g_{n}{\mathfrak{m}}_{E})(z)\neq 0\\}.$

Clearly,

$\displaystyle\ \mathcal{R}(\Lambda)\cap\mathcal{F}(\Lambda)=\emptyset\text{
and
}\mathcal{R}(\Lambda)\cup\mathcal{F}(\Lambda)\approx\mathbb{C},~{}\gamma-a.a..$
(5.1)

The concept of non-removable boundary and removable set was first introduced
by [4] for a string of beads set. Conway and Yang [5] extended the concept to
an arbitrary compact subset $K\subset\mathbb{C}$ and $\mu\in M_{0}^{+}(K).$ We
will see that $\mathcal{R}(\Lambda)$ and $\mathcal{F}(\Lambda)$ are more
appropriate than that of the envelope $E$ in studying $R^{\infty}(K,\mu).$

######  Lemma 5.4.

If $g\in R(\overline{E})^{\perp}\cap L^{1}({\mathfrak{m}}_{E}),$ then

$\
\mathcal{C}(g{\mathfrak{m}}_{E})(\lambda)=0,~{}\gamma|_{\mathcal{F}(\Lambda)}-a.a..$

###### Proof.

Assume there exists a compact subset $F_{1}\subset\mathcal{F}(\Lambda)$ such
that $\gamma(F_{1})>0$ and

$\displaystyle\ \mathcal{C}(g_{n}{\mathfrak{m}}_{E})(\lambda)=0\text{ for
}n\geq 1\text{ and
}Re(\mathcal{C}(g{\mathfrak{m}}_{E}))(\lambda)>0,~{}\lambda\in F_{1}.$ (5.2)

Applying Lemma 2.7 for $\\{g\\}\cup\Lambda\subset L^{1}({\mathfrak{m}}_{E})$
and $F_{1},$ we get $\eta\in M_{0}^{+}(F_{1})$ and $f(z)$ satisfy (1)-(3) of
Lemma 2.7. By (2.6), for $n\geq 1,$

$\ \int\mathcal{C}(g_{n}{\mathfrak{m}}_{E})(z)d\eta(z)=-\int
f(z)g_{n}(z)d{\mathfrak{m}}_{E}(z)=0,$

which implies $f\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$ by the Hahn-
Banach Theorem. Applying (2.6) to $g,$ we get

$\ \int Re(\mathcal{C}(g{\mathfrak{m}}_{E})(z))d\eta(z)=-Re\left(\int
f(z)g(z)d{\mathfrak{m}}_{E}(z)\right)=0,$

which implies $Re(\mathcal{C}(g\mu)(z))=0,~{}\eta-a.a..$ This contradicts
(5.2). ∎

The following corollary follows from Lemma 5.4.

######  Corollary 5.5.

Let $\Lambda=\\{g_{n}\\},~{}\Lambda^{\prime}=\\{g_{n}^{\prime}\\}\subset
R(\overline{E})^{\perp}\cap L^{1}({\mathfrak{m}}_{E})$ be two dense subsets.
Then

$\ \ \mathcal{F}(\Lambda)\approx\ \mathcal{F}(\Lambda^{\prime})\text{ and
}\mathcal{R}(\Lambda)\approx\ \mathcal{R}(\Lambda^{\prime}),~{}\gamma-a.a..$

Hence, $\mathcal{F}(\Lambda)$ and $\mathcal{R}(\Lambda)$ are independent of
choices of $\Lambda$ up to a set of zero analytic capacity. We will simply use
$\mathcal{F}$ for $\mathcal{F}(\Lambda)$ and $\mathcal{R}$ for
$\mathcal{R}(\Lambda).$

######  Lemma 5.6.

Let $g\in L^{1}({\mathfrak{m}}_{E}).$ Then there exists $\mathcal{Q}$ with
$\gamma(\mathcal{Q})=0$ such that for
$\lambda\in\mathbb{C}\setminus\mathcal{Q},$
$\Theta_{g{\mathfrak{m}}_{E}}(\lambda)=0$ and
${\mathcal{C}}(g{\mathfrak{m}}_{E})$ is $\gamma$-continuous at $\lambda.$

###### Proof.

By Lemma 2.8 and Corollary 2.3, there exists $\mathcal{Q}$ with
$\gamma(\mathcal{Q})=0$ such that for
$\lambda\in\mathbb{C}\setminus\mathcal{Q},$
$\Theta_{g{\mathfrak{m}}_{E}}(\lambda)=0$ and the principle values of
${\mathcal{C}}(g{\mathfrak{m}}_{E})(\lambda)$ exists. From Lemma 2.11, we see
that ${\mathcal{C}}(g{\mathfrak{m}}_{E})(z)$ is $\gamma$-continuous at
$\lambda.$ ∎

The following corollary for $\mathcal{R}$ is a generalization of Proposition
5.2 (b) for $E.$

######  Corollary 5.7.

The removable set $\mathcal{R}$ is strong $\gamma$-open.

