# DISTILLING THE KNOWLEDGE OF BERT FOR CTC-BASED ASR

###### Abstract

Connectionist temporal classification (CTC) -based models are attractive
because of their fast inference in automatic speech recognition (ASR).
Language model (LM) integration approaches such as shallow fusion and
rescoring can improve the recognition accuracy of CTC-based ASR by taking
advantage of the knowledge in text corpora. However, they significantly slow
down the inference of CTC. In this study, we propose to distill the knowledge
of BERT for CTC-based ASR, extending our previous study for attention-based
ASR. CTC-based ASR learns the knowledge of BERT during training and does not
use BERT during testing, which maintains the fast inference of CTC. Different
from attention-based models, CTC-based models make frame-level predictions, so
they need to be aligned with token-level predictions of BERT for distillation.
We propose to obtain alignments by calculating the most plausible CTC paths.
Experimental evaluations on the Corpus of Spontaneous Japanese (CSJ) and TED-
LIUM2 show that our method improves the performance of CTC-based ASR without
the cost of inference speed.

Index Terms—  speech recognition, CTC, BERT, knowledge distillation

##  1 Introduction

End-to-end automatic speech recognition (ASR) that directly maps acoustic
features into text sequences has shown remarkable results. There are some
variants for its modeling: CTC-based models [1], attention-based sequence-to-
sequence models [2, 3], and neural network transducers [4, 5]. Among them,
CTC-based models have the advantage of lightweight and fast inference. They
consist of an encoder followed by a compact linear layer only and can predict
all tokens in parallel, which is called non-autoregressive generation. For
these advantages, a lot of efforts have continuously been made to improve the
ASR performance of CTC-based models [6, 7]. In terms of output unit, CTC-based
models and transducers are categorized as frame-synchronous models that makes
frame-level predictions, while attention-based models are categorized as
label-synchronous models that makes token-level predictions.

End-to-end ASR models including CTC-based models are trained on paired speech
and transcripts. On the other hand, much larger amount of text-only data is
often available, and the most popular way to leverage it in end-to-end ASR is
the integration of external language models (LMs). In $n$-best rescoring,
$n$-best hypotheses obtained from an ASR model are re-scored by an LM, and
then the hypothesis of the highest score is selected. In shallow fusion [8],
the interpolated score of the ASR model and the LM is calculated at each ASR
decoding step. These two LM integration approaches are simple and effective
and therefore widely used in the CTC-based ASR. However, they degrade the fast
inference, which is the most important advantage of CTC over other variants of
end-to-end ASR. Specifically, beam search [9] to obtain multiple hypotheses
makes CTC lose its non-autoregressive nature. In addition to beam search, the
inference of LM takes much time during testing.

Recently, knowledge distillation [10] -based LM integration has been proposed
[11, 12, 13]. In this approach, an LM serves as a teacher model, and an
attention-based ASR model serves as a student model. The knowledge of the LM
is transferred to the ASR model during ASR training, and the LM is not
required during testing. However, in the formulation of existing studies, the
student ASR model has been limited to the attention-based model that makes
token-level predictions. In this study, we propose an extension of this
knowledge distillation to the frame-synchronous CTC-based models, so as to
integrate the LM while maintaining fast inference of CTC. We use BERT [14] as
a teacher LM that predicts each masked word on the basis of both its left and
right context. We have shown that BERT outperforms conventionally-used
unidirectional LMs that predicts each word on the basis of only its left
context in distillation for attention-based ASR [12]. In addition, as recent
successful CTC-based models mostly consist of a bidirectional encoder that
looks at both left and right context, a bidirectional LM, BERT is suited for a
teacher LM.

BERT and attention-based models give token-by-token predictions, while CTC-
based models give frame-by-frame predictions. For distillation from BERT to
attention-based ASR [11, 12, 13], it is obvious that the teacher BERT’s
prediction for the $i$-th token becomes the soft target for the student
attention-based model’s one for the $i$-th token. However, to distill the
knowledge of BERT for CTC-based ASR, it is not trivial how to correspond the
teacher BERT’s token-level predictions to the student CTC-model’s frame-level
predictions. In this study, we propose to leverage forced alignment from the
CTC forward-backward (or the Viterbi) algorithm [1] to solve the problem.
During ASR training, the most plausible CTC path attributed to the label
sequence is calculated to determine the correspondence between tokens and time
frames. The proposed method improves the performance of CTC-based ASR, even
with greedy decoding, without any additional inference steps related to BERT.

##  2 Preliminaries and related work

###  2.1 End-to-end ASR

####  2.1.1 CTC-based ASR

Let $\bm{X}=(\bm{x}_{1},...,\bm{x}_{t},...,\bm{x}_{T})$ denote the acoustic
features in an utterance and $\bm{y}=(y_{1},...,y_{i},...,y_{L})$ denote the
label sequence of tokens corresponding to $\bm{X}$. An encoder network that
consists of RNN, Transformer, or Conformer [15] transforms $\bm{X}$ into
higher-level representations of length $T$. A CTC-based model predicts CTC
path $\bm{\pi}=(\pi_{1},...,\pi_{T})$ using the encoded representations. Let
$\mathcal{V}$ denote the vocabulary and $\phi$ denote a blank token. Then, we
define the probability of predicting $v\in\mathcal{V}\cup\\{\phi\\}$ for the
$t$-th time frame as

$\displaystyle P_{\rm CTC}^{(t,v)}=p(v\,|\,\bm{X}).$ (1)

The output sequence $\bm{y}$ is obtained by $\bm{y}=\mathcal{B}(\bm{\pi})$,
where the mapping $\mathcal{B}$ removes blank tokens after removing repeated
ones. The CTC loss function is defined over all possible paths that can be
reduced to $\bm{y}$:

$\displaystyle\mathcal{L}_{\rm CTC}=-\log
p(\bm{y}|\bm{X})=-\sum_{\bm{\pi}\in\mathcal{B}^{-1}(\bm{y})}p(\bm{\pi}|\bm{X}).$
(2)

####  2.1.2 Attention-based ASR

An attention-based ASR model consists of encoder and decoder networks. The
decoder network predicts each token using the encoded representations and
previously decoded tokens. We define the probability of predicting
$v\in\mathcal{V}$ for the $i$-th token as

$\displaystyle P_{\rm Att}^{(i,v)}=p(v\,|\,\bm{X},\bm{y}_{<i}).$ (3)

The loss function is defined as the cross-entropy:

$\displaystyle\mathcal{L}_{\rm
Att}=-\sum_{i=1}^{L}\sum_{v\in\mathcal{V}}\delta(v,y_{i})\log P_{\rm
Att}^{(i,v)},$ (4)

where $\delta(v,y_{i})$ becomes $1$ when $v=y_{i}$, and $0$ otherwise.

###  2.2 BERT

BERT [14] that consists of Transformer encoders was originally proposed as a
pre-training method for downstream NLP tasks such as question answering and
language understanding. BERT is pre-trained on large text corpora for masked
language modeling (MLM) objective, where some of the input tokens are masked
and the original tokens are predicted given unmasked tokens. After this pre-
training, BERT can serve as an LM that predicts each masked word given both
its left and right context. BERT as an LM has been applied to ASR via $n$-best
rescoring [16, 17] and knowledge distillation [12, 13]. BERT has been reported
to perform better than conventional LMs in ASR thanks to the use of the
bidirectional context.

###  2.3 Distilling the knowledge of BERT for attention-based ASR

We have proposed to apply BERT to attention-based ASR via knowledge
distillation in [12]. BERT provides soft labels for attention-based ASR
training to encourage more syntactically or semantically likely hypotheses. To
generate better soft labels, context beyond the current utterance is used as
input to BERT. We define BERT’s prediction of $v\in\mathcal{V}$ for the $i$-th
target as

$\displaystyle P_{\rm BERT}^{(i,v)}$
$\displaystyle=p(v\,|\,[\bm{y}^{\rm(pre)};\bm{y}_{\backslash
i};\bm{y}^{\rm(suc)}]),$ (5)

where $\bm{y}_{\backslash i}$ is obtained by masking the $i$-th token, that
is, $\bm{y}_{\backslash i}=(y_{1},...,y_{i-1},$[MASK]$,y_{i+1},...,y_{L})$.
$\bm{y}_{\backslash i}$ is concatenated with tokens from the preceding
utterances $\bm{y}^{\rm(pre)}$ and tokens from the succeeding utterances
$\bm{y}^{\rm(suc)}$ to make an input sequence of fixed length
$[\bm{y}^{\rm(pre)};\bm{y}_{\backslash i};\bm{y}^{\rm(suc)}]$.

The knowledge distillation (KD) loss function is formulated by minimizing KL
divergence between $P_{\rm Att}^{(i)}$ and $P_{\rm BERT}^{(i)}$, which is
equivalent to minimizing the cross-entropy between them as

$\displaystyle\mathcal{L}_{\rm KD}=-\sum_{i=1}^{L}\sum_{v\in\mathcal{V}}P_{\rm
BERT}^{(i,v)}\log P_{\rm Att}^{(i,v)}.$ (6)

The work in [13] also performed knowledge distillation from BERT to an
attention-based non-autoregressive model [18], which is a label-synchronous
model different from CTC.

###  2.4 Knowledge distillation for CTC-based ASR

Knowledge distillation (KD) [10] between two CTC-based models has been
investigated [19, 20, 21, 22]. The simplest way is to minimize KL divergence
between the distributions of the student CTC and those of the teacher CTC
frame-by-frame [19]. However, it assumes the student and teacher models share
the same frame-wise alignment. This is not true in KD between CTC-based models
with different topologies such as KD from bidirectional RNN-based CTC to
unidirectional RNN-based CTC [20, 21], which is oriented for streaming ASR
applications. In [20], a guiding CTC model encourages student and teacher
models to share the same alignment. Sequence-level KD [23] was also proposed
to address the issue, where $n$-best hypotheses from the teacher CTC are used
as targets for the student CTC training [21, 22].

KD from an attention-based model to a CTC-based model has also been proposed
[24, 25]. Token-level predictions from the attention-based model need to be
aligned with frame-level predictions from the CTC, which is similar to our KD
from BERT to CTC. Attention weights of $L\times T$ are used for that purpose
in [24, 25]. However, for KD from BERT, BERT does not attend acoustic features
of length $T$, thus such attention weights cannot be obtained. Note that KD
from an LM including BERT to a CTC-based model is proposed in this study for
the first time.

##  3 Proposed method: Distilling the knowledge of BERT for CTC-based ASR

Fig. 1: Illustration of our proposed method. The forced alignment path with
forward-backward calculation determines which frames $t$ correspond to each
token $i$. For $i=2$ ($y_{2}$), corresponding frames are $t=4$ and $t=5$, so
the $4$-th and $5$-th CTC predictions $P_{\rm CTC}^{(4)}$ and $P_{\rm
CTC}^{(5)}$ are trained for the $2$-nd soft label from BERT $P_{\rm
BERT}^{(2)}$.

In this study, we propose to apply BERT to CTC-based ASR via knowledge
distillation. With this method, we expect CTC-based models to further learn
the syntactic or semantic relationship between tokens from BERT. CTC-based
models have difficulty in capturing it because they cannot learn it explicitly
from the output tokens because of the conditional independence assumption.
Soft labels from BERT help CTC-based models learn it implicitly from acoustic
features and intermediate representations in an encoder. BERT provides token-
level soft labels $P_{\rm BERT}^{(i)}(i=1,...,L)$. To distill the knowledge of
BERT for attention-based ASR that makes token-level predictions $P_{\rm
Att}^{(i)}(i=1,...,L)$, we just minimized KL divergence between $P_{\rm
Att}^{(i)}$ and $P_{\rm BERT}^{(i)}$ as in Eq. (6). On the other hand, CTC
makes frame-level predictions $P_{\rm CTC}^{(t)}(t=1,...,T)$, so the alignment
between $i$ and $t$ is necessary for using token-level soft labels from BERT.

To solve the problem, we propose to use the forced alignment result, that is,
the most plausible CTC path $\bm{\hat{\pi}}=(\hat{\pi}_{1},...,\hat{\pi}_{T})$
of $\bm{\pi}\in\mathcal{B}^{-1}(\bm{y})$ like [26], where the CTC path
$\bm{\hat{\pi}}$ is used to enable a monotonic chunkwise attention (MoChA)
model to learn optimal alignments for streaming ASR. $\bm{\hat{\pi}}$ can be
obtained by tracking the path that has the maximum products of forward and
backward variables, which can be obtained in the process of calculating the
CTC loss function $\mathcal{L}_{\rm CTC}$ from Eq. (2) with the forward-
backward algorithm [1]. This method does not introduce any additional
architectures or external alignment information (e.g. HMM-based one) for
alignment. For each $i$, $\bm{\hat{\pi}}$ has one or more $\hat{\pi}_{t}$ that
corresponds to $y_{i}$, and we define the alignment as an one-to-many mapping
$\mathcal{A}$ from token index $i$ to frame indices $\bm{t}$. Note that frame
$t$ assigned to $\phi$ in $\bm{\hat{\pi}}$ $(\hat{\pi}_{t}=\phi)$ does not
appear in $\mathcal{A}(i)$. For example,
$\bm{\hat{\pi}}=(y_{1},\phi,\phi,y_{2},y_{2},\phi,y_{3},\phi)$ defines
$\mathcal{A}(1)=\\{1\\}$, $\mathcal{A}(2)=\\{4,5\\}$, and
$\mathcal{A}(3)=\\{7\\}$.

