## 数学代写|实分析作业代写Real analysis代考|MATH2023

2023年2月7日

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## 数学代写|实分析作业代写Real analysis代考|Lebesgue Measure and Lebesgue–Stieltjes Measures

We now return to Lebesgue measure considered in Examples 2.3.10 and 2.4.8. Let $I$ be a cube in $\mathbb{R}^n$ of the form $[a, b]^n$ and let $\mathcal{A}_0$ be the algebra of finite unions of parallelepipeds in $I$ with edges parallel to the coordinate axes. We know that the usual volume $\lambda_n$ is countably additive on $\mathcal{A}_0$ and extends to a countably additive measure, also denoted by $\lambda_n$ and called Lebesgue measure on the $\sigma$-algebra $\mathcal{L}_n(I)$ of all $\lambda_n$-measurable sets in $I$, containing the Borel $\sigma$-algebra. Let us write $\mathbb{R}^n$ as the union of increasing cubes $I_k=\left{\left|x_i\right| \leqslant k, i=1, \ldots, n\right}$ and set
$$\mathcal{L}n:=\left{E \subset \mathbb{R}^n: E \cap I_k \in \mathcal{L}_n\left(I_k\right) \text { for all } k \in \mathbb{N}\right} .$$ It is clear that $\mathcal{L}_n$ is a $\sigma$-algebra. The formula $$\lambda_n(E)=\lim {k \rightarrow \infty} \lambda_n\left(E \cap I_k\right)$$
defines on it the $\sigma$-finite measure $\lambda_n$.
2.5.1. Definition. The measure $\lambda_n$ introduced above on $\mathcal{L}_n$ is called Lebesgue measure on $\mathbb{R}^n$. Sets in $\mathcal{L}_n$ are called Lebesgue measurable.

In those cases where a subset of $\mathbb{R}^n$ is considered with Lebesgue measure, it is customary to use the terms “a set of measure zero”, “a measurable set”, etc., without explicit mentioning Lebesgue measure. We shall also follow this tradition. For defining Lebesgue measure of a set $E \in \mathcal{L}n$ one can also use the formula $$\lambda_n(E)=\sum{j=1}^{\infty} \lambda_n\left(E \cap Q_j\right),$$
where $Q_j$ are pairwise disjoint cubes that are shifts of $[0,1)^n$ and give all of $\mathbb{R}^n$ in their union. Since the $\sigma$-algebra generated by the parallelepipeds of the form indicated above is the Borel $\sigma$-algebra $\mathcal{B}(I)$ of the cube $I$, all Borel sets in the cube $I$, hence also in all of $\mathbb{R}^n$, are Lebesgue measurable.