###### Proof.

From Lemma 5.6, for $\lambda\in\mathcal{R},~{}\gamma-a.a.,$ there exists
$g_{n}$ such that $\Theta_{g_{n}{\mathfrak{m}}_{E}}(\lambda)=0,$
${\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(\lambda)\neq 0,$ and
${\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)$ is $\gamma$-continuous at
$\lambda.$ Therefore,

$\displaystyle\lim_{\delta\rightarrow
0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\setminus\mathcal{R})}{\delta}$
$\displaystyle\ \leq$ $\displaystyle\lim_{\delta\rightarrow
0}\dfrac{\gamma\left(\mathbb{D}(\lambda,\delta)\cap\left\\{|{\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)-{\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(\lambda)|>\frac{|{\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(\lambda)|}{2}\right\\}\right)}{\delta}$
$\displaystyle\ =$ $\displaystyle 0.$

So $\mathcal{R}$ is strong $\gamma$-open. ∎

Recall $\tilde{\nu}$ is defined by (2.3) for $\nu\in M_{0}(\mathbb{C})$. For
$g\in R(\overline{E})^{\perp}\cap L^{1}({\mathfrak{m}}_{E}),$ define

$\displaystyle\
E_{0}(g{\mathfrak{m}}_{E})=\\{\lambda:~{}\widetilde{g{\mathfrak{m}}_{E}}(\lambda)<\infty,~{}\mathcal{C}(g{\mathfrak{m}}_{E})(\lambda)\neq
0\\}.$ (5.3)

Clearly, $E_{0}(g{\mathfrak{m}}_{E})\subset E$ because for $\lambda\in
E_{0}(g{\mathfrak{m}}_{E}),$
$\mu_{\lambda}=\frac{g{\mathfrak{m}}_{E}(z)}{{\mathcal{C}}(g{\mathfrak{m}}_{E})(\lambda)(z-\lambda)}$
satisfies Definition 5.1.

######  Proposition 5.8.

There exists a sequence $\\{g_{n}\\}\subset R(\overline{E})^{\perp}\cap
L^{1}({\mathfrak{m}}_{E})$ and there ecists a doubly indexed sequence of open
balls $\\{\mathbb{D}(\lambda_{ij},\delta_{ij})\\}$ such that

$\
E\approx\bigcup_{n=1}^{\infty}E_{0}(g_{n}{\mathfrak{m}}_{E})\approx\bigcap_{j=1}^{\infty}\bigcup_{i=1}^{\infty}\mathbb{D}(\lambda_{ij},\delta_{ij}),~{}{\mathfrak{m}}-a.a.$

###### Proof.

For $\lambda\in E$, let $\mu_{\lambda}$ be as in Definition 5.1. Let
$g_{\lambda}(z)=(z-\lambda)\frac{d\mu_{\lambda}}{d{\mathfrak{m}}_{E}},$ then
$g_{\lambda}\in R(\overline{E})^{\perp}\cap L^{1}({\mathfrak{m}}_{E})$ such
that $\widetilde{g_{\lambda}{\mathfrak{m}}_{E}}(\lambda)<\infty$ and
$\mathcal{C}(g_{\lambda}{\mathfrak{m}}_{E})(\lambda)\neq 0$. By Lemma 2.11 and
(2.1), we have

$\displaystyle\begin{aligned} \ &\lim_{\delta\rightarrow
0}\dfrac{{\mathfrak{m}}(\mathbb{D}(\lambda,\delta)\setminus
E_{0}(g_{\lambda}{\mathfrak{m}}_{E}))}{\delta^{2}}\\\ \
\leq&4\pi\lim_{\delta\rightarrow
0}\dfrac{\left(\gamma\left(\mathbb{D}(\lambda,\delta)\cap\left\\{|\mathcal{C}(g_{\lambda}{\mathfrak{m}}_{E})(z)-\mathcal{C}(g_{\lambda}{\mathfrak{m}}_{E})(\lambda)|>\frac{\mathcal{C}(g_{\lambda}{\mathfrak{m}}_{E})(\lambda)|}{2}\right\\}\right)\right)^{2}}{\delta^{2}}\\\
\ =&0.\end{aligned}$ (5.4)

There is $\delta_{\lambda}^{n}>0$ such that from (5.4), we have

$\ {\mathfrak{m}}(\mathbb{D}(\lambda,\delta)\setminus
E_{0}(g_{\lambda}{\mathfrak{m}}_{E}))\leq\dfrac{1}{n}\delta^{2},~{}\delta<\delta_{\lambda}^{n}.$

By $5r$-covering theorem (Theorem 2.2 in [12]), we can find a sequence of
disjoint disks
$\\{\mathbb{D}(\lambda_{mn},\frac{1}{5}\delta_{\lambda_{mn}}^{n})\\}$ such
that

$\ \bigcup_{m=1}^{\infty}E_{0}(g_{\lambda_{mn}}{\mathfrak{m}}_{E})\subset
E\subset\bigcup_{\lambda\in
E}\mathbb{D}(\lambda,\frac{1}{5}\delta_{\lambda}^{n})\subset\bigcup_{m=1}^{\infty}5\mathbb{D}(\lambda_{mn},\frac{1}{5}\delta_{\lambda_{mn}}^{n}).$

Hence,

$\
\bigcup_{n=1}^{\infty}\bigcup_{m=1}^{\infty}E_{0}(g_{\lambda_{mn}}{\mathfrak{m}}_{E})\subset
E\subset\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}5\mathbb{D}(\lambda_{mn},\frac{1}{5}\delta_{\lambda_{mn}}^{n})$

and

$\displaystyle{\mathfrak{m}}\left(\bigcup_{m=1}^{\infty}\mathbb{D}(\lambda_{mn},\delta_{\lambda_{mn}}^{n})\setminus\bigcup_{m=1}^{\infty}E_{0}(g_{\lambda_{mn}}{\mathfrak{m}}_{E})\right)$
$\displaystyle\ \leq$
$\displaystyle\sum_{m=1}^{\infty}{\mathfrak{m}}\left(\mathbb{D}(\lambda_{mn},\delta_{\lambda_{mn}}^{n})\setminus
E_{0}(g_{\lambda_{mn}}{\mathfrak{m}}_{E})\right)$ $\displaystyle\ \leq$
$\displaystyle\dfrac{1}{n}\sum_{m=1}^{\infty}(\delta_{\lambda_{mn}}^{n})^{2}.$