Once the alignment $\mathcal{A}$ is obtained, $P_{\rm BERT}^{(i)}$ is used as
a soft label for $P_{\rm CTC}^{(t)}$ where $t\in\mathcal{A}(i)$. For example,
given $\mathcal{A}(2)=\\{4,5\\}$, the CTC-based ASR is trained to make $P_{\rm
CTC}^{(4)}$ and $P_{\rm CTC}^{(5)}$ close to $P_{\rm BERT}^{(2)}$, as
illustrated in Fig. 1. The KD loss function is formulated as

$\displaystyle\mathcal{L}_{\rm
KD}=-\frac{1}{\sum_{i=1}^{L}|\mathcal{A}(i)|}\sum_{i=1}^{L}\sum_{t\in\mathcal{A}(i)}\sum_{v\in\mathcal{V}}P_{\rm
BERT}^{(i,v)}\log P_{\rm CTC}^{(t,v)}.$ (7)

Finally, $\mathcal{L}_{\rm KD}$ is interpolated with $\mathcal{L}_{\rm CTC}$
from Eq. (2) as

$\displaystyle\mathcal{L}=(1-\alpha)\,\mathcal{L}_{\rm
CTC}+\alpha\,\mathcal{L}_{\rm KD},$ (8)

where $\alpha$ $(0\leq\alpha\leq 1)$ is a tunable hyperparameter.

The alignment is calculated at each training step on the fly. To mitigate the
negative effects of unreliable alignments in early steps, the CTC-based model
is pre-trained with Eq. (2). The soft labels from BERT can be pre-computed for
all the training set. For memory efficiency, top-$K$ distillation [27] is
applied, where the top-$K$ probabilities of BERT are normalized and smoothed
by temperature parameter $T$ to generate soft labels for distillation. In this
study, $K=8$ and $T=3.0$ are used.

There is a clear advantage of our method over existing LM integration methods
for CTC-based ASR such as $n$-best rescoring and shallow fusion. Our method
introduces the knowledge of LM (BERT) during training, so it has no change in
terms of inference time during testing. Rescoring and shallow fusion require
LM modules during testing, which significantly increases the inference time.
Our method benefits from LM just with greedy decoding without the runtime use
of LM, whereas rescoring and shallow fusion requires time-consuming beam
search decoding [9] with LM inference.

##  4 Experimental evaluations

###  4.1 Experimental conditions

We evaluated our method using the Corpus of Spontaneous Japanese (CSJ) [28]
and the TED-LIUM2 corpus [29]. CSJ has two subcorpora of oral presentations,
CSJ-APS on academic and CSJ-SPS on general topics. In CSJ experiments, $240$
hours of transcribed speech of CSJ-APS was used for training ASR. $7$M-word
transcripts of CSJ (both CSJ-APS and CSJ-SPS) and additional $56$M-word text
of the Balanced Corpus of Contemporary Written Japanese (BCCWJ) [30] were used
for training LMs. The ASR model and LMs shared the same BPE vocabulary of
$10872$ entries. In TED-LIUM2 experiments, $207$ hours of transcribed speech
was used for training ASR, and $250$M-word text in official LM data was used
for training LMs. The BPE vocabulary has $1001$ entries.

CTC-based ASR models consist of Transformer encoder with $12$ layers, $256$
hidden units, and $4$ attention heads. Adam optimizer with Noam learning rate
scheduling [3] of $warmup\\_n=25,000,k=5$ was used for training the ASR
models. SpecAugment [31] was applied to acoustic features, and speed
perturbation [32] was also applied in the TED-LIUM2 experiments. When applying
knowledge distillation, CTC-based ASR was pre-trained for $50$ epochs with Eq.
(2) and then trained for $50$ epochs with Eq. (8). For a fair comparison,
baseline CTC-based ASR without knowledge distillation was pre-trained for $50$
epochs and then further trained for $50$ epochs.

We compared three types of LMs: BERT and Transformer LM (TLM) that consist of
$6$ layers, $512$ hidden units, and $8$ attention heads, and RNN LM that
consists of $4$-layer LSTM with $512$ hidden units. Adam optimizer of the
learning rate of $10^{-4}$ with learning rate warmup over the first $10\%$ of
total steps and linear decay was used for training LMs. During training,
sequences of $256$ tokens were fed into LMs, and $8\%$ of tokens in a sequence
were masked for BERT.

###  4.2 Experimental results

Table 1 shows the ASR results of our proposed method on CSJ. First of all,
CTC-based ASR trained with our proposed knowledge distillation (KD) method
(A2) outperformed a baseline without KD (A1) in terms of both word error rate
(WER) and PPL. PPL denotes the pseudo perplexity [33] of BERT on the resulting
hypotheses:

$\displaystyle{\rm
PPL}(\bm{y})=\frac{1}{|\bm{y}|}\sum_{i=1}^{|\bm{y}|}-\log{p(y_{i}|\bm{y}_{\backslash
i})}.$ (9)

The improvement in both PPL and WER with our method suggests that the
knowledge of BERT was indeed incorporated into CTC-based ASR and helped
improve WER. $\alpha$ in Eq. (8) was determined using the development set, and
$\alpha=0.5$ was found to achieve the best WER for the proposed method on CSJ.
We also observed that increasing $\alpha$ from $0.5$ to $0.7$ improved PPL
($12.87$ to $12.70$) but degraded WER ($9.05$ to $9.12$). We also trained CTC-
based ASR with the recently proposed regularization method, InterCTC [6] (B1).
Further WER improvement was obtained by training CTC with the combination of
InterCTC and our method (B2).

Here, we explored a few different KD strategies to find out if there is a
better way. In (A3), only the leftmost indices of non-blank tokens in the most
plausible path $\bm{\hat{\pi}}$ were used for alignment, while all indices
were used in the proposed method. For example,
$\bm{\hat{\pi}}=(y_{1},\phi,\phi,y_{2},y_{2},\phi,y_{3},\phi)$ corresponds to
$\mathcal{A}(1)=\\{1\\}$, $\mathcal{A}(2)=\\{4\\}$, and
$\mathcal{A}(3)=\\{7\\}$. In (A4), only the rightmost indices were used
instead. In (A5), CTC-based ASR was trained with KD from scratch, while KD was
applied after pre-training without KD in the proposed method. In (A6), soft
labels from TLM were used in KD, that is, $P_{\rm BERT}^{(i,v)}$ in Eq. (7)
was replaced with $P_{\rm TLM}^{(i,v)}$. In (A7), one-hot labels were used,
that is, $P_{\rm BERT}^{(i,v)}$ in Eq. (7) was replaced with $\delta(v,y_{i})$
that becomes $1$ when $v=y_{i}$, and $0$ otherwise. This does not use the
knowledge of any LM but just encourage CTC’s predictions to be aligned with
the most plausible path. Among them, our proposed way described in Section 3
performed the best, which demonstrates the effectiveness of our alignment
allocation, pre-training, and the use of BERT.

Table 2 shows the results with other LM integration methods: rescoring and
shallow fusion. Inference times relative to a plain CTC (A1) measured on CPU
are shown in the table as “InferTime”. Overall, these methods improved WER
more than our KD-based method (A2), but they increased inference time far more
than the baseline (A1). They took much time for beam search and LM inference.
Note that RNNLM (C3,D3) can carry over states so is faster than TLM (C4,D4) in
shallow fusion and that TLM (C5,D5) scores a hypothesis in a single step so is
faster than BERT (C6,D6) in rescoring [17]. On the other hand, our method did
not affect inference time, and WER improvement was obtained with greedy
decoding. Our method also improved the oracle WER (D7), and combinations of
our method and rescoring or shallow fusion (D3-D6) further improved WER
compared to rescoring or shallow fusion alone (C3-C6).

Table 1: ASR results on CSJ with proposed knowledge distillation (KD) -based LM integration. Different KD strategies are compared. |  WER(%) $\downarrow$ |  PPL (BERT) $\downarrow$
---|---|---
| eval1 | eval1
(A1)CTC | $9.34$ | $13.70$
(A2)+KD (BERT) | $\bm{9.05}$ | $\bm{12.87}$
(A3)+KD (BERT) (leftmost) | $9.15$ | $13.28$
(A4)+KD (BERT) (rightmost) | $9.14$ | $13.19$
(A5)+KD (BERT) (scratch) | $9.59$ | $14.03$
(A6)+KD (TLM) | $9.23$ | $13.25$
(A7)+Alignment | $9.32$ | $13.79$
(B1)InterCTC [6] | $9.22$ | $13.66$
(B2)+KD (BERT) | $\bm{8.89}$ | $\bm{12.77}$
Table 2: Comparison and combinations with rescoring (Resc) and shallow fusion (SF). “BS” means beam search without LM, and $b$ denotes beam width. $n$ denotes the number of hypotheses to rescore. |  WER(%) $\downarrow$ |  InferTime $\downarrow$
---|---|---
| eval1 | eval1
(A1)CTC | $9.34$ | $\times 1.0$
(C2)+BS ($b=5$) | $9.29$ | $\times 2.2$
(C3)+SF (RNNLM, $b=5$) | $8.95$ | $\times 20$
(C4)+SF (TLM, $b=5$) | $8.71$ | $\times 120$
(C5)+Resc (TLM, $n=5$ / $50$) |  $8.89$ / $8.61$ |  $\times 3.1$ / $\times 600$
(C6)+Resc (BERT, $n=5$ / $50$) |  $8.78$ / $\bm{8.49}$ |  $\times 32$ / $\times 890$
(C7)Oracle ($n=5$ / $50$) |  $7.25$ / $5.45$ | -
(A2)CTC+KD (BERT) | $9.05$ | $\times 1.0$
(D2)+BS ($b=5$) | $9.01$ | $\times 2.2$
(D3)+SF (RNNLM, $b=5$) | $8.65$ | $\times 20$
(D4)+SF (TLM, $b=5$) | $8.59$ | $\times 120$
(D5)+Resc (TLM, $n=5$ / $50$) |  $8.61$ / $8.45$ |  $\times 3.1$ / $\times 600$
(D6)+Resc (BERT, $n=5$ / $50$) |  $8.40$ / $\bm{8.20}$ |  $\times 32$ / $\times 890$
(D7)Oracle ($n=5$ / $50$) |  $7.03$ / $5.36$ | -
Table 3: ASR results on TED-LIUM2. |  WER(%) $\downarrow$ |  PPL (BERT) $\downarrow$
---|---|---
| test | dev | test
CTC | $12.13$ | $12.57$ | $5.53$
CTC+KD (BERT) | $\bm{11.40}$ | $\bm{11.79}$ | $\bm{4.92}$
CTC+KD (TLM) | $11.72$ | $12.28$ | $5.15$
Fig. 2: An example of decoded hypotheses and top-$5$ predictions (the upper
row is more probable) from two CTC-based models on TED-LIUM2. “_” denotes a
word boundary.

Table 3 summarizes ASR results on TED-LIUM2, and it shows that our proposed KD
with BERT improved WER and PPL of CTC-based ASR as well. Fig. 2 shows a
decoding example for an utterance from TED-LIUM2. The box at the top of the
figure shows decoded hypotheses from two CTC-based models, (a) one trained
without KD and (b) the other trained with our method. While “feed ourselves”
was erroneously recognized as “fe our cell” without KD, it was correctly
recognized with KD. The lower part of the figure shows the top-$5$ frame-level
predictions from the two models for the utterance. We see that the
probabilities of semantically plausible subwords “ed” and “selves” become
higher with KD, leading to correct recognition. It is also interesting to see
that semantically plausible subwords such as “why” ($t=40$) and “_kids”
($t=56$) appear in the top-$5$ predictions with KD, which indicates the model
considers the relationship between tokens in the context much more than the
baseline.

##  5 Conclusions

In this study, we have proposed knowledge distillation-based BERT integration
for CTC-based ASR. For knowledge distillation, BERT provides token-level soft
labels, while CTC-based ASR makes frame-level predictions. We obtained the
alignment between them by calculating the most plausible CTC paths. Our method
does not add any computational costs during testing, which maintains the fast
inference of CTC. We demonstrated that our method improved the performance of
CTC-based ASR on CSJ and TED-LIUM2 by exploiting the knowledge of BERT. For
future work, we will investigate applying BERT to neural network transducers
[4, 5] that are frame-synchronous models and have an autoregressive nature.

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# Mean Rational Approximation for Some Compact Planar Subsets

John B. Conway1 and Liming Yang2 1Department of Mathematics, The George
Washington University, Washington, DC 20052<EMAIL_ADDRESS>2Department of
Mathematics, Virginia Polytechnic and State University, Blacksburg, VA 24061.
<EMAIL_ADDRESS>
###### Abstract.

In 1991, J. Thomson [11] obtained celebrated structural results for
$P^{t}(\mu).$ Later, J. Brennan [3] generalized Thomson’s theorem to
$R^{t}(K,\mu)$ when the diameters of the components of $\mathbb{C}\setminus K$
are bounded below. The results indicate that if $R^{t}(K,\mu)$ is pure, then
$R^{t}(K,\mu)\cap L^{\infty}(\mu)$ is the “same as” the algebra of bounded
analytic functions on $\mbox{abpe}(R^{t}(K,\mu)),$ the set of analytic bounded
point evaluations. We show that if the diameters of the components of
$\mathbb{C}\setminus K$ are allowed to tend to zero, then even though
$\text{int}(K)=\mbox{abpe}(R^{t}(K,\mu))$ and $K=\overline{\text{int}(K)},$
the algebra $R^{t}(K,\mu)\cap L^{\infty}(\mu)$ may “be equal to” a proper sub-
algebra of bounded analytic functions on $\text{int}(K),$ where functions in
the sub-algebra are “continuous” on certain portions of the inner boundary of
$K.$

###### Key words and phrases:

Analytic Capacity, Analytic Bounded Point Evaluations, Bounded Point
Evaluations, and Mean Rational Approximation

###### 2010 Mathematics Subject Classification:

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

##  1\. Introduction

For a Borel subset $E$ of the complex plane $\mathbb{C},$ let $M_{0}(E)$
denote the set of finite complex-valued Borel measures that are compactly
supported in $E$ and let $M_{0}^{+}(E)$ be the set of positive measures in
$M_{0}(E).$ The support of $\nu\in M_{0}(\mathbb{C}),$ $\text{spt}(\nu),$ is
the smallest closed set that has full $|\nu|$ measure. For a compact subset
$K\subset\mathbb{C},$ $\mu\in M_{0}^{+}(K),$ and $1\leq t\leq\infty$, the
analytic polynomials and functions in

$\ \mbox{Rat}(K):=\\{q:\mbox{$q$ is a rational function with poles off
$K$}\\}$

are members of $L^{t}(\mu)$. If $1\leq t<\infty,$ we let $P^{t}(\mu)$ denote
the closure of the (analytic) polynomials in $L^{t}(\mu)$ and let
$R^{t}(K,\mu)$ denote the closure of $\mbox{Rat}(K)$ in $L^{t}(\mu)$. Define

$\ R^{t}(K,\mu)^{\perp}=\left\\{g\in L^{s}(\mu):~{}\int f(z)g(z)d\mu=0,\text{
for }f\in R^{t}(K,\mu)\right\\},$

where $\frac{1}{t}+\frac{1}{s}=1.$ Let $R^{\infty}(K,\mu)$ be the weak-star
closure of $\mbox{Rat}(K)$ in $L^{\infty}(\mu).$ Denote

$\ R^{t,\infty}(K,\mu)=R^{t}(K,\mu)\cap L^{\infty}(\mu).$

We let $\partial_{e}K$ (the exterior boundary of $K$) denote the union of the
boundaries of all the components of $\mathbb{C}\setminus K.$ The inner
boundary of $K$ is denoted $\partial_{i}K:=\partial K\setminus\partial_{e}K.$

Let $\mathbb{D}$ denote the unit disk $\\{z:|z|<1\\}$, let $\mathbb{T}$ denote
the unit circle $\\{z:|z|=1\\}$, and let $\mathbb{R}$ denote the real line
$\\{z:\text{Im}(z)=0\\}.$ For $z_{0}\in\mathbb{C}$ and $r>0,$ let
$\mathbb{D}(z_{0},r):=\\{z:~{}|z-z_{0}|<r\\}.$ Let $l_{B}$ be the arc-length
measure on a rectifiable curve $\Gamma$ restricted to a subset
$B\subset\Gamma.$ We use $C_{1},C_{2},...$ and $c_{1},c_{2},...$ for absolute
constants that may change from one step to the next.

