We call a set $A$ to be a lebesgue measurable if it belongs to the lebesgue $\sigma$-field (the completion of Borel $\sigma$-field with respect to Lesbesgue measure). We call a function $f: \mathbb{R}\to \mathbb{R}$ Lebesgue measurable if $f^{-1}(B)$ is lebesgue measurable for all $B\in \mathcal{B}(\mathbb{R})$. For any $\sigma$-field $\mathcal{F}$ on $\Omega$ and a measure $\mu$ on $\Omega$, we denote $\mathcal{F}_{\mu}$ as the completion $\sigma$-field.

Let $\lambda$ be the lebesgue measure on $\mathbb{R}$

Consider $A\subset \mathbb{R}^2$ be a lebesgue measurable set. Let us suppose $A^{x}=\{y\mid (x,y)\in A\}$. Is $A^{x}$ lebesgue measurable? Then is $x\to \lambda(A^{x})$ Lebesgue measurable?

I know have shown that if $A$ is borel then the conclusion holds. If $A$ is lebsegue measurable, Let $A=B\cup M$ where $B$ is Borel and $M\subset N$ and $(\lambda\otimes \lambda)(N)=0$. Then $A^{x}=(B\cup M)^{x}=B^{x}\cup M^{x}$. It can be shown $B^{x}$ is Borel. $M^{x} \subset N^{x}$ and $\lambda(N^{x})=0$. Hence $A^{x}$ is lebesgue measurable for each $x$.

For the next part, $A=B\cup M$. $B$ Borel and $M\subset N$ and $\lambda(N)=0$. Wlog Assume $B$ and $M$ are disjoint (Can we do this? I thought about taking $M_{new}=M-B$). $$\lambda(A^{x})=\lambda(B^{x})+\lambda(M^{x})$$ $x\to \lambda(B^{x})$ is measurable since $B$ is borel. For each $x$ $\lambda(M^{x})=0$. Hence $x\to \lambda(M^{x})$ is measurable. Since sum of two lebesgue measurable function is measurable, we are done.

Is my proof correct? Can someone verify?


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