Notice that $\sum_{m=1}^{\infty}(\delta_{\lambda_{mn}}^{n})^{2}$ is bounded,
therefore,

$\
E\approx\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}\mathbb{D}(\lambda_{mn},\delta_{\lambda_{mn}}^{n}),~{}{\mathfrak{m}}-a.a.$

since

$\
{\mathfrak{m}}\left(\bigcap_{n=1}^{\infty}\bigcup_{m=1}^{\infty}\mathbb{D}(\lambda_{mn},\delta_{\lambda_{mn}}^{n})\setminus\bigcup_{n=1}^{\infty}\bigcup_{m=1}^{\infty}E_{0}(g_{\lambda_{mn}}\mu)\right)=0.$

∎

######  Corollary 5.9.

$E\approx\mathcal{R},~{}{\mathfrak{m}}-a.a.$ and
$\overline{E}=\overline{\mathcal{R}}.$

###### Proof.

Let $\Lambda=\\{g_{n}\\}$ be as in Proposition 5.8. We may assume that
$\Lambda$ is dense in $R(\overline{E})^{\perp}\cap L^{1}({\mathfrak{m}}_{E}).$
If $\lambda\in E_{0}(g_{n}{\mathfrak{m}}_{E}),$ then
$\lim_{\epsilon\rightarrow 0}{\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(\lambda)$
exists since $\widetilde{g_{n}{\mathfrak{m}}_{E}}(\lambda)<\infty.$ Hence,
$E_{0}(g_{n}{\mathfrak{m}}_{E})\subset\mathcal{R}.$ The inclusion
$\mathcal{R}\subset\cup_{n=1}^{\infty}E_{0}(g_{n}{\mathfrak{m}}_{E}),~{}{\mathfrak{m}}-a.a..$
follows from Fubini’s theorem.

The inclusion $\overline{E}\subset\overline{\mathcal{R}}$ follows from Lemma
5.4, while $\overline{\mathcal{R}}\subset\overline{E}$ follows from Corollary
5.7. ∎

##  6\. A criterion for functions in
$R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$

In this section, we prove the following criterion.

######  Theorem 6.1.

Let $f\in L^{\infty}({\mathfrak{m}}_{E})$ be given with
$\|f\|_{L^{\infty}({\mathfrak{m}}_{E})}\leq 1$. Then the following statements
are equivalent.

(1) $f\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$

(2) There exists a constant $C_{f}>0$ (only depending on $f$) and an integer
$k_{1}\geq 2$ so that for all $\lambda\in\mathbb{C}$, $\delta>0$, and for all
choices of a smooth non-negative function $\varphi$ with support in
$\mathbb{D}(\lambda,\delta)$ satisfying $\varphi\leq 1$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{45}}{\delta},$
we have

$\
\left|\int(z-\lambda)^{n}f(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{E}(z)\right|\leq
C_{f}\delta^{n}\gamma(\mathbb{D}(\lambda,k_{1}\delta)\setminus\mathcal{R})$

for all integers $n\geq 0.$

(3) $f\in H^{\infty}(\mathcal{R}).$

###### Proof.

(1) $\Rightarrow$ (2): For $f\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}),$
we find $\\{f_{n}\\}\subset R(\overline{E})$ such that $\|f_{n}\|\leq C_{f}$
and $f_{n}(z)\rightarrow f(z),~{}{\mathfrak{m}}_{E}-a.a.$ (see [3, Proposition
VI.22.3]). Using [8, Theorem VIII.8.1], for a smooth $\varphi$ with support in
$\mathbb{D}(\lambda,\delta),$ we get

$\ \begin{aligned} \ &\left|\int
f_{n}(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{E}(z)\right|\\\
\
\leq&\left|\int_{\overline{E}}f_{n}(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|+\left|\int_{\overline{E}\setminus
E}f_{n}(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{E}(z)\right|\\\
\
\leq&C_{46}C_{f}\delta\left\|\dfrac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\left(\gamma(\mathbb{D}(\lambda,k_{1}\delta)\setminus\overline{E})+\sqrt{{\mathfrak{m}}(\mathbb{D}(\lambda,k_{1}\delta)\cap(\overline{E}\setminus
E))}\right).\end{aligned}$

From Corollary 5.9, we get ${\mathfrak{m}}_{\mathcal{R}}={\mathfrak{m}}_{E}.$
Using (2.1), we have

$\
\sqrt{{\mathfrak{m}}(\mathbb{D}(c_{ij},k_{1}\delta)\cap(\overline{E}\setminus
E))}\leq\sqrt{{\mathfrak{m}}(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{R})}\leq\sqrt{4\pi}\gamma(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{R}).$

Hence, since $\mathcal{R}\subset\overline{E},$

$\ \left|\int
f_{n}(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{E}(z)\right|\leq
C_{46}C_{f}(1+\sqrt{4\pi})\delta\left\|\dfrac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\gamma(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{R}).$

Taking weak-star limits for $n\rightarrow\infty$ and replacing $\varphi$ by
$(z-\lambda)^{n}\varphi,$ we prove (2).

(2) $\Rightarrow$ (3) follows from Theorem 4.3 since $\mathcal{R}$ is
$\gamma$-open from Corollary 5.7.