A point $z_{0}$ in $\mathbb{C}$ is called a bounded point evaluation for
$P^{t}(\mu)$ (resp., $R^{t}(K,\mu)$) if $f\mapsto f(z_{0})$ defines a bounded
linear functional, denoted $\phi_{z_{0}},$ for the analytic polynomials
(resp., functions in $\mbox{Rat}(K)$) with respect to the $L^{t}(\mu)$ norm.
The collection of all such points is denoted $\mbox{bpe}(P^{t}(\mu))$ (resp.,
$\mbox{bpe}(R^{t}(K,\mu)$)). If $z_{0}$ is in the interior of
$\mbox{bpe}(P^{t}(\mu))$ (resp., $\mbox{bpe}(R^{t}(K,\mu)$)) and there exist
positive constants $r$ and $M$ such that $|f(z)|\leq M\|f\|_{L^{t}(\mu)}$,
whenever $z\in\mathbb{D}(z_{0},r)$ and $f$ is an analytic polynomial (resp.,
$f\in\mbox{Rat}(K)$), then we say that $z_{0}$ is an analytic bounded point
evaluation for $P^{t}(\mu)$ (resp., $R^{t}(K,\mu)$). The collection of all
such points is denoted $\mbox{abpe}(P^{t}(\mu))$ (resp.,
$\mbox{abpe}(R^{t}(K,\mu))$). Actually, it follows from Thomson’s Theorem [11]
(or see Theorem 1.1, below) that $\mbox{abpe}(P^{t}(\mu))$ is the interior of
$\mbox{bpe}(P^{t}(\mu))$. This also holds in the context of $R^{t}(K,\mu)$ as
was shown by J. Conway and N. Elias in [7]. Now, $\mbox{abpe}(P^{t}(\mu))$ is
the largest open subset of $\mathbb{C}$ to which every function $f\in
P^{t}(\mu)$ has an analytic continuation

$\ \rho(f)(z):=\phi_{z}(f),~{}z\in\mbox{abpe}(P^{t}(\mu)),$

that is,

$\displaystyle\ f(z)=\rho(f)(z)~{}\mu|_{\mbox{abpe}(P^{t}(\mu))}-a.a.,$ (1.1)

and similarly in the context of $R^{t}(K,\mu)$. The map $\rho$ is called the
evaluation map.

Let $S_{\mu}$ denote the multiplication by $z$ on $P^{t}(\mu)$ (resp.,
$R^{t}(K,\mu)$). The operator $S_{\mu}$ is pure if $P^{t}(\mu)$ (resp.,
$R^{t}(K,\mu)$) does not have a direct $L^{t}$ summand, and is irreducible if
$P^{t}(\mu)$ (resp., $R^{t}(K,\mu)$) contains no non-trivial characteristic
functions. For an open subset $U\subset\mathbb{C},$ we let $H^{\infty}(U)$
denote the bounded analytic functions on $U$. This is a Banach algebra in the
maximum modulus norm, but it is important to endow it also with a weak-star
topology.

We shall use $\overline{E}$ to denote the closure of the set
$E\subset\mathbb{C}$ and $\text{int}(E)$ to denote the interior of $E.$ Our
story begins with celebrated results of J. Thomson, in [11].

######  Theorem 1.1.

(Thomson 1991) Let $\mu\in M_{0}^{+}(\mathbb{C})$ and suppose that $1\leq
t<\infty$. Then there is a Borel partition $\\{\Delta_{i}\\}_{i=0}^{\infty}$
of $\mbox{spt}(\mu)$ such that

$\
P^{t}(\mu)=L^{t}(\mu|_{\Delta_{0}})\oplus\bigoplus_{i=1}^{\infty}P^{t}(\mu|_{\Delta_{i}})$

and the following statements are true:

(a) If $i\geq 1$, then $S_{\mu|_{\Delta_{i}}}$ on $P^{t}(\mu|_{\Delta_{i}})$
is irreducible.

(b) If $i\geq 1$ and $U_{i}:=abpe(P^{t}(\mu|_{\Delta_{i}}))$, then $U_{i}$ is
a simply connected region and $\Delta_{i}\subset\overline{U_{i}}$.

(c) If $S_{\mu}$ is pure (that is, if $\Delta_{0}=\emptyset$) and
$U=\text{abpe}(P^{t}(\mu))$, then the evaluation map $\rho$ is an isometric
isomorphism and a weak-star homeomorphism from $P^{t}(\mu)\cap
L^{\infty}(\mu)$ onto $H^{\infty}(U)$.

The next result is due to Brennan [3], which extends Theorem 1.1.

######  Theorem 1.2.

(Brennan 2008) Let $\mu\in M_{0}^{+}(K)$ for a compact set $K$. Suppose that
$1\leq t<\infty$ and the diameters of the components of $\mathbb{C}\setminus
K$ are bounded below. Then there is a Borel partition
$\\{\Delta_{i}\\}_{i=0}^{\infty}$ of $\mbox{spt}(\mu)$ and compact subsets
$\\{K_{i}\\}_{i=1}^{\infty}$ such that $\Delta_{i}\subset K_{i}$ for $i\geq
1$,

$\
R^{t}(K,\mu)=L^{t}(\mu|_{\Delta_{0}})\oplus\bigoplus_{i=1}^{\infty}R^{t}(K_{i},\mu|_{\Delta_{i}}),$

and the following statements are true:

(a) If $i\geq 1$, then $S_{\mu|_{\Delta_{i}}}$ on
$R^{t}(K_{i},\mu|_{\Delta_{i}})$ is irreducible.

(b) If $i\geq 1$ and $U_{i}:=\mbox{abpe}(R^{t}(K_{i},\mu|_{\Delta_{i}}))$,
then $K_{i}=\overline{U_{i}}$.

(c) If $i\geq 1$, then the evaluation map $\rho_{i}$ is an isometric
isomorphism and a weak-star homeomorphism from
$R^{t,\infty}(K_{i},\mu|_{\Delta_{i}})$ onto $H^{\infty}(U_{i})$.

In this paper, we show that when the diameters of the components of
$\mathbb{C}\setminus K$ are allowed to tend to $0$, then Theorem 1.2 (c) is no
long valid, even though $\text{int}(K)=\text{abpe}(R^{t}(K,\mu))$ and
$K=\overline{\text{int}(K)}.$ In fact, the evaluation map is an isometric
isomorphism and a weak-star homeomorphism from $R^{t,\infty}(K,\mu)$ onto a
proper sub-algebra of $H^{\infty}(\text{int}(K)),$ where functions in the sub-
algebra have equal non-tangential limits on certain portions of
$\partial_{i}K.$

######  Definition 1.3.

Let $\\{\lambda_{n}\\}_{n=1}^{\infty}$ be a subset of $\mathbb{R}$ such that
$\mathbb{D}(\lambda_{n},r_{n})$ is a subset of $\mathbb{D}$ and
$\overline{\mathbb{D}(\lambda_{i},r_{i})}\cap\overline{\mathbb{D}(\lambda_{j},r_{j})}=\emptyset$
for $i\neq j$. Define

$\ K\ :=\
\overline{\mathbb{D}}\setminus\left(\bigcup_{n=1}^{\infty}\mathbb{D}(\lambda_{n},r_{n})\right).$

Assume $K\cap\mathbb{R}$ has no interior as a subset of $\mathbb{R}$, and that
the length of $K\cap\mathbb{R}$ is greater than zero (i.e.
$\sum_{n=1}^{\infty}r_{n}<1$). A set of this form is called a string of beads
set. Set $\Omega=int(K)$. Define
$\Omega_{U}=\\{z\in\Omega:~{}\text{Im}(z)>0\\},~{}\Omega_{L}=\Omega\setminus\Omega_{U}$,
$K_{U}=\overline{\Omega_{U}}$, and $K_{L}=\overline{\Omega_{L}}$.

Clearly, $\partial_{i}K=K\cap\mathbb{R},~{}l_{\mathbb{R}}-a.a..$ For
$\lambda\in\mathbb{R}$ and $0<\alpha\leq 1$, define the upper cone

$\displaystyle\
U_{n}(\lambda,\alpha)=\left\\{z:~{}|Re(z-\lambda)|<\alpha(Im(z-\lambda))\right\\}\cap\mathbb{D}\left(\lambda,\frac{1}{n}\right)$
(1.2)

and the lower cone

$\displaystyle\
L_{n}(\lambda,\alpha)=\left\\{z:~{}|Re(z-\lambda)|<-\alpha(Im(z-\lambda))\right\\}\cap\mathbb{D}\left(\lambda,\frac{1}{n}\right).$
(1.3)

It is easy to verify that for
$\lambda\in\partial_{i}K,~{}l_{\partial_{i}K}-a.a.,$ there exists
$n_{\lambda}$ such that the non-tangential limit region

$\displaystyle\ U_{n_{\lambda}}(\lambda,\alpha)\cup
L_{n_{\lambda}}(\lambda,\alpha)\subset\Omega.$ (1.4)

Fatou’s theorem says that bounded analytic functions in the unit disk have
non-tangential limits almost everywhere. Precomposing with a Riemann map from
$\mathbb{D}$ onto $\Omega_{U}$ we get that, for $f\in H^{\infty}(\Omega),$ the
limits

$\displaystyle\ f_{+}(\lambda):=\lim_{z\in
U_{n}(\lambda,\alpha)\rightarrow\lambda}f(z)$ (1.5)

exist $l_{\partial_{i}K}-a.a.$ since the Riemann map is conformal from
$\overline{\mathbb{D}}$ onto $\overline{\Omega_{U}}$ (see [9, pp 208-218] and
[10, pp 80-82]), and likewise the limits

$\displaystyle\ f_{-}(\lambda):=\lim_{z\in
L_{n}(\lambda,\alpha)\rightarrow\lambda}f(z)$ (1.6)

exist $l_{\partial_{i}K}-a.a.$

For a subset $E_{0}\subset\partial_{i}K,$ $H^{\infty}(\Omega,E_{0})$ is the
sub-algebra of $f\in H^{\infty}(\Omega)$ for which
$f_{+}(z)=f_{-}(z),~{}l_{E_{0}}-a.a..$ The set $E_{0}$ is called the removable
boundary for $H^{\infty}(\Omega,E_{0}).$

The aim of this paper is to prove our main theorem below. By (1.1), we will
also use $f|_{\text{abpe}(R^{t}(K,\mu))}$ for $\rho(f).$

######  Main Theorem.

Let $1\leq t<\infty$ and let $K$ be a string of beads set defined as in
Definition 1.3. Suppose that $\mu\in M_{0}^{+}(K),$ $S_{\mu}$ on
$R^{t}(K,\mu)$ is pure, and $\text{abpe}(R^{t}(K,\mu))=\Omega.$ Then there
exists a Borel subset $E_{0}\subset\partial_{i}K$ such that the following
statements are true:

(1) If $f\in R^{t}(K,\mu)$, then the non-tangential limit from the top on
$E_{0}$, $f_{+}(z)$, and the non-tangential limit from the bottom on $E_{0}$,
$f_{-}(z)$, exist $l_{E_{0}}-a.a.$, and

$\ f_{+}(z)=f_{-}(z),~{}l_{E_{0}}-a.a..$

Consequently, if $f\in R^{t,\infty}(K,\mu),$ then $\rho(f)\in
H^{\infty}(\Omega,E_{0}).$

(2) If $F\in H^{\infty}(\Omega,E_{0}),$ then there exists $f\in
R^{t,\infty}(K,\mu)$ such that $F=\rho(f).$

(3) The evaluation map $\rho$ is an isometric isomorphism and a weak-star
homeomorphism from $R^{t,\infty}(K,\mu)$ onto $H^{\infty}(\Omega,E_{0}).$

Our methods can be applied to a compact subset $K\subset\mathbb{C}$ and
$\mu\in M_{0}^{+}(K)$ if $S_{\mu}$ on $R^{t}(K,\mu)$ is pure,
$\text{abpe}(R^{t}(K,\mu))=\text{int}(K),$ $\partial_{i}K$ is a subset of a
Lipschitz graph, and $\text{int}(K)$ contains non-tangential limit regions
(similar to (1.4)) whose vertices are on $\partial_{i}K$ almost everywhere
with respect to the arc-length measure restricted to $\partial_{i}K.$

In section 2, we review some recent results of analytic capacity and Cauchy
transform that are needed in our analysis. In the context of
$R^{\infty}(K,\mu),$ Proposition 3.1 is a simple version of Main Theorem (1)
where $E_{0}$ is always equal to $\partial_{i}K.$ However, for $R^{t}(K,\mu)$
when $1\leq t<\infty,$ $E_{0}$ can be an arbitrary subset of $\partial_{i}K$
as shown in Proposition 3.4. section 4 constructs the removable boundary
$E_{0}$ and proves Main Theorem (1). Main Theorem (2) and (3) are proved in
section 5 and section 6.

