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# 数学代写|数学分析作业代写Mathematical Analysis代考|MATH1013 Abstract Integration

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## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Abstract Integration

In this section, we examine Lebesgue’s revolutionary approach to the definition of the integral. The motivation below is imprecise and does not rigorously develop any particular set of ideas. For the sake of simplicity, we assume that $f$ is a positive continuous function on a compact interval.

The Riemann integral is based on the simple geometrical idea of dividing the region below the graph of $f$ into thin vertical strips, where the area below the graph is approximated by the integral of a step function (the Riemann sum). Lebesgue’s idea was to divide the range of $f$ by points $y_0, \ldots, y_n$, and, for $k=1, \ldots, n$, we consider the sets $E_k=f^{-1}\left(\left[y_k, y_{k+1}\right)\right)$. Even for an uncomplicated function, the set $E_k$ may come in several fragments, as shown in figure 8.1, where $E_k$ has three fragments. When $y_{k+1}-y_k$ is small, the approximate combined area of the three shaded strips is the approximate common height, $y_k$, times the sum of the lengths of the three fragments that comprise the set $E_k$ or, more precisely, the measure of $E_k$. Thus the approximate area below the graph is $\sum_{k=1}^n y_k \mu\left(E_k\right)$, which, by definition, is the integral of a simple function. Needless to say, as the partition of the range of $f$ gets finer, we expect the integrals of the simple functions to converge to the integral of $f$. This is the overarching idea in Lebesgue integration. As it turns out, we can integrate far more functions under the Lebesgue definition than under the Riemann definition. For example, the integral of any positive measurable function is defined, although it may not be finite. Additionally, the definition of the integral extends seamlessly to abstract measure spaces. The section results capture the above ideas. First we define the integral of a positive measurable function $f$, then we show that $f$ is the limit of simple functions, $s_n$, and then we show that $\int_X f d \mu=\lim _n \int_X s_n d \mu$. Extending the definition of the integral to complex functions follows without difficulty. The section concludes with three important convergence theorems.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Convergence Theorems

Theorem 8.3.8 (Fatou’s theorem). Let $f_n: X \rightarrow[0, \infty]$ be a sequence of measurable functions. Then
$$\int_X \liminf n f_n d \mu \leq \liminf _n \int_X f_n d \mu .$$ Proof. Let $g_n=\inf {k \geq n} f_k$. Then $0 \leq g_1 \leq g_2 \leq \ldots$, and let $f(x)=\lim _n g_n(x)$. Note that $f(x)=\liminf _n f_n(x)$. If $s$ is a simple function such that $0 \leq s \leq f$, then, by lemma 8.3.3, $\int_X s d \mu \leq \lim _n \int_X g_n d \mu$. Hence $\int_X f d \mu=\sup \left{\int_X s d \mu: s \leq f\right} \leq \lim _n \int_X g_n d \mu$. Since $g_n \leq f_n, \int_X g_n d \mu \leq \int_X f_n d \mu$, and $\lim _n \int_X g_n d \mu \leq \liminf _n \int_X f_n d \mu$.

Example 5. Let $\left(f_n\right)$ be a convergent sequence in $\mathfrak{Q}^1(\mu)$, and let $f$ be its $\mathfrak{Q}^1$-limit. Then $\left(f_n\right)$ contains a subsequence that converges to $f$ for almost every $x \in X$.
Choose a subsequence $\left(f_{n_i}\right)$ of $\left(f_n\right)$ such that, for $i \in \mathbb{N},\left|f_{n_i}-f\right|_1<2^{-i}$. Let $g_k=\sum_{i=1}^k\left|f_{n_i}-f\right|$. The functions $g_k$ are in $\mathfrak{Q}^1$ and, by construction, $0 \leq g_1 \leq$ $g_2 \leq \ldots$, and $\left|g_k\right|_1 \leq 1$. Let $g(x)=\lim k g_k(x)$. By Fatou’s theorem, $\int_X g d \mu \leq$ $\liminf _n\left|g_k\right|_1 \leq 1$. This shows that $g \in \mathfrak{Q}^{1.5}$. Since $g(x)=\sum{i=1}^{\infty}\left|f_{n_i}(x)-f(x)\right|$, it follows that the series $\sum_{i=1}^{\infty}\left|f_{n_i}(x)-f(x)\right|$ is convergent for a.e. $x \in X$ (by example 1). In particular, $\lim {i \rightarrow \infty}\left|f{n_i}(x)-f(x)\right|=0$ for a.e. $x \in X$.

Theorem 8.3.9 (the monotone convergence theorem). If $f_n: X \rightarrow[0, \infty]$ is an increasing sequence of measurable functions such that $f(x)=\lim _n f_n(x)$ exists for every $x \in X$, then
$$\int_X f d \mu=\lim _n \int_X f_n d \mu .$$

Proof. Since $f_n$ is increasing, $\liminf _n f_n=f$, and since $\int_X f_n d \mu$ is increasing, $\liminf f_n \int_X f_n=\lim _n \int_X f_n d \mu$. By Fatou’s theorem, $\int_X f d \mu \leq \lim _n \int_X f_n d \mu$. Since $f_n \leq f, \int_X f_n d \mu \leq \int_X f d \mu$, and $\lim _n \int_X f_n d \mu \leq \int_X f d \mu$, and the proof is complete.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Abstract Integration

$\sum_{k=1}^n y_k \mu\left(E_k\right)$ ，根据定义，它是简单函数的积分。不用说，作为范围的划分 $f$ 变得更好，我们期望简单函数的积分收敛到积分 $f$. 这是 Lebesgue 整合的首要思想。事实证明，我们可以在勒贝格定义下集成比在黎自定义下更多的函数。例呰，定义了任何正 义一个正可测函数的积分 $f$ ，然后我们证明 $f$ 是简单函数的极限， $s_n$, 然后我们证明 $\int_X f d \mu=\lim n \int_X s_n d \mu$. 将积分的定义扩展 到复函数没有困难。本节以三个重要的收敛定理结束。

## 数学代写数学分析代写MATHEMATICAL ANALYSIS代考|Convergence Theorems

$$\int_X f d \mu=\lim _n \int_X f_n d \mu .$$
$\int_X f d \mu \leq \lim _n \int_X f_n d \mu$. 自从 $f_n \leq f, \int_X f_n d \mu \leq \int_X f d \mu ，$ 和 $\lim _n \int_X f_n d \mu \leq \int_X f d \mu$, 证明完成。

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