(3) $\Rightarrow$ (1): Let $f$ be a bounded and analytic function outside of a
compact subset $E_{f}\subset\mathbb{C}\setminus\mathcal{R}$ with $\|f\|\leq
1.$ For $\lambda\in\mathbb{C}$, $\delta>0$, and a smooth function $\varphi$
with support in $\mathbb{D}(\lambda,\delta)$, we see that
$\frac{T_{\varphi}f}{\|T_{\varphi}f\|}$ is analytic off
$\text{supp}(\varphi)\cap E_{f}.$ Using (3.1), we have the following
calculation.

$\ \begin{aligned} \
|(T_{\varphi}f)^{\prime}(\infty)|=&\dfrac{1}{\pi}\left|\int
f(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|\\\ \
\leq&\|T_{\varphi}f\|\gamma(\text{supp}(\varphi)\cap E_{f})\\\ \
\leq&C_{47}\delta\left\|\dfrac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\gamma(\mathbb{D}(\lambda,\delta)\setminus\mathcal{R}).\
\end{aligned}$

Since, by (2.1),

$\ \begin{aligned} \
\left|\int_{\mathbb{C}\setminus\mathcal{R}}f(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|\leq&C_{48}\left\|\dfrac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}{\mathfrak{m}}(\mathbb{D}(\lambda,\delta)\setminus\mathcal{R})\\\
\
\leq&C_{49}\delta\left\|\dfrac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\gamma(\mathbb{D}(\lambda,\delta)\setminus\mathcal{R}),\
\end{aligned}$

we get

$\ \left|\int
f(z)\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{E}(z)\right|\leq
C_{50}\delta\left\|\dfrac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\gamma(\mathbb{D}(\lambda,\delta)\setminus\mathcal{R}).$

Therefore, the assumption of Theorem 4.3 holds if we replace $\varphi$ by
$(z-\lambda)^{n}\varphi$ above. We now use the notation in the proof of
Theorem 4.3 for $\mathcal{D}=\mathcal{R}.$ Applying Lemma 2.7, we get
$\eta_{ij}^{0}\in
M_{0}^{+}(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{R})$ for
$\Lambda=\\{g_{n}\\},$ a dense subset of $R(\overline{E})^{\perp}\cap
L^{1}({\mathfrak{m}}_{E})$, such that
$\|\eta_{ij}^{0}\|\geq\frac{1}{2C_{5}}\alpha_{ij},$ where
$\alpha_{ij}=\gamma(\mathbb{D}(c_{ij},k_{1}\delta)\setminus\mathcal{R}).$ By
(2.6), $\int{\mathcal{C}}(\eta_{ij}^{0})g_{n}d{\mathfrak{m}}_{E}=0$ for $n\geq
1$ since ${\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)=0,~{}\eta_{ij}^{0}-a.a.$
by Lemma 2.4. Hence, using the Hahn-Banach Theorem, we see that
${\mathcal{C}}(\eta_{ij}^{0})\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$
Set $\eta_{ij}=\frac{\eta_{ij}^{0}}{\|\eta_{ij}^{0}\|}\alpha_{ij}.$ Then
$\eta_{ij}$ satisfies (4.5) and $f_{\delta}\in
R^{\infty}(\overline{E},{\mathfrak{m}}_{E}),$ where $f_{\delta}$ is defined as
in (4.6). Applying Lemma 4.4 (3), we see that $f\in
R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$ Thus,
$H^{\infty}(\mathcal{R})\subset R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$
since those functions $f$ above are weak-star dense in
$H^{\infty}(\mathcal{R}).$ ∎

######  Corollary 6.2.

Let $\eta\in M_{0}^{+}(\mathbb{C})$ and
$\|{\mathcal{C}}\eta\|_{L^{\infty}(\mathbb{C})}\leq 1.$ Suppose that
${\mathcal{C}}\eta\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$ If $w\in
L^{\infty}(\eta)$ such that $\|w\|_{L^{\infty}(\eta)}\leq 1$ and
$\|{\mathcal{C}}(w\eta)\|_{L^{\infty}(\mathbb{C})}\leq 1,$ then
${\mathcal{C}}(w\eta)\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$

###### Proof.

For a non-negative smooth function $\phi$ supported in
$\mathbb{D}(\lambda,2\delta)$ with $0\leq\phi\leq 1,$ $\phi(z)=1$ for
$z\in\mathbb{D}(\lambda,\delta),$ and
$\|\bar{\partial}\phi\|\leq\frac{C_{51}}{\delta},$ we infer that, from Theorem
6.1 (2), (2.4), and (2.1),

$\displaystyle\ \eta(\mathbb{D}(\lambda,\delta))\leq$ $\displaystyle
C_{52}\left|\int{\mathcal{C}}\eta(z)\dfrac{\partial\phi(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|$
$\displaystyle\leq$ $\displaystyle
C_{52}\left|\int{\mathcal{C}}\eta(z)\dfrac{\partial\phi(z)}{\partial\bar{z}}d{\mathfrak{m}}_{E}(z)\right|+C_{52}\left\|\dfrac{\partial\phi(z)}{\partial\bar{z}}\right\|{\mathfrak{m}}(\mathbb{D}(\lambda,2\delta)\setminus\mathcal{R})$
$\displaystyle\ \leq$ $\displaystyle
C_{\eta}\gamma(\mathbb{D}(\lambda,2k_{1}\delta)\setminus\mathcal{R}),$

since ${\mathcal{C}}\eta\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}),$
where $C_{\eta}$ is a constant (depending on ${\mathcal{C}}\eta$).