##  2\. Prelimaries

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

$\displaystyle\
\mathcal{C}_{\epsilon}(\nu)(z)=\int_{|w-z|>\epsilon}\dfrac{1}{w-z}d\nu(w).\ $
(2.1)

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 $\lim_{r\rightarrow 0}\frac{|\nu|(\mathbb{D}(\lambda,r))}{r}=0$ and
$\lim_{\epsilon\rightarrow 0}\mathcal{C}_{\epsilon}(\nu)(\lambda)$ 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 almost all $z$ with respect to the area measure
${\mathfrak{m}}$ on $\mathbb{\mathbb{C}}$, and clearly $\mathcal{C}(\nu)$ is
analytic in $\mathbb{C}_{\infty}\setminus\mbox{spt}(\nu)$, where
$\mathbb{C}_{\infty}:=\mathbb{C}\cup\\{\infty\\}.$ In fact, from Corollary 2.2
below, we see that (2.2) is defined for all $z$ except for a set of zero
analytic capacity. Throughout this paper, the Cauchy transform of a measure
always means the principal value of the transform.

The maximal Cauchy transform is defined by

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

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
$\mathbb{C}_{\infty}\setminus B$ 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 $E$ of $\mathbb{C}$ is given by:

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

Good sources for basic information about analytic capacity are Chapter VIII of
[8], Chapter V of [6], and [15].

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

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

where the supremum is taken over $\mu\in M_{0}^{+}(E)$ 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:

$\ \gamma_{+}(E)\leq\gamma(E)$

for all subsets $E$ of $\mathbb{C}$. X. Tolsa has established the following
astounding results. See [14] (also Theorem 6.1 and Corollary 6.3 in [15]) for
(1) and (2). See [13, Proposition 2.1] (also [15, 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

$\displaystyle\ \gamma(E)\leq A_{T}\gamma_{+}(E)$ (2.4)

for all $E\subset\mathbb{C}.$

(2) Semiadditivity of analytic capacity:

$\displaystyle\ \gamma\left(\bigcup_{i=1}^{m}E_{i}\right)\leq
A_{T}\sum_{i=1}^{m}\gamma(E_{i})$ (2.5)

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

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

$\displaystyle\ \gamma(\\{\mathcal{C}_{*}(\nu)\geq
a\\})\leq\dfrac{C_{T}}{a}\|\nu\|.$ (2.6)

Combining Theorem 2.1 (1), [15, Theorem 4.14], and [14] (or [15, Theorem
8.1]), we get the following corollary. The reader may also see [1, Corollary
3.1].

######  Corollary 2.2.

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
$z\in\mathbb{C}\setminus\mathcal{Q}$.

######  Proposition 2.3.

Let $X\subset\mathbb{C}$ and $a_{1},a_{2}\in\mathbb{C}.$ Let $\mathcal{Q}$ be
any set in $\mathbb{C}$ with $\gamma(\mathcal{Q})=0$. Let $f_{1}(z)$ and
$f_{2}(z)$ be functions on
$\mathbb{D}(\lambda,\delta_{0})\setminus\mathcal{Q}$ for some $\delta_{0}>0.$
If

$\ \lim_{\delta\rightarrow 0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\cap
X\cap\\{|f_{i}(z)-a_{i}|>\epsilon\\})}{\delta}=0,~{}i=1,2,$

for all $\epsilon>0,$ then the following properties hold:

(1)

$\ \lim_{\delta\rightarrow 0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\cap
X\cap\\{|f_{1}(z)+f_{2}(z)-(a_{1}+a_{2})|>\epsilon\\})}{\delta}=0;$

(2)

$\ \lim_{\delta\rightarrow 0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\cap
X\cap\\{|f_{1}(z)f_{2}(z)-a_{1}a_{2}|>\epsilon\\})}{\delta}=0;$

(3) If $a_{2}\neq 0,$ then

$\ \lim_{\delta\rightarrow 0}\dfrac{\gamma(\mathbb{D}(\lambda,\delta)\cap
X\cap\\{|\frac{f_{1}(z)}{f_{2}(z)}-\frac{a_{1}}{a_{2}})|>\epsilon\\})}{\delta}=0.$

###### Proof.

We have the following set inclusions.

$\displaystyle\\{|f_{1}(z)+f_{2}(z)-(a_{1}+a_{2})|>\epsilon\\}$
$\displaystyle\ \subset$
$\displaystyle\\{|f_{1}(z)-a_{1}|>\frac{\epsilon}{2}\\}\cup\\{|f_{2}(z)-a_{2}|>\frac{\epsilon}{2}\\}$

and

$\displaystyle\\{|f_{1}(z)f_{2}(z)-a_{1}a_{2}|>\epsilon\\}$ $\displaystyle\
\subset$
$\displaystyle\\{|f_{2}(z)||f_{1}(z)-a_{1}|>\frac{\epsilon}{2}\\}\cup\\{|a_{1}||f_{2}(z)-a_{2}|>\frac{\epsilon}{2}\\}$
$\displaystyle\ \subset$
$\displaystyle\\{\frac{3}{2}|a_{2}||f_{1}(z)-a_{1}|>\frac{\epsilon}{2}\\}\cup\\{|f_{2}(z)-a_{2}|>\frac{1}{2}|a_{2}|\\}\cup\\{|a_{1}||f_{2}(z)-a_{2}|>\frac{\epsilon}{2}\\}.$

(1) and (2) follows from Theorem 2.1 (2).

For (3): From (2), we may assume that $f_{1}(z)=1$ and $a_{1}=1.$ Because

$\displaystyle\left\\{\left|\frac{1}{f_{2}(z)}-\frac{1}{a_{2}})\right|>\epsilon\right\\}$
$\displaystyle\ \subset$
$\displaystyle\left\\{\frac{2}{|a_{2}|^{2}}\left|f_{2}(z)-a_{2}\right|>\epsilon\right\\}\cup\left\\{|f_{2}(z)-a_{2}|>\frac{|a_{2}|}{2}\right\\}.$

(3) follows from (2) and Theorem 2.1 (2). ∎

######  Definition 2.4.

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 corollary follows from Proposition 2.3.

######  Corollary 2.5.

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 2.6.

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

  * (a)

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

  * (b)

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

Then:

(1) $\mathcal{C}(\nu)(\lambda)=\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(\nu)(\lambda)$ exists for every $\lambda$ in
$\mathbb{C}$ except perhaps for some set $\mathcal{Q}$ with analytic capacity
$0$.

(2) The Cauchy transform $\mathcal{C}(\nu)(\lambda)$ is $\gamma$-continuous at
$\lambda_{0}$.

For $\nu\in M_{0}(\mathbb{C}),$ define

$\ \mathcal{ND}(\nu)=\left\\{\lambda:\underset{\delta\rightarrow
0}{\overline{\lim}}\dfrac{|\nu|(\mathbb{D}(\lambda,\delta))}{\delta}>0\right\\}$

and $\mathcal{ZD}(\nu)=\mathbb{C}\setminus\mathcal{ND}(\nu).$ The following
lemma is from [1, Lemma 3.5] (also see [15, Lemma 8.12]).

######  Lemma 2.7.

Let $\nu\in M_{0}(\mathbb{C})$. Suppose that $|\nu|\perp l_{E}$ for some Borel
set $E\subset\mathbb{R}$. Then

$\ l_{E}(\mathcal{ND}(\nu))=0.$

The following Lemma is from Lemma B in [2]. In its statement we let
${\mathfrak{m}}$ denote the area measure (i.e., two-dimensional Lebesgue
measure) on $\mathbb{C}$.

######  Lemma 2.8.

There are absolute constants $\epsilon_{1}>0$ and $C_{1}<\infty$ with the
following property. If $R>0$ and $E\subset\overline{\mathbb{D}(0,R)}$ with
$\gamma(E)<R\epsilon_{1}$, then

$\ |p(\lambda)|\leq\dfrac{C_{1}}{\pi
R^{2}}\int_{\overline{\mathbb{D}(0,R)}\setminus E}|p|\,d{\mathfrak{m}}$

for all $\lambda$ in $\mathbb{D}(0,\frac{R}{2})$ and all analytic polynomials
$p$.

The following lemma is from [15, Proposition 6.5].

######  Lemma 2.9.

Let $\Gamma$ be a rectifiable curve. Let $E\subset\Gamma$ be compact with
$l_{\Gamma}(E)<\infty.$ Let $f:\mathbb{C}_{\infty}\setminus
E\rightarrow\mathbb{C}$ be analytic such that $\|f\|_{\infty}\leq 1$ and
$f(\infty)=0$. Then there is a measure $\nu=bl_{\Gamma}|_{E}$, where $b$ is a
measurable function with $|b(z)|\leq 1$ for all $z\in E$, such that
$f(z)=\mathcal{C}(\nu)(z)$ for all $z\notin E$.

Plemelj’s formula for Lipschitz graphs as in X. Tolsa [15, Theorem 8.8] is
applied to the following lemma.

######  Lemma 2.10.

For all $f\in L^{1}(l_{\mathbb{R}})$, the non-tangential limits of
$\mathcal{C}(fdz|_{\mathbb{R}})(z)$ exists for $l_{\mathbb{R}}-a.a.$ and the
following identities hold $l_{\mathbb{R}}-a.a.$ for $\lambda\in\mathbb{R}$:

$\ \dfrac{1}{2\pi i}\lim_{z\in
U_{n}(\lambda,\alpha)\rightarrow\lambda}\mathcal{C}(fdz|_{\Gamma})(z)=\dfrac{1}{2\pi
i}\mathcal{C}(fdz|_{\mathbb{R}})(\lambda)+\dfrac{1}{2}f(\lambda),$

and

$\ \dfrac{1}{2\pi i}\lim_{z\in
L_{n}(\lambda,\alpha)\rightarrow\lambda}\mathcal{C}(fdz|_{\Gamma})(z)=\dfrac{1}{2\pi
i}\mathcal{C}(fdz|_{\mathbb{R}})(\lambda)-\dfrac{1}{2}f(\lambda).$

##  3\. Examples

For a compact subset $K\subset\mathbb{C}$ and $\mu\in M_{0}^{+}(K),$ the
envelope $E=E(K,\mu)$ is the subset of $z\in K$ that satisfies the property:
there exists $\mu_{z}\in M_{0}(K)$, with $\mu_{z}$ absolutely continuous with
respect to $\mu$, and such that $\mu_{z}(\\{z\\})=0$ and $\int fd\mu_{z}=f(z)$
for each $f\in\text{Rat}(K).$ Chaumat’s Theorem says that if
$R^{\infty}(K,\mu)$ does not contain a $L^{\infty}$ summand, then the map
$\rho(f)=f$ for $f\in\text{Rat}(K)$ extends an isometric isomorphism and a
weak-star homeomorphism from $R^{\infty}(K,\mu)$ onto
$R^{\infty}(K,{\mathfrak{m}}|_{E})$ (see [4] or [6, Chapter VI.3]). The map
$\rho$ is called Chaumat’s map. We start with a simple version of Main Theorem
(1) for $R^{\infty}(K,\mu)$ as the following proposition.

######  Proposition 3.1.

For a string of beads set $K$ and $\mu\in M_{0}^{+}(K),$ assume that
$R^{\infty}(K,\mu)$ has no $L^{\infty}$ summand and the envelope $E$ contains
$\Omega.$ Then $\rho(f)\in H^{\infty}(\Omega,\partial_{i}K)$ for $f\in
R^{\infty}(K,\mu).$

###### Proof.

Let $a=\frac{1}{2}i$ and $b=-\frac{1}{2}i$, where $i=\sqrt{-1}$. Let
$\omega_{a}$ and $\omega_{b}$ be the harmonic measures of $\Omega_{U}$ and
$\Omega_{L}$ evaluated at $a$ and $b$, respectively. Let
$\omega=\omega_{a}+\omega_{b}.$ It is clear that $\int
r(z)(z-a)(z-b)d\omega(z)=0$ for $r\in Rat(K)$. So $R^{\infty}(K,\omega)$ has
no $L^{\infty}$ summand and the envelope $E(K,\omega)\supset\Omega.$ From
Chaumat’s Theorem, the Chaumat map $\rho$ is an isometric isomorphism and a
weak-star homeomorphism from $R^{\infty}(K,\omega)$ onto
$R^{\infty}(K,{\mathfrak{m}}_{K}).$

Let $\\{r_{n}\\}\subset Rat(K)$ be such that $r_{n}\rightarrow f$ in
$L^{\infty}(\omega)$ weak-star topology. Hence, $f\in H^{\infty}(\Omega_{U})$
and $f\in H^{\infty}(\Omega_{L})$. Thus, $f_{+}(z)$ exists,
$f_{+}(z)=f(z),~{}\omega_{a}|_{\partial_{i}K}-a.a.$; and $f_{-}(z)$ exists,
$f_{-}(z)=f(z),~{}\omega_{b}|_{\partial_{i}K}-a.a.$. Since
$\partial\Omega_{U}$ and $\partial\Omega_{L}$ are rectifiable Jordan curves,
the measures $\omega_{a}|_{\partial_{i}K}$, $\omega_{b}|_{\partial_{i}K}$, and
$l_{\partial_{i}K}$ are mutually absolutely continuous. Therefore,

$\ f_{+}(z)=f_{-}(z)=f(z),~{}l_{\partial_{i}K}-a.a.$

Hence, $R^{\infty}(K,{\mathfrak{m}}_{K})\subset
H^{\infty}(\Omega,\partial_{i}K).$ Now the proof follows from Chaumat’s
Theorem. ∎

However, for $1\leq t<\infty$, the situations for $R^{t}(K,\mu)$ are very
different. The following example constructs a measure $\mu$ such that
$S_{\mu}$ on $R^{2}(K,\mu)$ is pure, $\text{abpe}(R^{2}(K,\mu))=\Omega$, and
$R^{2}(K,\mu)$ splits.

