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$f$可测 $f=g$ a.e. 则$g$可测

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hbghlyj 发表于 2022-12-9 06:40 |阅读模式
Mfunct.pdf
Theorem 1.3
Let $f : E → \mathbb R$ be measurable and let $g : E → \mathbb R$. If $f = g$ a.e. on $E$, then $g$ is measurable.
Proof: For $r \in \mathbb{R}$, let $A=\{x \in E: f(x)>r\}$ and $B=\{x \in E: g(x)>r\}$. Then $A$ is measurable and $A-B, B-A$ are subsets of $\{x \in E: f(x) \neq g(x)\}$ so that they have zero measures. Since
\[
B=(B-A) \cup(B \cap A)=(B-A) \cup(A-(A-B))
\]
is measurable. Thus, $g$ is measurable.


蓝色句未加证明使用了»$A-B,B-A$ are measurable«
$A-B, B-A$ are subsets of $\{x \in E: f(x) \neq g(x)\}$如何推出$A-B,B-A$ are measurable
未假定$E$是完备测度空间

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 楼主| hbghlyj 发表于 2022-12-9 06:44
chap5.pdf中, 假定了完备测度空间
Theorem 3.1. Let $(X, S, μ)$ be a complete measure space
(1) For two functions $f$ and $g$ from $X$ to $\mathbb R$, if $f$ is measurable and $g=f$ almost everywhere, then $g$ is measurable.
Proof. (1) Let $E:=\{x \in X: f(x)=g(x)\}$. Then $E^c$ is a null set. For $a \in \mathbb{R}$, we have
\[
\{x \in X: g(x)>a\}=\{x \in E: g(x)>a\} \cup\left\{x \in E^c: g(x)>a\right\} .
\]
Moreover,
\[
\{x \in E: g(x)>a\}=\{x \in E: f(x)>a\}=\{x \in X: f(x)>a\} \backslash\left\{x \in E^c: f(x)>a\right\} .
\]
Since $\mu$ is a complete measure and $E^c$ is a null set, both sets $\left\{x \in E^c: g(x)>a\right\}$ and $\left\{x \in E^c: f(x)>a\right\}$ are measurable. Furthermore, the set $\{x \in X: f(x)>a\}$ is measurable, because $f$ is measurable. This shows that $\{x \in X: g(x)>a\}$ is measurable for every $a \in \mathbb{R}$. Therefore, $g$ is a measurable function.

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 楼主| hbghlyj 发表于 2022-12-9 07:02
MSE @saz的证明与2#类似.
a complete measure is a measure space in which every subset of every null set is measurable

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