For $\lambda\in\mathbb{C}$, $\delta>0$, a smooth non-negative function
$\varphi$ with support in $\mathbb{D}(\lambda,\delta),$ $\varphi\leq 1,$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{45}}{\delta},$
using (2.1), we have

$\ \begin{aligned} \
&\left|\int(z-\lambda)^{n}{\mathcal{C}}(w\eta)\bar{\partial}\varphi
d{\mathfrak{m}}_{E}\right|\\\ \
\leq&\left|\int(z-\lambda)^{n}{\mathcal{C}}(w\eta)\bar{\partial}\varphi
d{\mathfrak{m}}\right|+\delta^{n}\|{\mathcal{C}}(w\eta)\|\|\bar{\partial}\varphi\|{\mathfrak{m}}(\mathbb{D}(\lambda,\delta)\setminus
E)\\\ \
\leq&\pi\delta^{n}\eta(\mathbb{D}(\lambda,\delta))+\delta^{n-1}{\mathfrak{m}}(\mathbb{D}(\lambda,\delta)\setminus\mathcal{R}))\\\
\
\leq&C_{53}C_{\eta}\delta^{n}\gamma(\mathbb{D}(\lambda,2k_{1}\delta)\setminus\mathcal{R}).\
\end{aligned}$

Now the proof follows from Theorem 6.1. ∎

######  Corollary 6.3.

Let $\eta\in M_{0}^{+}(\mathbb{C})$ and
$\|{\mathcal{C}}\eta\|_{L^{\infty}(\mathbb{C})}\leq 1.$ If
${\mathcal{C}}\eta\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}),$ then
$\eta(\mathcal{R})=0.$

###### Proof.

Suppose that there exists a compact subset $B_{0}\subset\mathcal{R}$ such that
$\eta(B_{0})>0.$ Then from Corollary 2.4, we see that $\gamma(B_{0})>0.$ By
Proposition 2.2 (1)-(3), there exists $B_{1}\subset B_{0}$ with
$\eta(B_{1})>0$ and $N_{2}(\eta_{B_{1}})<\infty.$ Without loss of generality,
we assume $N_{2}(\eta_{B_{1}})\leq 1.$ Using Proposition 2.2 (4) and Theorem
2.1 (1), we have $\eta_{B_{1}}(B_{2})\leq C_{54}\gamma(B_{2})$ for
$B_{2}\subset B_{1}$ since $N_{2}(\eta_{B_{2}})\leq 1.$ Therefore, applying
Lemma 2.6, we may assume ${\mathcal{C}}_{*}(g_{n}{\mathfrak{m}}_{E})\in
L^{\infty}(\eta_{B_{1}}).$ Using Proposition 2.2 (4), we find $0\leq w(z)\leq
1$ supported on $B_{1}$ such that $\eta(B_{1})\leq 2\int wd\eta_{B_{1}},$ and
$\|{\mathcal{C}}_{\epsilon}(w\eta_{B_{1}})\|\leq C_{55}.$ Thus, for a non-
negative smooth function $\psi$ with compact support,
$T_{\psi}({\mathcal{C}}(w\eta_{B_{1}}))={\mathcal{C}}(\psi w\eta_{B_{1}})\in
L^{\infty}(\mathbb{C}).$ From Corollary 6.2, we see that ${\mathcal{C}}(\psi
w\eta_{B_{1}})\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$ Using the
Lebesgue dominated convergence theorem and taking $\epsilon\rightarrow 0$ for

$\ \int{\mathcal{C}}_{\epsilon}(g_{n}{\mathfrak{m}}_{E})\psi
wd\eta_{B_{1}}=-\int{\mathcal{C}}_{\epsilon}(\psi
w\eta_{B_{1}})g_{n}d{\mathfrak{m}}_{E},$

we obtain $\int{\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})\psi
wd\eta_{B_{1}}=0,~{}n\geq 1$, which implies
${\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)=0,~{}w\eta_{B_{1}}-a.a..$ This is a
contradiction. The proof completes. ∎

##  7\. Proof of Theorem 1.1

For a dense subset $\Lambda=\\{g_{n}\\}\subset R(\overline{E})^{\perp}\cap
L^{1}({\mathfrak{m}}_{E})$ and $N\geq 1,$ define

$\ \mathcal{E}_{N}=\left\\{\lambda:~{}\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(g_{n}{\mathfrak{m}}_{E})(\lambda)\text{ exists,
}\max_{1\leq n\leq
N}|\mathcal{C}(g_{n}{\mathfrak{m}}_{E})(\lambda)|\leq\frac{1}{N}\right\\}.$

######  Lemma 7.1.

There exists an absolute constant $C>0$ such that

$\
\lim_{N\rightarrow\infty}\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N})\leq
C\gamma(\mathbb{D}(\lambda,2\delta)\cap\mathcal{F}).$

###### Proof.

Set

$\
\epsilon_{0}=\lim_{N\rightarrow\infty}\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N})(=\inf_{N\geq
1}\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N})).$

We assume $\epsilon_{0}>0$. The case that $\epsilon_{0}=0$ is trivial. From
Lemma 2.6, there exists a Borel subset $F\subset\mathbb{D}(\lambda,\delta)$
such that (a) $\gamma(\mathbb{D}(\lambda,\delta)\setminus
F)<\frac{1}{2A_{T}}\epsilon_{0}$ and (b) $\mathcal{C}_{*}(g_{n}\mu)(z)\leq
M_{n}<\infty$ for $z\in F.$ By Theorem 2.1 (2) and (a),

$\ \gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N})\leq
A_{T}\gamma(\mathbb{D}(\lambda,\delta)\cap
F\cap\mathcal{E}_{N})+\frac{1}{2}\epsilon_{0}.$