######  Example 3.2.

Let ${\mathfrak{m}}_{W}$ be the weighted planar Lebesgue measure
$W(z)d{\mathfrak{m}}(z)$ on $\Omega$, where

$\
W(z)=\begin{cases}\text{dist}(z,\mathbb{R})^{4},&z\in\Omega_{U}\cup\Omega_{L};\\\
0.&z\in(\Omega_{U}\cup\Omega_{L})^{c}.\end{cases}$

Then,

$\displaystyle\
R^{2}(K,{\mathfrak{m}}_{W})=R^{2}(K_{U},{\mathfrak{m}}_{W}|_{\Omega_{U}})\oplus
R^{2}(K_{L},{\mathfrak{m}}_{W}|_{\Omega_{L}}).$ (3.1)

###### Proof.

It is easy to verify that $S_{{\mathfrak{m}}_{W}}$ is pure and
$\text{abpe}(R^{2}(K,{\mathfrak{m}}_{W}))=\Omega$. Let
$\lambda\in\partial_{i}K$. Choose a subsequence $\\{\lambda_{n_{k}}\\}$ such
that $\lambda_{n_{k}}\rightarrow\lambda$. For $g\perp
R^{2}(K,{\mathfrak{m}}_{W})$, we have

$\ \begin{aligned} \
&\int\left|\dfrac{1}{z-\lambda_{n_{k}}}-\dfrac{1}{z-\lambda}\right||g|d{\mathfrak{m}}_{W}\\\
\
\leq&|\lambda_{n_{k}}-\lambda|\left(\int_{G}\dfrac{W(z)}{|z-\lambda_{n_{k}}|^{2}|z-\lambda|^{2}}d{\mathfrak{m}}(z)\right)^{\frac{1}{2}}\|g\|_{L^{2}({\mathfrak{m}}_{W})}\\\
\ \leq&\sqrt{\pi}\|g\|_{L^{2}({\mathfrak{m}}_{W})}|\lambda_{n_{k}}-\lambda|\\\
\ &\rightarrow 0.\ \end{aligned}$

Thus, $\mathcal{C}(g{\mathfrak{m}}_{W})(\lambda)=0$ for
$\lambda\in\partial_{i}K$. Let $\nu$ be $\dfrac{1}{2\pi i}dz$ restricted to
upper circle of $|z|=2$ and the interval $[-2,2]$. By Fubini’s theorem, we see
that

$\
\int\mathcal{C}(\nu)(z)g(z)d{\mathfrak{m}}_{W}(z)=-\int\mathcal{C}(g{\mathfrak{m}}_{W})(\lambda)d\nu(\lambda)=0.$

Hence, $\chi_{\Omega_{U}}(z)\in R^{2}(K,{\mathfrak{m}}_{W})$, where $\chi_{A}$
is the characteristic function for the set $A$. (3.1) is proved. ∎

Now let us look at a simple example of $\mu$ such that $S_{\mu}$ is
irreducible.

######  Example 3.3.

Let $a,$ $b,$ $\omega_{a},$ $\omega_{b},$ and $\omega$ be as in the beginning
of the proof of Proposition 3.1. Then, for every $f\in R^{2}(K,\omega)$, the
non-tangential limit from the top on $\partial_{i}K$, $f_{+}(z)$, and the non-
tangential limit from the bottom on $\partial_{i}K$, $f_{-}(z)$, exist
$l_{\partial_{i}K}-a.a.$, and
$f_{+}(z)=f_{-}(z)=f(z),~{}l_{\partial_{i}K}-a.a..$ Consequently, $S_{\omega}$
is irreducible.

###### Proof.

The same argument in the proof of Proposition 3.1 applies. ∎

The examples above suggest that, in order to ensure that $S_{\mu}$ on
$R^{t}(K,\mu)$ is irreducible, the values of every function $f\in
R^{t}(K,\mu)$ on $\Omega_{U}$ and $\Omega_{L}$ must be associated in some
ways. Main Theorem points out the relationship.

Assuming our Main Theorem has been proved, we have the following proposition
that combines Example 3.2 and 3.3.

######  Proposition 3.4.

Let $K$ be a string of beads set. If $F\subset\partial_{i}K$ is a compact
subset with $l_{\partial_{i}K}(F)>0$, then there exists $\mu\in M_{0}^{+}(K)$
such that $S_{\mu}$ on $R^{2}(K,\mu)$ is irreducible,
$\text{abpe}(R^{2}(K,\mu))=\Omega$, and $E_{0}=F.$

###### Proof.

Let $\Gamma_{U}$ be a closed rectifiable Jordan curve in $K_{U}$ such that
$\Gamma_{U}\cap\partial\Omega_{U}=F$. Let $\Gamma_{L}$ be a closed rectifiable
Jordan curve in $K_{L}$ such that $\Gamma_{L}\cap\partial\Omega_{L}=F$. Let
$G_{U}$ and $G_{L}$ be the regions surrounded by $\Gamma_{U}$ and
$\Gamma_{L}$, respectively. Let $\omega_{a}$ and $\omega_{b}$ be the harmonic
measures of $G_{U}$ and $G_{L}$, respectively, where $a\in G_{U}$ and $b\in
G_{L}$. Let ${\mathfrak{m}}_{W}$ be the weighted planar Lebesgue measure as in
Example 3.2. Let $\mu=\omega_{a}+\omega_{b}+{\mathfrak{m}}_{W}$.

Similarly to the proof of Example 3.3, we see that $f_{+}(z)=f_{-}(z)=f(z)$ on
$F$ $l_{\partial_{i}K}-a.a.$. Hence, $S_{\mu}$ is irreducible. Obviously,
$\text{abpe}(R^{2}(K,\mu))=\Omega$.

Suppose there exists a compact subset $F_{0}\subset\partial_{i}K\setminus F$
that is a part of $E_{0}$ and $l_{\partial_{i}K}(F_{0})>0$. Let $f$ be a
bounded analytic function on $\mathbb{C}_{\infty}\setminus F_{0}$ with
$\|f\|_{\infty}=1.$ Using Lemma 2.9, we have
$f=\mathcal{C}(wl_{\partial_{i}K})$, where $w$ is bounded and supported in
$F_{0}$. Let $\\{g_{j}\\}$ be a dense subset of $R^{2}(K,\mu)^{\perp}.$ As in
the proof of Example 3.2, we see that $\dfrac{1}{z-\lambda}\in R^{2}(K,\mu)$
for $\lambda\in F_{0}$. Moreover,

$\ \begin{aligned} \ &\int\dfrac{1}{|z-\lambda|}|g_{j}|d\mu\\\ \
\leq&\int_{\Gamma_{U}}\dfrac{1}{|z-\lambda|}|g_{j}|d\omega_{a}+\int_{\Gamma_{L}}\dfrac{1}{|z-\lambda|}|g_{j}|d\omega_{b}+\int_{\Omega}\dfrac{W(z)}{|z-\lambda|}|g_{j}|d{\mathfrak{m}}\\\
\
\leq&\dfrac{\|g_{j}\|}{\text{dist}(\lambda,\Gamma_{U})}+\dfrac{\|g_{j}\|}{\text{dist}(\lambda,\Gamma_{L})}+\|g_{j}\|\\\
\ \end{aligned}$

for $\lambda\in F_{0}$. Using Fubini’s theorem, we have

$\ \int
f(z)g_{j}(z)d\mu(z)=-\int\mathcal{C}(g_{j}\mu)(\lambda)w(\lambda)dl_{\partial_{i}K}(\lambda)=0.$

Hence, $f\in R^{2}(K,\mu)$. This contradicts Main Theorem because, by Lemma
2.10, $f_{+}$ and $f_{-}$ are not identically equal with respect to
$l_{\partial_{i}K}$ on $F_{0}$. ∎

##  4\. Construction of $E_{0}$ and proof of Main Theorem (1)

We assume that $1\leq t<\infty,$ $K$ is a string of beads set, $\mu\in
M_{0}^{+}(K),$ $S_{\mu}$ on $R^{t}(K,\mu)$ is pure, and
$\text{abpe}(R^{t}(K,\mu))=\Omega(=\text{int}(K)).$

Let $\mu=hl_{\partial_{i}K}+\mu_{s}$ be the Radon-Nikodym decomposition with
respect to $l_{\partial_{i}K}$ on $\mathbb{R}$, where $h\in
L^{1}(l_{\partial_{i}K})$ and $\mu_{s}\perp l_{\partial_{i}K}$. We define
$\mathcal{N}(h)=\\{z\in\partial_{i}K,~{}h(z)\neq 0\\}$ and
$\mathcal{Z}(h)=\\{z\in\partial_{i}K,~{}h(z)=0\\}.$

Let $\\{g_{j}\\}\subset R^{t}(K,\mu)^{\perp}$ be a dense set. For $f\in
R^{t}(K,\mu)$, we see that $f(z)$ is analytic on
$\text{abpe}(R^{t}(K,\mu))=\Omega$ and $\frac{f(z)-f(\lambda)}{z-\lambda}\in
R^{t}(K,\mu)$ for $\lambda\in\Omega.$ Therefore, by Corollary 2.2,

$\displaystyle
f(\lambda)\mathcal{C}(g_{j}\mu)(\lambda)=\mathcal{C}(fg_{j}\mu)(\lambda),~{}\lambda\in\Omega,~{}\gamma-a.a.$
(4.1)

for all $j\geq 1.$

######  Lemma 4.1.

Let $f(z)$ be analytic on $U_{n_{\lambda}}(\lambda,\alpha)$ for some
$\lambda\in\partial_{i}K$ and $n_{\lambda}\geq 1$ such that (1.4) holds. If
for all $\epsilon>0$,

$\
\lim_{n\rightarrow\infty}n\gamma(U_{n}(\lambda,\alpha)\cap\\{|f(z)-a|>\epsilon\\})=0,$

then for $0<\theta<1$,

$\ \underset{z\in
U_{n}(\lambda,\theta\alpha)}{\lim_{n\rightarrow\infty}}f(z)=a.$

The same result holds if we replace $U_{n}(\lambda,\alpha)$ by
$L_{n}(\lambda,\alpha).$

###### Proof.

Let $n>n_{\lambda}.$ There exists a constant $0<\theta_{0}<1$ only depending
on $\theta$ such that $\mathbb{D}(w,\theta_{0}\frac{1}{n})\subset
U_{n}(\lambda,\alpha)$ for $w\in U_{2n}(\lambda,\theta\alpha)$ and
$\frac{1}{2(n+1)}<|w-\lambda|\leq\frac{1}{2n}$. From the assumption, we find
$n_{\epsilon}$ such that for $n>n_{\epsilon},$

$\
\gamma(U_{n}(\lambda,\alpha)\cap\\{|f(z)-a|>\epsilon\\})<\epsilon_{1}\theta_{0}\frac{1}{n},$

where $\epsilon_{1}$ is as in Lemma 2.8. Applying Lemma 2.8, we conclude that

$\
|f(w)-a|\leq\dfrac{C_{1}n^{2}}{\pi}\int_{\mathbb{D}(w,\theta_{0}\frac{1}{n})\setminus\\{|f(z)-a|>\epsilon\\}}|f(z)-a|d{\mathfrak{m}}(z)\leq
C_{2}\epsilon$

for $w\in U_{2n}(\lambda,\theta\alpha)$ and
$\frac{1}{2(n+1)}<|w-\lambda|\leq\frac{1}{2n}$. The lemma follows. ∎

######  Lemma 4.2.

The principle value $\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(g\mu)(z)=\mathcal{C}(g\mu)(z)$ exists
$l_{\partial_{i}K}-a.a.$ for $g\in L^{1}(\mu)$.

###### Proof.

The proof follows from Corollary 2.2 and the fact that
$\gamma|_{\partial_{i}K}\approx l_{\partial_{i}K}$ (see [8, VIII.2.2]). ∎

######  Definition 4.3.

Define

$\
F_{0}(\\{g_{j}\mu\\})=\bigcap_{j=1}^{\infty}\\{z\in\mathcal{Z}(h),~{}\mathcal{C}(g_{j}\mu)(z)=0\\}$

and

$\
R_{0}(\\{g_{j}\mu\\})=\bigcup_{j=1}^{\infty}\\{z\in\mathcal{Z}(h),~{}\mathcal{C}(g_{j}\mu)(z)\neq
0\\}$

Clearly, from Lemma 4.2, we see that

$\displaystyle\begin{aligned} \ &l_{\partial_{i}K}(F_{0}(\\{g_{j}\mu\\})\cap
R_{0}(\\{g_{j}\mu\\}))=0,\\\ \
&l_{\partial_{i}K}(\mathcal{Z}(h)\setminus(F_{0}(\\{g_{j}\mu\\})\cup
R_{0}(\\{g_{j}\mu\\})))=0.\end{aligned}$ (4.2)

######  Lemma 4.4.

If $f\in R^{t}(K,\mu),$ then

$\ f_{+}(z)=f_{-}(z),~{}z\in R_{0}(\\{g_{j}\mu\\}),~{}l_{\partial_{i}K}-a.a..$

###### Proof.

Let $\lambda\in R_{0}(\\{g_{j}\mu\\}).$ We find $j_{0}$ such that
$\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(g_{j_{0}}\mu)(\lambda)=\mathcal{C}(g_{j_{0}}\mu)(\lambda)$
exists and $C(g_{j_{0}}\mu)(\lambda)\neq 0.$ Since
$l_{\partial_{i}K}|_{\mathcal{Z}(h)}\perp|g_{j_{0}}\mu|,$ by Lemma 2.7, we see
that

$\
l_{\partial_{i}K}(\mathcal{Z}(h)\cap(\mathcal{ND}(g_{j_{0}}\mu)\cup\mathcal{ND}(fg_{j_{0}}\mu)))=0.$