## 数学代写|实分析作业代写Real analysis代考|Signed Measures

In this section we consider signed measures. The next theorem enables us in many cases to reduce signed measures to nonnegative measures.
2.6.1. Theorem. (THE HAHN DECOMPOSITION) Let $\mu$ be a countably additive real measure on a measurable space $(X, \mathcal{A})$. Then there exists a set $X^{-} \in \mathcal{A}$ such that, letting $X^{+}=X \backslash X^{-}$, for all sets $A \in \mathcal{A}$, we have
$$\mu\left(A \cap X^{-}\right) \leqslant 0 \text { and } \mu\left(A \cap X^{+}\right) \geqslant 0 .$$
Proof. A set $E \in \mathcal{A}$ will be called negative if $\mu(A \cap E) \leqslant 0$ for all $A \in \mathcal{A}$. Similarly we define positive sets. Let $\alpha=\inf \mu(E)$, where the infimum is taken over all negative sets. Let $\left{E_n\right}$ be a sequence of negative sets for which $\lim {n \rightarrow \infty} \mu\left(E_n\right)=\alpha$. It is clear that $X^{-}=\bigcup{n=1}^{\infty} E_n$ is a negative set and $\mu\left(X^{-}\right)=\alpha$, since $\alpha \leqslant \mu\left(X^{-}\right) \leqslant \mu\left(E_n\right)$. We show that $X^{+}=X \backslash X^{-}$is a positive set. Suppose the contrary. Then there exists a set $A_0 \in \mathcal{A}$ such that $A_0 \subset X^{+}$and $\mu\left(A_0\right)<0$. The set $A_0$ cannot be negative, since in that case we would take the negative set $X^{-} \cup A_0$ for which $\mu\left(X^{-} \cup A_0\right)<\alpha$, which is impossible. Hence there exists a set $A_1 \subset A_0$ and a natural number $k_1$ such that $A_1 \in \mathcal{A}, \mu\left(A_1\right) \geqslant 1 / k_1$, and $k_1$ is the smallest possible natural number $k$ for which $A_0$ contains a subset of measure at least $1 / k$. We observe that $\mu\left(A_0 \backslash A_1\right)<0$. Repeating the reasoning given for $A_0$ and applying it to $A_0 \backslash A_1$, we obtain a set $A_2 \subset A_0 \backslash A_1$ from $\mathcal{A}$ for which $\mu\left(A_2\right) \geqslant 1 / k_2$ with the smallest possible natural number $k_2$. We continue this process inductively. This will give pairwise disjoint sets $A_i \in \mathcal{A}$ with the following property: $A_{n+1} \subset A_0 \backslash \bigcup_{i=1}^n A_i$ and $\mu\left(A_n\right) \geqslant 1 / k_n$, where $k_n$ is the smallest natural number $k$ for which $A_0 \backslash \bigcup_{i=1}^{n-1} A_i$ contains a subset of measure at least $1 / k$. We observe that $k_n \rightarrow+\infty$, since otherwise by the disjointness of the sets $A_n$ we would obtain $\mu\left(A_0\right)=+\infty$. Let $B=A_0 \backslash \bigcup_{i=1}^{\infty} A_i$. Then $\mu(B)<0$, since $\mu\left(A_0\right)<0, \mu\left(\bigcup_{i=1}^{\infty} A_i\right)>0$ and $\bigcup_{i=1}^{\infty} A_i \subset A_0$. Moreover, $B$ is a negative set. Indeed, if $C \subset B, C \in \mathcal{A}$ and $\mu(C)>0$, then there exists a natural number $k$ with $\mu(C)>1 / k$, which in case $k_n>k$ contradicts our choice of $k_n$, because $C \subset A_0 \backslash \bigcup_{i=1}^n A_i$. Thus, joining $B$ to $X^{-}$, we arrive at a contradiction with the definition of $\alpha$. Therefore, the set $X^{+}$is positive.

# 实分析代写

## 数学代写|实分析作业代写Real analysis代考|Lebesgue Measure and Lebesgue–Stieltjes Measures

Imathcal{L}n:=\left } { \text { E Isubset } \backslash m a t h b b { R } \wedge n : \text { E Icap I_k \ }

$$\lambda_n(E)=\lim k \rightarrow \infty \lambda_n\left(E \cap I_k\right)$$

2.5.1. 定义。的措施 $\lambda_n$ 上面介绍了 $\mathcal{L}_n$ 称为勒贝格测度 $\mathbb{R}^n$. 设置在 $\mathcal{L}_n$ 称为勒贝格可测。

$$\lambda_n(E)=\sum j=1^{\infty} \lambda_n\left(E \cap Q_j\right),$$

## 数学代写|实分析作业代写Real analysis代考|Signed Measures

$\mu\left(A_n\right) \geqslant 1 / k_n$ ，在哪里 $k_n$ 是最小的自然数 $k$ 为了哪 个 $A_0 \backslash \bigcup_{i=1}^{n-1} A_i$ 至少包含测量的一个子集 $1 / k$. 我们观 察到 $k_n \rightarrow+\infty$ ， 因为否则由集合的不相交 $A_n$ 我们会得 到 $\mu\left(A_0\right)=+\infty$. 让 $B=A_0 \backslash \bigcup_{i=1}^{\infty} A_i$.然后 $\mu(B)<0$ ，自从 $\mu\left(A_0\right)<0, \mu\left(\bigcup_{i=1}^{\infty} A_i\right)>0$ 和 $\bigcup_{i=1}^{\infty} A_i \subset A_0$. 而且， $B$ 是负集。的确，如果 $C \subset B, C \in \mathcal{A}$ 和 $\mu(C)>0$, 那么存在一个自然数 $k$ 和 $\mu(C)>1 / k$ ，万 $k_n>k$ 与我们的选择相矛盾 $k_n$ ，因为 $C \subset A_0 \backslash \bigcup_{i=1}^n A_i$. 因此，加入 $B$ 到 $X^{-}$，我们 得出与的定义矛盾 $\alpha$. 因此，集合 $X^{+}$是积极的。

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