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# 数学代写|实分析代写Real Analysis代考|MTH408 Proof of Lebesgue’s Theorem

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## 数学代写|实分析代写Real Analysis代考|Proof of Lebesgue’s Theorem

In this final section, we present a self-contained proof of Lebesgue’s characterization of the Riemann integrable functions on $[a, b]$. Recall that a subset $E$ of $\mathbb{R}$ has measure zero if for any $\epsilon>0$, there exists a finite or countable collection $\left{I_n\right}$ of open intervals with $E \subset \bigcup_n I_n$ and $\sum_n \ell\left(I_n\right)<\epsilon$, where $\ell\left(I_n\right)$ denotes the length of the interval $I_n$. We begin with several preparatory lemmas.

LEMMA 6.7.1 A finite or countable union of sets of measure zero has measure zero.

Proof. We will prove the lemma for the case of a countable sets of measure zero. The result for a finite union is an immediate consequence.

Suppose $\left{E_n\right}_{n \in \mathbb{N}}$ is a countable collection of sets of measure zero. Set $E=\bigcup_n E_n$ and let $\epsilon>0$ be given. Since each set $E_n$ is a set of measure zero, for each $n \in \mathbb{N}$ there exists a finite or countable collection $\left{I_{n, k}\right}_k$ of open intervals such that $E_n \subset \bigcup_k I_{n, k}$ and $\sum_k \ell\left(I_{n, k}\right)<\epsilon / 2^k$. Since we can always take $I_{n, k}$ to be the empty set, there is no loss of generality in assuming that the collection $\left{I_{n, k}\right}_k$ is countable. Then $\left{I_{n, k}\right}_{n, k}$ is again a countable collection of open intervals with $E \subset \bigcup_{n, k} I_{n, k}$.

Since $\mathbb{N} \times \mathbb{N}$ is countable, there exists a one-to-one function $f$ from $\mathbb{N}$ onto $\mathbb{N} \times \mathbb{N}$. For each $m \in \mathbb{N}$, set $J_m=I_{f(m}$. Then $\left{J_m\right}_{m \in \mathbb{N}}$ is a countable collection of open intervals with $E \subset \bigcup_m J_m$. Since $f$ is one-to-one, for each $N \in \mathbb{N}$, the set $F_N=f({1, \ldots, N})$ is a finite subset of $\mathbb{N} \times \mathbb{N}$. Hence there exists positive integers $N_1$ and $K_1$ such that for all $(n, k) \in F_N$ we have $1 \leq n \leq N_1$ and $1 \leq k \leq K_1$. Hence
$$\sum_{m=1}^N \ell\left(J_m\right)=\sum_{(n, k) \in F_N} \ell\left(I_{n, k}\right) \leq \sum_{(n, k) \in N_1 \times K_1} \ell\left(I_{n, k}\right)$$
But
$$\sum_{(n, k) \in N_1 \times K_1} \ell\left(I_{n, k}\right)=\sum_{n=1}^{N_1} \sum_{k=1}^{K_1} \ell\left(I_{n, k}\right) \leq \sum_{n=1}^{N_1} \sum_{k=1}^{\infty} \ell\left(I_{n, k}\right)<\sum_{n=1}^{N_1} \frac{\epsilon}{2^n}<\epsilon$$
Thus $\sum_{m=1}^{\infty} \ell\left(J_m\right)<\epsilon$. Therefore $E$ has measure zero.

## 数学代写|实分析代写Real Analysis代考|Convergence Tests

In Section 3.7, we provided a very brief introduction to the subject of infinite series. In the study of infinite series, it is very useful to have tests available by means of which one is able to determine whether a given series converges or diverges. For example, Corollary $3.7 .5$ is very useful in determining divergence of a series. If the sequence $\left{a_k\right}$ does not converge to zero, then the series $\sum a_k$ diverges. On the other hand however, if $\lim a_k=0$, then nothing can be ascertained concerning convergence or divergence of the series $\sum a_k$. In this section, we will state and prove several useful results that can be used to establish convergence or divergence of a given series. Additional tests for convergence will also be given in the exercises and subsequent sections. With the exception of Theorem 7.1.1, all of our results in this section will be stated for series of nonnegative terms.

As in Definition 3.7.1, given an infinite series $\sum_{k=1}^{\infty} a_k$ of real numbers, $\left{s_n\right}_{n=1}^{\infty}$ will denote the associated sequence of partial sums defined by
$$s_n=\sum_{k=1}^n a_k$$
The series $\sum a_k$ converges if and only if the sequence $\left{s_n\right}$ of $n$th partial sums converges. Furthermore, if $\lim {n \rightarrow \infty} s_n=s,(s \in \mathbb{R})$, then $s$ is called the sum of the series, and we write $$\sum{k=1}^{\infty} a_k=s$$

## 数学代写|实分析代写Real Analysis代考|Proof of Lebesgue’s Theorem

$E_n$ 是一组测量零，对于每个 $n \in \mathbb{N}$ 存在一个有限的或可数的集合 $l$ left 的分隔符缺失或无法识别 的开区间使得
$\begin{array}{ll}E_n \subset \bigcup_k I_{n, k} \text { 和 } \sum_k \ell\left(I_{n, k}\right)<\epsilon / 2^k \text {. 因为我们总能拿 } I_{n, k} \text { 作为空集，假设集合不失一般性 } \ \text { \left 的分隔符缺失或无法识别 } & \text { 是可数的。然后 left 的分隔符缺失或无法识别 }\end{array}$

\left 的分隔符缺失或无法识别 是开区间的可数集合 $E \subset \bigcup_m J_m$. 自从 $f$ 是一对一的，对于每个 $N \in \mathbb{N}$, 集合
$F_N=f(1, \ldots, N)$ 是的有限子集 $\mathbb{N} \times \mathbb{N}$. 因此存在正整数 $N_1$ 和 $K_1$ 这样对于所有人 $(n, k) \in F_N$ 我们有 $1 \leq n \leq N_1$ 和 $1 \leq k \leq K_1$. 因此
$$\sum_{m=1}^N \ell\left(J_m\right)=\sum_{(n, k) \in F_N} \ell\left(I_{n, k}\right) \leq \sum_{(n, k) \in N_1 \times K_1} \ell\left(I_{n, k}\right)$$

$$\sum_{(n, k) \in N_1 \times K_1} \ell\left(I_{n, k}\right)=\sum_{n=1}^{N_1} \sum_{k=1}^{K_1} \ell\left(I_{n, k}\right) \leq \sum_{n=1}^{N_1} \sum_{k=1}^{\infty} \ell\left(I_{n, k}\right)<\sum_{n=1}^{N_1} \frac{\epsilon}{2^n}<\epsilon$$

## 数学代写|实分析代写Real Analysis代考|Convergence Tests

\left 的分隔符缺失或无法识别 不收敛到零，则级数 $\sum a_k$ 分歧。然而另一方面，如果 $\lim a_k=0$ ，那么就无

$$s_n=\sum_{k=1}^n a_k$$

$\lim n \rightarrow \infty s_n=s,(s \in \mathbb{R})$ ，然后 $s$ 称为级数之和，我们写
$$\sum k=1^{\infty} a_k=s$$

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