So $\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N})\leq
2A_{T}\gamma(\mathbb{D}(\lambda,\delta)\cap F\cap\mathcal{E}_{N}).$ From Lemma
2.7, there exists $\eta_{N}\in M_{0}^{+}(\mathbb{D}(\lambda,\delta)\cap
F\cap\mathcal{E}_{N})$ with $1$-linear growth and
$\|\mathcal{C}_{\epsilon}(\eta_{N})\|_{\infty}\leq 1$ such that
$\gamma(\mathbb{D}(\lambda,\delta)\cap F\cap\mathcal{E}_{N})\leq
C_{5}\|\eta_{N}\|\leq C_{5}\delta.$ Applying (2.6), we get

$\displaystyle\
\left|\int{\mathcal{C}}\eta_{N}g_{n}d{\mathfrak{m}}_{E}\right|=\left|\int{\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})d\eta_{N}\right|\leq\dfrac{\|\eta_{N}\|}{N}$
(7.1)

By passing to a subsequence if necessary, we may assume that
$\eta_{N}\rightarrow\eta$ in $C(\overline{\mathbb{D}(\lambda,\delta)\cap
F})^{*}$ weak-star topology. Since $\mathcal{C}_{\epsilon}(\eta_{N})$
converges to $\mathcal{C}_{\epsilon}(\eta)$ in $L^{\infty}(\mathbb{C})$ weak-
star topology as $N\rightarrow\infty,$ we get
$\|\mathcal{C}_{\epsilon}(\eta)\|\leq 1.$ Clearly,
$\text{spt}(\eta)\subset\overline{F}$, $\eta$ is $1$-linear growth, and
$\lim_{N\rightarrow\infty}\|\eta_{N}\|=\|\eta\|.$

By passing to a subsequence, we may assume that $\mathcal{C}(\eta_{N})$
converges to $H(z)$ in $L^{\infty}(\mathbb{C})$ weak-star topology. Let
$\varphi$ be a smooth function with compact support. Then
${\mathcal{C}}(\varphi{\mathfrak{m}})$ is continuous and we have

$\displaystyle\ \int\varphi Hd{\mathfrak{m}}=$
$\displaystyle\lim_{N\rightarrow\infty}\int\varphi{\mathcal{C}}\eta_{N}d{\mathfrak{m}}=-\lim_{N\rightarrow\infty}\int{\mathcal{C}}(\varphi{\mathfrak{m}})d\eta_{N}$
$\displaystyle\ =$
$\displaystyle-\int{\mathcal{C}}(\varphi{\mathfrak{m}})d\eta=\int\varphi{\mathcal{C}}\eta
d{\mathfrak{m}}.$

Hence, $H={\mathcal{C}}\eta.$ Letting $N\rightarrow\infty$ to (7.1), we
conclude that $\int{\mathcal{C}}\eta g_{n}d{\mathfrak{m}}_{E}=0,~{}n\geq 1.$
So, ${\mathcal{C}}\eta\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$ by the
Hahn-Banach Theorem.

Applying Corollary 6.3, we see that $\eta(\mathcal{R})=0.$ There is an open
subset $O$ such that $\text{spt}(\eta)\cap\mathcal{R}\subset O$ and
$\eta(O)\leq\frac{1}{2}\|\eta\|$. If we define $W:=\text{spt}(\eta)\setminus
O$, then $W$ is compact, $W\subset\text{spt}(\eta)\cap\mathcal{F}$, and
$\|\eta\|\leq 2\eta(W).$ Hence, by (2.5) ,

$\ c^{2}(\eta|_{W})\leq c^{2}(\eta)\leq C_{56}\|\eta\|\leq
2C_{56}\|\eta|_{W}\|.$

Let $\eta_{1}=\frac{\eta|_{W}}{\sqrt{2C_{56}}}.$ Then $c^{2}(\eta_{1})\leq 1.$
Applying Theorem 2.1 (1) and Proposition 2.2 (3) & (4), we conclude that

$\ \|\eta|_{W}\|\leq C_{57}\gamma(W)\leq
C_{57}\gamma(\text{spt}(\eta)\cap\mathcal{F}).$

The lemma is proved. ∎

######  Lemma 7.2.

Let $f\in L^{\infty}({\mathfrak{m}}_{E})$ and
$\|f\|_{L^{\infty}({\mathfrak{m}}_{E})}\leq 1.$ Suppose that for $\epsilon>0,$
there exists $A_{\epsilon}\subset\mathcal{R}$ with
$\gamma(A_{\epsilon})<\epsilon$ and there exists $f_{\epsilon,N}\in
R(K_{\epsilon,N}),$ where
$\mathcal{R}_{\epsilon,N}=\mathcal{R}\setminus(A_{\epsilon}\cup\mathcal{E}_{N})$
and $K_{\epsilon,N}=\overline{\mathcal{R}_{\epsilon,N}},$ such that
$\|f_{\epsilon,N}\|\leq 2$ and

$\displaystyle\
f(z)=f_{\epsilon,N}(z),~{}{\mathfrak{m}}_{\mathcal{R}_{\epsilon,N}}-a.a..$
(7.2)

Then for $\varphi$ a smooth function with support in
$\mathbb{D}(\lambda,\delta)$,

$\ \left|\int f(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}_{E}(z)\right|\leq
C_{58}\delta\|\bar{\partial}\varphi\|\gamma(\mathbb{D}(\lambda,2\delta)\cap\mathcal{F}).$

###### Proof.

We extend $f_{\epsilon,N}$ as a continuous function on $\mathbb{C}$ with
$\|f_{\epsilon,N}\|_{\mathbb{C}}\leq 2$. Let $\varphi$ be a smooth function
with $\text{supp}(\varphi)\subset\mathbb{D}(\lambda,\delta)$. Using [8,
Theorem VIII.8.1] and Theorem 2.1 (2), we have the following calculation:

$\ \begin{aligned} \ &|(T_{\varphi}f_{\epsilon,N})^{\prime}(\infty)|\\\ \
=&\dfrac{1}{\pi}\left|\int
f_{\epsilon,N}(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}(z)\right|\\\ \
\leq&C_{59}\delta\|\bar{\partial}\varphi\|\gamma(\mathbb{D}(\lambda,\delta)\setminus
K_{\epsilon,N})\\\ \
\leq&C_{59}\delta\|\bar{\partial}\varphi\|\gamma(\mathbb{D}(\lambda,\delta)\cap(\mathcal{F}\cup
A_{\epsilon}\cup\mathcal{E}_{N}))\\\ \
\leq&C_{60}\delta\|\bar{\partial}\varphi\|(\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{F})+\gamma(A_{\epsilon})+\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N}))\end{aligned}$

Notice that $f(z)=0,~{}{\mathfrak{m}}|_{\mathcal{F}}-a.a.$, by (7.2), (2.1),
and Theorem 2.1 (2), we get,

$\ \begin{aligned} \ &\left|\int
f(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}_{E}(z)\right|\\\ \
\leq&\left|\int(f(z)-f_{\epsilon,N}(z))\bar{\partial}\varphi(z)d{\mathfrak{m}}(z)\right|+\left|\int
f_{\epsilon,N}(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}(z)\right|\\\ \
\leq&\int_{\mathcal{R}_{\epsilon,N}^{c}}|f(z)-f_{\epsilon,N}(z)||\bar{\partial}\varphi(z)|d{\mathfrak{m}}(z)+\pi|(T_{\varphi}f_{\epsilon,N})^{\prime}(\infty)|\\\
\
\leq&C_{61}\|\bar{\partial}\varphi\|{\mathfrak{m}}(\mathbb{D}(\lambda,\delta)\cap\mathcal{R}_{\epsilon,N}^{c})+\pi|(T_{\varphi}f_{\epsilon,N})^{\prime}(\infty)|\\\
\
\leq&C_{62}\delta\|\bar{\partial}\varphi\|\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{R}_{\epsilon,N}^{c})+\pi|(T_{\varphi}f_{\epsilon,N})^{\prime}(\infty)|\\\
\
\leq&C_{63}\delta\|\bar{\partial}\varphi\|(\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{F})+\gamma(A_{\epsilon})+\gamma(\mathbb{D}(\lambda,\delta)\cap\mathcal{E}_{N})).\end{aligned}$

Thus, using Lemma 7.1 and taking $\epsilon\rightarrow 0$ and
$N\rightarrow\infty$, we get

$\ \left|\int f(z)\bar{\partial}\varphi(z)d{\mathfrak{m}}_{E}(z)\right|\leq
C_{64}\delta\|\bar{\partial}\varphi\|\gamma(\mathbb{D}(\lambda,2\delta)\cap\mathcal{F}).$

This completes the proof. ∎

######  Lemma 7.3.

Let $f\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$ and
$\|f\|_{L^{\infty}({\mathfrak{m}}_{E})}\leq 1.$ Let
$\\{r_{m}\\}\subset\text{Rat}(\overline{E})$ such that
$\sup_{m}\|r_{m}\|_{L^{\infty}({\mathfrak{m}}_{E})}<\infty$ and
$r_{m}(z)\rightarrow f(z),~{}{\mathfrak{m}}_{E}-a.a..$ Then the following
properties hold:
(1) There exists $\mathcal{Q}_{f}$ with $\gamma(\mathcal{Q}_{f})=0$ such that
$f$ can be extended to $\mathcal{R}\setminus\mathcal{Q}_{f},$ $f$ is
$\gamma$-continuous at each point of $\mathcal{R}\setminus\mathcal{Q}_{f},$
and
$f(z){\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)={\mathcal{C}}(fg_{n}{\mathfrak{m}}_{E})(z)$
for $z\in\mathcal{R}\setminus\mathcal{Q}_{f}$ and $n\geq 1.$
(2) For $\epsilon>0,$ there exists $A_{\epsilon}\subset\mathcal{R}$ with
$\gamma(A_{\epsilon})<\epsilon$ and a subsequence $\\{r_{m_{k}}\\}$ such that
$r_{m_{k}}$ uniformly converges to $f$ on
$\mathcal{R}_{\epsilon,N}:=\mathcal{R}\setminus(A_{\epsilon}\cup\mathcal{E}_{N}).$
Consequently, there exists $f_{\epsilon,N}\in R(K_{\epsilon,N}),$ where
$K_{\epsilon,N}=\overline{\mathcal{R}_{\epsilon,N}},$ such that
$f(z)=f_{\epsilon,N}(z)$ for $z\in\mathcal{R}_{\epsilon,N}.$

###### Proof.

(1) For $r\in\text{Rat}(\overline{E}),$
$\frac{r(z)-r(\lambda)}{z-\lambda}\in\text{Rat}(\overline{E}).$ Hence, by
Corollary 2.3,