Hence, we may assume that
$\lambda\in\mathcal{ZD}(g_{j_{0}}\mu)\cap\mathcal{ZD}(fg_{j_{0}}\mu).$ From
Lemma 2.6, we infer that $\mathcal{C}(g_{j_{0}}\mu)$ and
$\mathcal{C}(fg_{j_{0}}\mu)$ are $\gamma$-continuous at $\lambda.$ Since
$\mathcal{C}(g_{j_{0}}\mu)(\lambda)\neq 0,$ by Corollary 2.5, we see that
$\frac{\mathcal{C}(fg_{j_{0}}\mu)(z)}{\mathcal{C}(g_{j_{0}}\mu)(z)}$ is
$\gamma$-continuous at $\lambda.$ From (4.1) and Lemma 4.1, we conclude that

$\
f_{+}(\lambda)=f_{-}(\lambda)=\frac{\mathcal{C}(fg_{j_{0}}\mu)(\lambda)}{\mathcal{C}(g_{j_{0}}\mu)(\lambda)}.$

The lemma is proved. ∎

######  Lemma 4.5.

Let $\nu\in M_{0}(\mathbb{C})$. Suppose that
$\nu=h_{0}l_{\partial_{i}K}+\sigma$ is the Radon-Nikodym decomposition with
respect to $l_{\partial_{i}K}$, where $h_{0}\in L^{1}(l_{\partial_{i}K})$ and
$\sigma\perp l_{\partial_{i}K}$. Then there exists a subset
$\mathcal{Q}\subset\mathbb{C}$ with $\gamma(\mathcal{Q})=0$, such that the
following hold:

(a) $\mathcal{C}(\nu)(\lambda)=\lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(\nu)(\lambda)$ exists for
$\lambda\in\mathbb{C}\setminus\mathcal{Q}$,

(b) for $\lambda\in\partial_{i}K\setminus\mathcal{Q}$ and $\epsilon>0$,
$v^{+}(\nu)(\lambda):=\mathcal{C}(\nu)(\lambda)+\pi ih_{0}(\lambda)$,

$\displaystyle\
\lim_{n\rightarrow\infty}n\gamma(U_{n}(\lambda,\alpha)\cap\\{|\mathcal{C}(\nu)(z)-v^{+}(\nu)(\lambda)|>\epsilon\\})=0,$
(4.3)

and

(c) for $\lambda\in\partial_{i}K\setminus\mathcal{Q}$ and $\epsilon>0$,
$v^{-}(\nu)(\lambda):=\mathcal{C}(\nu)(\lambda)-\pi ih_{0}(\lambda)$,

$\displaystyle\
\lim_{n\rightarrow\infty}n\gamma(L_{n}(\lambda,\alpha)\cap\\{|\mathcal{C}(\nu)(z)-v^{-}(\nu)(\lambda)|>\epsilon\\})=0.$
(4.4)

###### Proof.

(a) is trivial by Corollary 2.2. For (b): Because

$\displaystyle\\{|\mathcal{C}(\nu)(z)-v^{+}(\nu)(\lambda)|>\epsilon\\}$
$\displaystyle\ \subset$
$\displaystyle\\{|\mathcal{C}(\sigma)(z)-\mathcal{C}(\sigma)(\lambda)|>\frac{\epsilon}{2}\\}\cup\\{|\mathcal{C}(h_{0}l_{\partial_{i}K})(z)-v^{+}(h_{0}l_{\partial_{i}K})(\lambda)|>\frac{\epsilon}{2}\\},$

(b) follows from Lemma 2.6, Lemma 2.10 and Theorem 2.1 (2). The proof of (c)
is the same. ∎

######  Definition 4.6.

Define

$\
F_{+}(\\{g_{j}\mu\\})=\bigcap_{j=1}^{\infty}\\{z\in\mathcal{N}(h),~{}v^{+}(g_{j}\mu)(z)=0\\},$
$\
F_{-}(\\{g_{j}\mu\\})=\bigcap_{j=1}^{\infty}\\{z\in\mathcal{N}(h),~{}v^{-}(g_{j}\mu)(z)=0\\},$

and

$\displaystyle\ R_{1}(\\{g_{j}\mu\\})=$
$\displaystyle\left(\bigcup_{j=1}^{\infty}\\{z\in\mathcal{N}(h),~{}v^{+}(g_{j}\mu)(z)\neq
0\\}\right)$
$\displaystyle\bigcap\left(\bigcup_{j=1}^{\infty}\\{z\in\mathcal{N}(h),~{}v^{-}(g_{j}\mu)(z)\neq
0\\}\right).$

Clearly, from Lemma 4.2, we see that

$\displaystyle\begin{aligned} \ &l_{\partial_{i}K}((F_{+}(\\{g_{j}\mu\\})\cup
F_{-}(\\{g_{j}\mu\\}))\cap R_{1}(\\{g_{j}\mu\\}))=0,\\\ \
&l_{\partial_{i}K}(\mathcal{N}(h)\setminus(F_{+}(\\{g_{j}\mu\\})\cup
F_{-}(\\{g_{j}\mu\\}))\cup R_{1}(\\{g_{j}\mu\\}))))=0.\end{aligned}$ (4.5)

######  Lemma 4.7.

If $f\in R^{t}(K,\mu),$ then

$\
f(z)\mathcal{C}(g_{j}\mu)(z)=\mathcal{C}(fg_{j}\mu)(z),~{}z\in\mathcal{N}(h),~{}l_{\partial_{i}K}-a.a..$

###### Proof.

There exists a sequence $\\{f_{n}\\}\subset Rat(K)$ such that

$\ \|f_{n}-f\|_{L^{t}(\mu)}\rightarrow 0,~{}f_{n}\rightarrow f,~{}\mu-a.a..$

Let $g\in\\{g_{j}\\}.$ From Corollary 2.2, we let
$\mathcal{Q}_{1}\subset\mathbb{C}$ with $\gamma(\mathcal{Q}_{1})=0$ such that
the principal values of $\mathcal{C}(g\mu)(z),$ $\mathcal{C}(f_{n}g\mu)(z)$
for $n\geq 1,$ and $\mathcal{C}(fg\mu)(z)$ exist for
$z\in\mathbb{C}\setminus\mathcal{Q}_{1}$. Define

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

Since

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

applying Theorem 2.1 (3), we get

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

Choose $n_{m}$ so that $\|f_{n_{m}}-f\|_{L^{t}(\mu)}\leq\frac{1}{m\|g\|2^{m}}$
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), we have

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

Set $\mathcal{Q}_{2}=\cap_{k=1}^{\infty}B_{k}$. Clearly,
$\gamma(\mathcal{Q}_{2})=0$. If
$\mathcal{Q}=\mathcal{Q}_{1}\cup\mathcal{Q}_{2}$, then on
$\mathbb{C}\setminus\mathcal{Q}$, $\mathcal{C}(f_{n_{m}}g\mu)(\lambda)$
converges to $\mathcal{C}(fg\mu)(\lambda)$. Therefore,

$\
f_{n_{m}}(z)\mathcal{C}(g\mu)(z)=\mathcal{C}(f_{n_{m}}g\mu)(z)\rightarrow\mathcal{C}(fg\mu)(z),~{}\gamma-a.a..$

The lemma follows since $\gamma|_{\partial_{i}K}\approx l_{\partial_{i}K}.$ ∎

######  Lemma 4.8.

If $f\in R^{t}(K,\mu),$ then

$\ f_{+}(z)=f_{-}(z)=f(z),~{}z\in
R_{1}(\\{g_{j}\mu\\}),~{}l_{\partial_{i}K}-a.a..$

###### Proof.

Let $\lambda\in R_{1}(\\{g_{j}\mu\\}).$ We find $j_{1}$ and $j_{2}$ such that
$v^{+}(g_{j_{1}}\mu)(\lambda)$ and $v^{-}(g_{j_{2}}\mu)(\lambda)$ exist,
$v^{+}(g_{j_{1}}\mu)(\lambda)\neq 0,$ and $v^{-}(g_{j_{2}}\mu)(\lambda)\neq
0.$ From (4.3) and (4.4), we assume that

$\
\lim_{n\rightarrow\infty}n\gamma\left(U_{n}(\lambda,\alpha)\cap\\{|\mathcal{C}(g_{j_{1}}\mu)(z)-v^{+}(g_{j_{1}}\mu)(\lambda)|>\epsilon\\}\right)=0,$
$\
\lim_{n\rightarrow\infty}n\gamma\left(U_{n}(\lambda,\alpha)\cap\\{|\mathcal{C}(fg_{j_{1}}\mu)(z)-v^{+}(fg_{j_{1}}\mu)(\lambda)|>\epsilon\\}\right)=0$
$\
\lim_{n\rightarrow\infty}n\gamma\left(L_{n}(\lambda,\alpha)\cap\\{|\mathcal{C}(g_{j_{2}}\mu)(z)-v^{-}(g_{j_{2}}\mu)(\lambda)|>\epsilon\\}\right)=0,$

and

$\
\lim_{n\rightarrow\infty}n\gamma\left(L_{n}(\lambda,\alpha)\cap\\{|\mathcal{C}(fg_{j_{2}}\mu)(z)-v^{-}(fg_{j_{2}}\mu)(\lambda)|>\epsilon\\}\right)=0.$

Applying Proposition 2.3 (3) for $X=U_{n_{\lambda}}(\lambda,\alpha)$ for some
$n_{\lambda}\geq 1,$ we conclude that

$\
\lim_{n\rightarrow\infty}n\gamma\left(U_{n}(\lambda,\alpha)\cap\left\\{\left|\dfrac{\mathcal{C}(fg_{j_{1}}\mu)(z)}{\mathcal{C}(g_{j_{1}}\mu)(z)}-\dfrac{v^{+}(fg_{j_{1}}\mu)(\lambda)}{v^{+}(g_{j_{1}}\mu)(\lambda)}\right|>\epsilon\right\\}\right)=0.$

Since by (4.1),
$f(z)=\frac{\mathcal{C}(fg_{j_{1}}\mu)(z)}{\mathcal{C}(g_{j_{1}}\mu)(z)}$ for
$z\in U_{n_{\lambda}}(\lambda,\alpha),~{}\gamma-a.a.$ and by Lemma 4.7,
$f(z)=\frac{v^{+}(fg_{j_{1}}\mu)(z)}{v^{+}(g_{j_{1}}\mu)(z)},~{}z\in\mathcal{N}(h)\cap\\{z:~{}v^{+}(g_{j_{1}}\mu)(z)\neq
0\\},~{}l_{\partial_{i}K}-a.a.,$ we get

$\
\lim_{n\rightarrow\infty}n\gamma\left(U_{n}(\lambda,\alpha)\cap\\{|f(z)-f(\lambda)|>\epsilon\\}\right)=0.$

Thus, from Lemma 4.1, we conclude that $f_{+}(\lambda)=f(\lambda).$ Similarly,
$f_{-}(\lambda)=f(\lambda).$ The lemma is proved. ∎

###### Proof.

(Main Theorem (1)) Let $E_{0}=R_{0}(\\{g_{j}\mu\\})\cup
R_{1}(\\{g_{j}\mu\\}).$ Then the proof follows from Lemma 4.4 and Lemma 4.8. ∎

##  5\. Building block functions and an integral estimate

We will continue to use the assumptions and notations of the last section. Set

$\ F(\\{g_{j}\mu\\})=F_{0}(\\{g_{j}\mu\\})\cup F_{+}(\\{g_{j}\mu\\})\cup
F_{-}(\\{g_{j}\mu\\}).$

Clearly,

$\ l_{\partial_{i}K}(E_{0}\cap
F(\\{g_{j}\mu\\}))=0,~{}l_{\partial_{i}K}(\partial_{i}K\setminus(E_{0}\cup
F(\\{g_{j}\mu\\}))=0.$

A measure $\eta\in M_{0}^{+}(\mathbb{C})$ is $c$-linear growth if

$\ \eta(\mathbb{D}(\lambda,\delta))\leq c\delta,\text{ for
}\lambda\in\mathbb{C}\text{ and }\delta>0.$

The following lemma is critical for proving Main Theorem (2).

######  Lemma 5.1.

Let $E_{1}\subset F(\\{g_{j}\mu\\})$ be a compact subset with
$\gamma(E_{1})>0$. Then there exist $f\in R^{t,\infty}(K,\mu),$ $\eta\in
M_{0}^{+}(E_{1}),$ and $\|\mathcal{C}_{\epsilon}(\eta)\|\leq C_{3}$ such that
$f(z)=\mathcal{C}(\eta)(z)$ for $z\in E_{1}^{c}$,

$\ \|f\|_{L^{\infty}(\mu)}\leq
C_{3},~{}f(\infty)=0,~{}f^{\prime}(\infty)=\gamma(E_{1}),$

and $\rho(f)=\mathcal{C}(\eta)|_{\Omega}\in H^{\infty}(\Omega,E_{0}).$

###### Proof.

From Theorem 2.1 (2), we have

$\ \gamma(E_{1})\leq A_{T}(\gamma(E_{1}\cap
F_{0}(\\{g_{j}\mu\\}))+\gamma(E_{1}\cap
F_{+}(\\{g_{j}\mu\\}))+\gamma(E_{1}\cap F_{-}(\\{g_{j}\mu\\}))).$

Let $E_{1}^{\prime}$ denote one of the sets $E_{1}\cap F_{0}(\\{g_{j}\mu\\}),$
$E_{1}\cap F_{+}(\\{g_{j}\mu\\}),$ or $E_{1}\cap F_{-}(\\{g_{j}\mu\\})$ that
has the biggest analytic capacity. Then
$\gamma(E_{1}^{\prime})\geq\frac{1}{3A_{T}}\gamma(E_{1}).$

Let $A_{j}=\\{\mathcal{C}_{*}(g_{j}\mu)(z)\leq M_{j}\\}$ for some constant
$M_{j}>0.$ By Theorem 2.1 (3), we can select $M_{j}$ so that
$\gamma(A_{j}^{c})<\frac{\gamma(E_{1})}{2^{j}(6A_{T}^{3})}$. Set
$E_{2}=\cap_{j=1}^{\infty}A_{j}\cap E_{1}^{\prime}$. Then applying Theorem 2.1
(2), we get

