This is a proof from measure and integral by Richard and Antoni in page

90

Theorem: Let $f(x,y)$ be a measurable function on $\mathbb{R}^{n+m}$. Then for almost every $x \in$ $\mathbb{R}^{n}$, $f(x,y)$ is a measurable function of $y \in\mathbb{R}^{n}$. In particular, if $E$ is a measurable subset of $\mathbb{R}^{n+m}$, then the set $E_x=${$y:(x,y) \in E$} is a measurable in $\mathbb{R}^{n}$ for almost every $x \in$ $\mathbb{R}^{n}$

Proof:

First, if $f$ is a characteristic function $\chi_E$ of a measurable set $E$ in $\mathbb{R}^{n+m}$, then the two statements above are equal. In this case, write $E=H$ $\cup$ $Z$ ,where $H$ is of type $F_{\sigma}$ (ie: the union of closed set) in $\mathbb{R}^{n}$ and $Z$ is measure $0$,

Claim 1: $E_x$=$H_x$ $\cup$ $Z_x$, where $H_x$ is of type $F_{\sigma}$ and for almost every $x$, $Z_x$ is measure $0$

suppose the claim hold, then the result follow in this case.

Claim2:

If $f$ is any measurable function on $\mathbb{R}^{n+m}$, consider the set $E(a)$={$(x,y):f(x,y)>a$}, clearly $E(a)$ is a measurable in $\mathbb{R}^{n+m}$, show that the set $E(a)_x$= {$y:(x,y) \in{E(a)}$} is a measurable set in $\mathbb{R}^{n}$ for almost every $x$ $\in$ $\mathbb{R}^{n}$

For claim $1$: that is to say the projection of a

$F_{\sigma}$ is also$F_{\sigma}$, I have a question that why it is closed, (ie: the projection of the closed set is not general true), but why this hold?

For claim two: it says that the section of a measurable set is measurable, but I have no idea to prove this fact

can someone help me, thanks

## Best Answer

sorry, everyone, I have an answer as follows.

Sorry for disturbing you, I am apologizing for my stupid, sincerely.