$\displaystyle\
r_{m}(z){\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)={\mathcal{C}}(r_{m}g_{n}{\mathfrak{m}}_{E})(z),~{}\gamma-a.a.$
(7.3)

for $n\geq 1.$ Therefore,
$f(z){\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)={\mathcal{C}}(fg_{n}{\mathfrak{m}}_{E})(z),~{}{\mathfrak{m}}_{E}-a,a,$
for $n\geq 1.$ Using Lemma 2.8 and Corollary 2.3, we conclude that there
exists $\mathcal{Q}_{f}$ with $\gamma(\mathcal{Q}_{f})=0$ such that for
$z\in\mathcal{R}\setminus\mathcal{Q}_{f},$
$\Theta_{g_{n}{\mathfrak{m}}_{E}}(z)=0,$ $\lim_{\epsilon\rightarrow
0}{\mathcal{C}}_{\epsilon}(g_{n}{\mathfrak{m}}_{E})(z)$ exists,
$\Theta_{fg_{n}{\mathfrak{m}}_{E}}(z)=0,$ and $\lim_{\epsilon\rightarrow
0}{\mathcal{C}}_{\epsilon}(fg_{n}{\mathfrak{m}}_{E})(z)$ exists. It is
straightforward to verify (1) from Lemma 2.10 and Lemma 2.11. Therefore, for
$n\geq 1,$

$\displaystyle\
f(z){\mathcal{C}}(g_{n}{\mathfrak{m}}_{E})(z)={\mathcal{C}}(fg_{n}{\mathfrak{m}}_{E})(z),~{}\gamma-a.a..$
(7.4)

(2): Using Lemma 2.5, we find $A^{1}_{\epsilon}$ and a subsequence
$\\{r_{m,1}\\}$ of $\\{r_{m}\\}$ such that
$\gamma(A^{1}_{\epsilon})<\frac{\epsilon}{2A_{T}}$ and
$\\{\mathcal{C}(r_{m,1}g_{1}{\mathfrak{m}}_{E})\\}$ uniformly converges to
$\mathcal{C}(fg_{1}{\mathfrak{m}}_{E})$ on $\mathbb{C}\setminus
A^{1}_{\epsilon}$. Then we find $A^{2}_{\epsilon}$ and a subsequence
$\\{r_{m,2}\\}$ of $\\{r_{m,1}\\}$ such that
$\gamma(A^{2}_{\epsilon})<\frac{\epsilon}{2^{2}A_{T}}$ and
$\\{\mathcal{C}(r_{m,2}g_{2}{\mathfrak{m}}_{E})\\}$ uniformly converges to
$\mathcal{C}(fg_{2}{\mathfrak{m}}_{E})$ on $\mathbb{C}\setminus
A^{2}_{\epsilon}$. Therefore, we have a subsequence $\\{r_{m,m}\\}$ such that
$\\{\mathcal{C}(r_{m,m}g_{n}{\mathfrak{m}}_{E})\\}$ uniformly converges to
$\mathcal{C}(fg_{n}{\mathfrak{m}}_{E})$ on $\mathbb{C}\setminus A_{\epsilon}$
for all $n\geq 1$, where
$A_{\epsilon}=\mathcal{Q}_{f}\cup\cup_{n}A^{n}_{\epsilon}$ and
$\gamma(A_{\epsilon})<\epsilon$ by Theorem 2.1 (2). From (7.3) and (7.4), we
infer that $\\{r_{m,m}\\}$ uniformly tends to $f$ on
$\mathcal{R}_{\epsilon,N}.$ Thus, $\\{r_{m,m}\\}$ uniformly tends to
$f_{\epsilon,N}\in R(K_{\epsilon,N})$ on $K_{\epsilon,N}.$ ∎

Now we are ready to finish the proof of Theorem 1.1 as the following.

###### Proof.

(Theorem 1.1): $\Rightarrow$: If $f\in
R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$ is invertible, then there exists
$f_{0}\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$ such that $ff_{0}=1.$
Therefore, $|f(z)|\geq\frac{1}{\|f_{0}\|},~{}{\mathfrak{m}}_{E}-a.a..$

$\Leftarrow$: Suppose that $f\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E})$
and there exists $\epsilon_{f}>0$ such that
$|f(z)|\geq\epsilon_{f},~{}{\mathfrak{m}}_{E}-a.a..$ Using Lemma 7.3 (2), for
$\epsilon>0,$ there exists $f_{\epsilon,N}\in R(K_{\epsilon,N})$ such that
$f(z)=f_{\epsilon,N}(z)$ for $z\in\mathcal{R}_{\epsilon,N}.$ Hence,
$|f_{\epsilon,N}(z)|\geq\epsilon_{f},~{}z\in K_{\epsilon,N}.$ Thus,
$f_{\epsilon,N}$ is invertible in $R(K_{\epsilon,N}).$ Using Lemma 7.2 for
$\frac{\epsilon_{f}}{f(z)}$ and $\frac{\epsilon_{f}}{f_{\epsilon,N}(z)}$, we
conclude that for $\varphi$ a smooth function with support in
$\mathbb{D}(\lambda,\delta)$, we have

$\
\left|\int\dfrac{\epsilon_{f}}{f(z)}\bar{\partial}\varphi(z)d{\mathfrak{m}}_{E}(z)\right|\leq
C_{65}\delta\|\bar{\partial}\varphi\|\gamma(\mathbb{D}(\lambda,2\delta)\cap\mathcal{F}).$

Replacing $\varphi$ by $(z-\lambda)^{n}\varphi$ and applying Theorem 6.1, we
conclude that $\frac{1}{f}\in R^{\infty}(\overline{E},{\mathfrak{m}}_{E}).$ ∎

######  Remark 7.4.

Using Lemma 7.3 and Lemma 7.2, we see that the constant $C_{f}$ in Theorem 6.1
(2) can be chosen as an absolute constant (independent of $f$).

Acknowledgments. The author would like to thank Professor John McCarthy for
carefully reading through the manuscript and providing many useful comments.
The author would also like to thank the referee for carefully reading the
manuscript and providing helpful comments.

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