$\
\gamma(E_{2})\geq\dfrac{1}{A_{T}}\gamma(E_{1}^{\prime})-\gamma(\cup_{j=1}^{\infty}A_{j}^{c})\geq\dfrac{\gamma(E_{1})}{3A_{T}^{2}}-A_{T}\sum_{j=1}^{\infty}\gamma(A_{j}^{c})\geq\dfrac{\gamma(E_{1})}{6A_{T}^{2}}$

and

$\displaystyle\ \mathcal{C}_{*}(g_{j}\mu)(z)\leq M_{j},\text{ for }z\in
E_{2}\text{ and }j\geq 1.$ (5.1)

From Theorem 2.1 (1) and [15, Theorem 4.14], there exists $\eta_{0}\in
M_{0}^{+}(E_{2})$ such that $c_{1}\gamma(E_{2})\leq\|\eta_{0}\|,$ $\eta_{0}$
is $1$-linear growth, and ${\mathcal{C}}_{*}(\eta_{0})\leq 1.$ Clearly,
$\eta_{0}$ is absolutely continuous with respect to $l_{E_{2}}.$ The Cauchy
transform of $l_{E_{2}}$ operator is $L^{2}(l_{E_{2}})$ bounded (see [5] or
[15, Theorem 3.11]). Hence, from [12] (or [15, Theorem 8.1]), we have, for
$j\geq 1,$

$\displaystyle\ \lim_{\epsilon\rightarrow
0}\mathcal{C}_{\epsilon}(g_{j}\mu)(z)=\mathcal{C}(g_{j}\mu)(z),~{}\eta_{0}-a.a..$
(5.2)

There exists a subsequence $f_{k}(z)=\mathcal{C}_{\epsilon_{k}}(\eta_{0})(z)$
such that $f_{k}$ converges to $f_{0}\in L^{\infty}(\mu)$ in weak-star
topology, $f_{k}(\lambda)$ converges to
$f_{0}(\lambda)=\mathcal{C}(\eta_{0})(\lambda)$ uniformly on any compact
subset of $(\text{spt}(\eta_{0}))^{c}$ as $\epsilon_{k}\rightarrow 0.$ Since

$\displaystyle\
\int\mathcal{C}_{\epsilon_{k}}(\eta_{0})(z)g_{j}(z)d\mu(z)=-\int\mathcal{C}_{\epsilon_{k}}(g_{j}\mu)(z)d\eta_{0}(z),$
(5.3)

Using (5.1), (5.2), and the Lebesgue dominated convergence theorem, we get

$\displaystyle\ \int
f_{0}(z)g_{j}(z)d\mu(z)=-\int\mathcal{C}(g_{j}\mu)(z)d\eta_{0}(z).$ (5.4)

Let $\lambda\notin E_{2}$ and $d=\text{dist}(\lambda,E_{2})$, for
$z\in\mathbb{D}(\lambda,\frac{d}{2})$ and $\epsilon<\frac{d}{2}$, we have

$\
\left|\dfrac{\mathcal{C}_{\epsilon}(\eta_{0})(z)-f_{0}(\lambda)}{z-\lambda}\right|\leq\left|\mathcal{C}_{\epsilon}\left(\dfrac{\eta_{0}(s)}{s-\lambda}\right)(z)\right|\leq\dfrac{2}{d^{2}}\|\eta_{0}\|.$

For $z\notin\mathbb{D}(\lambda,\frac{d}{2})$ and $\epsilon<\frac{d}{2}$,

$\
\left|\dfrac{\mathcal{C}_{\epsilon}(\eta_{0})(z)-f_{0}(\lambda)}{z-\lambda}\right|\leq\dfrac{4}{d}.$

Thus, we can replace the above proof for the measure
$\dfrac{\eta_{0}(s)}{s-\lambda}$. In fact, we can choose a subsequence
$\\{\mathcal{C}_{\epsilon_{k_{m}}}(\eta_{0})\\}$ such that
$e_{k_{m}}(z)=\dfrac{\mathcal{C}_{\epsilon_{k_{m}}}(\eta_{0})(z)-f_{0}(\lambda)}{z-\lambda}$
converges to $e(z)$ in weak-star topology. Clearly,
$(z-\lambda)e_{k_{m}}(z)+f_{0}(\lambda)=\mathcal{C}_{\epsilon_{k_{m}}}(\eta_{0})(z)$
converges to $(z-\lambda)e(z)+f_{0}(\lambda)=f_{0}(z)$ in weak-star topology.
On the other hand, the equation 5.3 becomes

$\displaystyle\
\int\mathcal{C}_{\epsilon_{k_{m}}}(\dfrac{\eta_{0}(s)}{s-\lambda})(z)g_{j}d\mu=-\int\mathcal{C}_{\epsilon_{k_{m}}}(g_{j}\mu)(z)\dfrac{d\eta_{0}(z)}{z-\lambda}$
(5.5)

and for $\epsilon_{k_{m}}<\frac{d}{2}$, we have

$\displaystyle\ \begin{aligned} \
&\left|\mathcal{C}_{\epsilon_{k_{m}}}(\dfrac{\eta_{0}(s)}{s-\lambda})(z)-e_{k_{m}}(z)\right|\\\
\ \leq&\begin{cases}0,&z\in\mathbb{D}(\lambda,\frac{d}{2}),\\\
\dfrac{2}{d^{2}}\eta_{0}(\mathbb{D}(z,\epsilon_{k_{m}})),&z\notin\mathbb{D}(\lambda,\frac{d}{2})\end{cases},\
\end{aligned}$ (5.6)

which goes to zero as $\epsilon_{k_{j}}\rightarrow 0$. Combining (5.1), (5.2),
(5.5), (5.6), and the Lebesgue dominated convergence theorem,

$\displaystyle\
\int\dfrac{f_{0}(z)-f_{0}(\lambda)}{z-\lambda}g_{j}(z)d\mu(z)=-\int\mathcal{C}(g_{j}\mu)(z)\dfrac{d\eta_{0}(z)}{z-\lambda}.$
(5.7)

Using [12] (or [15, Theorem 8.1]), we see that

$\ f_{0}(z)=\mathcal{C}(\eta_{0})(z),~{}\mu|_{\mathcal{N}(h)}-a.a..$

Case I: $E_{2}\subset E_{1}\cap F_{0}(\\{g_{j}\mu\\}),$ then from (5.4) and
(5.7), we see that, since $\\{g_{j}\\}$ is dense in $R^{t}(K,\mu)^{\perp},$
$f_{0}\in R^{t,\infty}(K,\mu)$ and
$\frac{f_{0}(z)-f_{0}(\lambda)}{z-\lambda}\in R^{t,\infty}(K,\mu),$ which
implies $\rho(f_{0})=f_{0}|_{\Omega}\in H^{\infty}(\Omega,E_{0}).$ Set

$\ f=\dfrac{f_{0}}{\|\eta_{0}\|}\gamma(E_{1})\text{ and
}\eta=\dfrac{\eta_{0}}{\|\eta_{0}\|}\gamma(E_{1}).$

Case II: $E_{2}\subset E_{1}\cap F_{+}(\\{g_{j}\mu\\}),$ then from (5.4) , we
get, since $v^{+}(g_{j}\mu)(z)=0,~{}z\in E_{2},~{}l_{\partial_{i}K}-a.a.,$

$\ \int f_{0}(z)g_{j}(z)d\mu(z)=-\int\mathcal{C}(g_{j}\mu)(z)d\eta_{0}(z)=\pi
i\int h(z)g_{j}(z)d\eta_{0}(z).$

The function $f_{0}-\pi i\frac{d\eta_{0}}{dl_{\partial_{i}K}}$ is the non-
tangential limit of $f_{0}$ from the bottom on $E_{2}$ (Lemma 2.10). Thus,
$f_{0}-\pi i\frac{d\eta_{0}}{dl_{\partial_{i}K}}\in L^{\infty}(\mu).$ Hence,

$\ \int(f_{0}(z)-\pi
i\dfrac{d\eta_{0}}{dl_{\partial_{i}K}}(z))g_{j}(z)d\mu(z)=0,$

which implies $f_{0}-\pi i\frac{d\eta_{0}}{dl_{\partial_{i}K}}\in
R^{t,\infty}(K,\mu).$

Using a similar argument with (5.7) instead of (5.4), we see that

$\ \int\dfrac{f_{0}(z)-\pi
i\frac{d\eta_{0}}{dl_{\partial_{i}K}}-f_{0}(\lambda)}{z-\lambda}g_{j}(z)d\mu(z)=0.$

Therefore, $\rho(f_{0}-\pi
i\frac{d\eta_{0}}{dl_{\partial_{i}K}})=f_{0}|_{\Omega}\in
H^{\infty}(\Omega,E_{0}).$

$\ f=\dfrac{f_{0}-\pi
i\frac{d\eta_{0}}{dl_{\partial_{i}K}}}{\|\eta_{0}\|}\gamma(E_{1})\text{ and
}\eta=\dfrac{\eta_{0}}{\|\eta_{0}\|}\gamma(E_{1}).$

Case III: Similarly to Case II, we set

$\ f=\dfrac{f_{0}+\pi
i\frac{d\eta_{0}}{dl_{\partial_{i}K}}}{\|\eta_{0}\|}\gamma(E_{1})\text{ and
}\eta=\dfrac{\eta_{0}}{\|\eta_{0}\|}\gamma(E_{1}).$

It is easy to verify that $f$ and $\eta$ satisfy the properties of the lemma.
∎

Define

$\displaystyle\
\mathcal{F}_{1}=F(\\{g_{j}\mu\\})\cup\mathbb{T}\cup\bigcup_{n=1}^{\infty}\partial\mathbb{D}(\lambda_{n},r_{n}).$
(5.8)

Clearly, $\mathcal{F}_{1}=\partial\Omega\setminus E_{0},~{}\gamma-a.a..$The
set $\mathcal{F}:=\ \mathcal{F}_{1}\cup K^{c}$ is called the non-removable
boundary for $R^{t}(K,\mu).$

In the remaining section, we prove the following integral estimate.

######  Lemma 5.2.

Let $S$ be a closed square whose sides are parallel to $x$-axis or $y$-axis
and let the length of $S$ is $\delta.$ Let $\varphi$ be a smooth function with
support in $S,$ $\|\varphi\|_{\infty}\leq 1,$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{4}}{\delta}.$
Then for $f\in H^{\infty}(\Omega,E_{0})$ with $\|f\|_{\Omega}\leq 1$ and
$f(z)=0,~{}z\in\mathbb{C}\setminus K,$ and $n\geq 1,$

$\displaystyle\begin{aligned} \ &\left|\int
f(z)(z-\lambda)^{n-1}\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|\\\
\
\leq&C_{4}\delta^{n-1}l_{\partial\Omega}(S\cap\mathcal{F}_{1}).\end{aligned}$
(5.9)

###### Proof.

For $f\in H^{\infty}(\Omega,E_{0})$ with $f(z)=0,~{}z\in\mathbb{C}\setminus K$
and $\|f\|_{\Omega}\leq 1,$ applying Lemma 2.9 for $\Gamma=\partial\Omega,$ we
find a function $b(z)\in L^{\infty}(l_{\partial\Omega})$ with $\|b\|\leq 1$
such that $f(z)={\mathcal{C}}(bl_{\partial\Omega})(z)$ for
$z\in\mathbb{C}\setminus\partial\Omega.$ From Lemma 4.5, we have

$\
f_{+}(z)=v^{+}(bl_{\partial\Omega})(z)={\mathcal{C}}(bl_{\partial\Omega})(z)+\pi
ib(z),~{}l_{\partial_{i}K}-a.a.$

and

$\
f_{-}(z)=v^{-}(bl_{\partial\Omega})(z)={\mathcal{C}}(bl_{\partial\Omega})(z)-\pi
ib(z),~{}l_{\partial_{i}K}-a.a..$

Therefore, $b(z)=0,~{}l_{E_{0}}-a.a.$ since
$f_{+}(z)=f_{-}(z),~{}l_{E_{0}}-a.a..$ For a smooth function $\varphi$ with
support in $S,$ $\|\varphi\|_{\infty}\leq 1,$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{4}}{\delta}.$
we get

$\displaystyle\left|\int
f(z)(z-\lambda)^{n-1}\dfrac{\partial\varphi(z)}{\partial\bar{z}}d{\mathfrak{m}}(z)\right|$
$\displaystyle\ \leq$ $\displaystyle\pi\left|\int_{\partial\Omega\setminus
E_{0}}(z-\lambda)^{n-1}b(z)\varphi(z)dl_{\partial\Omega}(z)\right|$
$\displaystyle\ \leq$ $\displaystyle
C_{5}\delta^{n-1}l_{\partial\Omega}(S\cap(\partial\Omega\setminus E_{0})).$

This completes the proof. ∎

##  6\. Proofs of Main Theorem (2) and (3)

We continue to use the assumptions and notation of sections 5 and 6. If a
compact subset $E$ is contained in the disk $\mathbb{D}(a,\delta)$ and $f$ is
bounded analytic on $\mathbb{C}_{\infty}\setminus E$ with $f(\infty)=0.$ We
consider the Laurent expansion of $f$ centered at $a$ for
$z\in\mathbb{C}\setminus\mathbb{D}(a,\delta),$

$\displaystyle\ f(z)=\sum_{n=1}^{\infty}\dfrac{c_{n}(f,a)}{(z-a)^{n}}.$ (6.1)

As $c_{1}(f,a)$ does not depend on the choice of $a$, we define
$c_{1}(f)=c_{1}(f,a).$ The coefficient $c_{2}(f,a)=c_{2}(f,a)$ does depend on
$a$.

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 $T_{\varphi}f.$

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

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

######  Lemma 6.1.

Let $S$ be a closed square whose sides are parallel to $x$-axis or $y$-axis
such that $S\subset\mathbb{D}$ and $S\cap\mathbb{R}\neq\emptyset$ and let
$\delta$ and $c_{S}$ be the length and the center of $S,$ respectively. Let
$\varphi$ be a smooth function with support in $S,$ $\|\varphi\|_{\infty}\leq
1,$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{6}}{\delta}.$
Let $f\in H^{\infty}(\Omega,E_{0})$ with $\|f\|_{\Omega}\leq 1$ and
$f(z)=0,~{}z\in\mathbb{C}\setminus K.$ Then there exists $f_{S}\in
R^{t,\infty}(K,\mu)$ that is bounded analytic on $\mathbb{C}_{\infty}\setminus
S$ satisfying the following properties hold

(1) $\|f_{S}\|_{\Omega}\leq C_{7},$

(2) $\|f_{S}\|_{L^{\infty}(\mu)}\leq C_{7}$ and $f_{S}(\infty)=0,$

(3) $c_{1}(f_{S})=c_{1}(T_{\varphi}f),$ and

(4) $\rho(f_{S})=f_{S}|_{\Omega}\in H^{\infty}(\Omega,E_{0}).$

###### Proof.

Let

$\ D_{S}=\bigcup_{\overline{\mathbb{D}(\lambda_{n},r_{n})}\subset
S}\mathbb{D}(\lambda_{n},r_{n}),~{}T_{S}=\bigcup_{\overline{\mathbb{D}(\lambda_{n},r_{n})}\subset
S}\partial{\mathbb{D}(\lambda_{n},r_{n})}.$

There are at most two disks $D_{l}=\mathbb{D}(\lambda_{n_{l}},r_{n_{l}})$ and
$D_{r}=\mathbb{D}(\lambda_{n_{r}},r_{n_{r}})$ with
$\lambda_{n_{l}}<\lambda_{n_{r}}$ such that $\overline{D_{l}}\cap(S\setminus
D_{S})\neq\emptyset$ and $\overline{D_{r}}\cap(S\setminus
D_{S})\neq\emptyset.$ Then we have

$\displaystyle l_{\partial\Omega}(S\cap\mathcal{F}_{1})\leq
l_{\partial\Omega}(S\cap F(\\{g_{j}\mu\\}))$
$\displaystyle+l_{\partial\Omega}(S\cap
T_{S})+l_{\partial\Omega}(S\cap\partial
D_{l})+l_{\partial\Omega}(S\cap\partial D_{r}).$

Case A: Assume that $l_{\partial\Omega}(S\cap
F(\\{g_{j}\mu\\}))\geq\frac{1}{4}l_{\partial\Omega}(S\cap\mathcal{F}_{1}).$
Let $f_{0}$ be the function in Lemma 5.1 for $E_{1}=S\cap F(\\{g_{j}\mu\\}).$
Set $f_{S}=\frac{f_{0}}{c_{1}(f_{0})}c_{1}(T_{\varphi}f).$ (1)-(4) follows
from Lemma 5.1 and Lemma 5.2.

Case B: Assume that $l_{\partial\Omega}(S\cap
F(\\{g_{j}\mu\\}))<\frac{1}{4}l_{\partial\Omega}(S\cap\mathcal{F}_{1}).$ Let
$E_{2}$ be one of $S\cap T_{S},$ $S\cap\partial D_{l},$ and $S\cap\partial
D_{r}$ with the largest $l_{\partial\Omega}$ measure. Then
$l_{\partial\Omega}(E_{2})\geq\frac{1}{4}l_{\partial\Omega}(S\cap\mathcal{F}_{1}).$
We can find a compact subset of a line segment

$\ E_{3}\subset S\cap(D_{S}\cup D_{l}\cup D_{r})$

such that $l_{\partial\Omega}(E_{3})\geq\frac{1}{2}l_{\partial\Omega}(E_{2}).$
Let $f_{0}$ be the admissible function for $E_{3},$ that is, $f_{0}$ is
analytic on $\mathbb{C}_{\infty}\setminus E_{3},$ $\|f_{0}\|=1,$ and
$f_{0}^{\prime}(\infty)=\gamma(E_{3}).$ Using [8, VIII.2.2] and Lemma 5.2, we
conclude that the function
$f_{S}=\frac{f_{0}}{f_{0}^{\prime}(\infty)}c_{1}(T_{\varphi}f)$ satisfies
(1)-(4). ∎

######  Lemma 6.2.

Let $S$ be a closed square whose sides are parallel to $x$-axis or $y$-axis
and let $\delta$ and $c_{S}$ be the length and the center of $S,$
respectively. Suppose that (a) $S\setminus\mathbb{D}\neq\emptyset$ or (b)
$dist(c_{S},\mathbb{R})\geq\delta$ and $S\cap\partial K\neq\emptyset.$ Let
$\varphi$ be a smooth function with support in $S,$ $\|\varphi\|_{\infty}\leq
1,$ and
$\left\|\frac{\partial\varphi(z)}{\partial\bar{z}}\right\|_{\infty}\leq\frac{C_{8}}{\delta}.$
Let $f\in H^{\infty}(\Omega,E_{0})$ with $\|f\|_{\Omega}\leq 1$ and
$f(z)=0,~{}z\in\mathbb{C}\setminus K.$ Then there exists $f_{S}\in
R^{t,\infty}(K,\mu)$ that is bounded analytic on
$\mathbb{C}_{\infty}\setminus(\frac{3}{2}S)$ satisfying the following
properties hold

(1) $\|f_{S}\|_{\Omega}\leq C_{9},$

(2) $\|f_{S}\|_{L^{\infty}(\mu)}\leq C_{9}$ and $f_{S}(\infty)=0,$

(3) $c_{1}(f_{S})=c_{1}(T_{\varphi}f),$

(4) $c_{2}(f_{S},c_{S})=c_{2}(T_{\varphi}f,c_{S}),$ and

(5) $\rho(f_{S})=f_{S}|_{\Omega}\in H^{\infty}(\Omega,E_{0}).$

###### Proof.

If (a) holds, then there exists a line segment $L$ with length greater than
$\frac{1}{4}\delta$ such that
$L\subset\frac{3}{2}S\setminus\overline{\mathbb{D}}.$ If $S\subset\mathbb{D}$
and (b) holds, then there exists $\mathbb{D}(\lambda_{n_{0}},r_{n_{0}})$ such
that $S\cap\partial\mathbb{D}(\lambda_{n_{0}},r_{n_{0}})\neq\emptyset.$ Hence,
$r_{n_{0}}\geq\frac{1}{2}\delta$ and there exists a line segment $L$ with
length greater than $\frac{1}{4}\delta$ such that
$L\subset\frac{3}{2}S\cap\mathbb{D}(\lambda_{n_{0}},r_{n_{0}}).$

There exists a function $f_{1}\in C(\mathbb{C}_{\infty})$ such that
$\|f_{1}\|\leq 1,$ $f_{1}$ is analytic on $\mathbb{C}_{\infty}\setminus L,$
$f_{1}(\infty)=0,$ and $c_{1}(f_{1})=\frac{\delta}{16}.$ Set
$f_{2}=\frac{\delta f_{1}}{16c_{1}(f_{1})}.$ Then $\|f_{2}\|\leq 1,$
$c_{1}(f_{2}^{2})=0,$ and $c_{2}(f_{2}^{2},c_{S})=\frac{\delta^{2}}{256}.$ By
[8, Theorem 2.5 on page 201], we get $|c_{2}(f_{2},c_{S})|\leq 3\delta^{2}.$
Let

$\ f_{3}=f_{2}-\dfrac{256c_{2}(f_{2},c_{S})}{\delta^{2}}f_{2}^{2}.$

Then $\|f_{3}\|\leq C_{10},$ $c_{1}(f_{3})=\frac{\delta}{16},$ and
$c_{2}(f_{3},c_{S})=0,$ Set

$\
f_{S}=\dfrac{16c_{1}(T_{\varphi}f)}{\delta}f_{3}+\dfrac{256c_{2}(T_{\varphi}f,c_{S})}{\delta^{2}}f_{2}^{2}.$

The proof now follows from Lemma 5.2. ∎

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) $\varphi_{ij}$ is a $C^{\infty}$ smooth function supported in $2S_{ij},$
and with values in $[0,1]$;

(iii)

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

Given a square grid $\\{S_{ij}\\}$, we shall let $c_{ij}$ denote the center of
$S_{ij}$ (See [8, VIII.7] for details).

###### Proof.

(Main Theorem (2)) Let $\\{\varphi_{ij},S_{ij},\delta\\}$ be a smooth
partition of unity. Let $f\in H^{\infty}(\Omega,E_{0}),$ $\|f\|\leq 1,$ and
set $f(z)=0$ for $z\in\mathbb{C}\setminus\Omega.$ Then

$\displaystyle\
f=\sum_{2S_{ij}\cap\partial\Omega\neq\emptyset}(f_{ij}:=T_{\varphi_{ij}}f).$
(6.2)

Let $J$ be the set of $(i,j)$ satisfying $i\in\\{-2,-1,0,1\\}$ and
$2S_{ij}\subset\mathbb{D}.$ We apply Lemma 6.1 when $(i,j)\in J$ and Lemma 6.2
when $(i,j)\notin J$ for $S=2S_{ij}$ and $\varphi=\varphi_{ij}$ to get
$f_{S_{ij}}.$ We rewrite (6.2) as the following

$\ f=\sum_{(i,j)\in
J,~{}2S_{ij}\cap\partial\Omega\neq\emptyset}(f_{ij}-f_{S_{ij}})+\sum_{(i,j)\notin
J,~{}2S_{ij}\cap\partial\Omega\neq\emptyset}(f_{ij}-f_{S_{ij}})+f_{\delta}$

where

$\displaystyle\
f_{\delta}=\sum_{2S_{ij}\cap\partial\Omega\neq\emptyset}f_{S_{ij}}\in
R^{t,\infty}(K,\mu).$ (6.3)

Clearly, $\rho(f_{\delta})=f_{\delta}|_{\Omega}\in H^{\infty}(\Omega,E_{0}).$
Using Vitushkin approximation scheme (see [8, VIII.7]), we get

$\displaystyle|f(z)-f_{\delta}(z)|$ $\displaystyle\ \leq$
$\displaystyle\sum_{(i,j)\in
J,~{}2S_{ij}\cap\partial\Omega\neq\emptyset}|f_{ij}(z)-f_{S_{ij}}(z)|+\sum_{(i,j)\notin
J,~{}2S_{ij}\cap\partial\Omega\neq\emptyset}|f_{ij}(z)-f_{S_{ij}}(z)|$
$\displaystyle\ \leq$ $\displaystyle C_{11}\sum_{\underset{-2\leq i\leq
1}{2S_{ij}\cap\partial\Omega\neq\emptyset}}\min\left(1,\dfrac{\delta^{2}}{dist(z,3S_{ij})^{2}}\right)+C_{9}\sum_{2S_{ij}\cap\partial\Omega\neq\emptyset}\min\left(1,\dfrac{\delta^{3}}{dist(z,3S_{ij})^{3}}\right)$
$\displaystyle\ \leq$ $\displaystyle
C_{12}\min\left(1,\dfrac{\delta}{dist(z,\partial\Omega)}\right).$

Therefore, $f_{\delta}(z)\rightarrow f(z)$ uniformly on compact subsets of
$\mathbb{C}\setminus\partial\Omega$ and $\|f_{\delta}\|_{L^{\infty}(\mu)}\leq
C_{13}.$ There exists a subsequence $\\{f_{\delta_{n}}\\}$ such that
$f_{\delta_{n}}(z)\rightarrow\tilde{f}(z)$ in $L^{\infty}(\mu)$ weak-star
topology. Hence, $\tilde{f}\in R^{t,\infty}(K,\mu)$ and
$\mathcal{C}(f_{\delta_{n}}g\mu)(z)\rightarrow\mathcal{C}(\tilde{f}g\mu)(z),~{}{\mathfrak{m}}-a.a..$
Thus,
$f(z)\mathcal{C}(g\mu)(z)=\mathcal{C}(\tilde{f}g\mu)(z),~{}{\mathfrak{m}}-a.a.$
for $g\perp R^{t}(K,\mu),$ which implies $\rho(\tilde{f})=f.$ ∎

We are now ready to prove Main Theorem (3).

###### Proof.

(Main Theorem (3)) From Main Theorem (1), we see that $\rho(f)\in
H^{\infty}(\Omega,E_{0})$ for $f\in R^{t,\infty}(K,\mu).$ Clearly,

$\ \rho(f_{1}f_{2})=\rho(f_{1})\rho(f_{2}),~{}f_{1},f_{2}\in
R^{t,\infty}(K,\mu).$

Therefore, for $f\in R^{t,\infty}(K,\mu),$ since $\lambda\in\Omega$ is an
analytic bounded point evaluation, we see that there exists a constant
$C_{\lambda}>0$ such that

$\ |\rho(f)(\lambda)|\leq
C_{\lambda}^{\frac{1}{n}}\|f^{n}\|_{L^{t}(\mu)}^{\frac{1}{n}}.$

Taking $n\rightarrow\infty,$ we get
$\|\rho(f)\|_{\Omega}\leq\|f\|_{L^{\infty}(\mu)}.$

From the last paragraph of the proof of Main Theorem (2), we get
$\|f^{n}\|_{L^{\infty}(\mu)}^{\frac{1}{n}}\leq
C_{13}^{\frac{1}{n}}\|\rho(f^{n})\|_{\Omega}^{\frac{1}{n}}.$ Again letting
$n\rightarrow\infty,$ we get
$\|f\|_{L^{\infty}(\mu)}\leq\|\rho(f)\|_{\Omega}.$ Thus,

$\ \|\rho(f)\|_{\Omega}=\|f\|_{L^{\infty}(\mu)},~{}f\in\in
R^{t,\infty}(K,\mu).$

On the other hand, $\rho$ is surjective by Main Theorem (2). Therefore, $\rho$
is a bijective isomorphism between two Banach algebras $R^{t,\infty}(K,\mu)$
and $H^{\infty}(\Omega,E_{0})$. Clearly $\rho$ is also weak-star sequentially
continuous, so an application of the Krein-Smulian Theorem shows that $\rho$
is a weak-star homeomorphism. ∎

Acknowledgments. The authors would like to thank Professor John McCarthy for
carefully reading through the manuscript and providing many useful comments.
They would also like to thank Professor John Akeroyd for providing useful
discussions.

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