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Let $A\subseteq[0,1]^2$ be a Lebesgue-measurable set. For $x\in [0,1]$, we define $A_x$ as $\{y:(x,y)\in A\}$.

Prove that $A_x$ is measurable for almost all $x\in[0,1]$.

I know that the "almost every'' is indeed needed, since we could have $A=V\times\{0\}$ for $V$ the Vitali set -- then $A$ has outer measure zero, so it is measurable, but $A_0$ is not measurable.

This question is similar, but it asks about product measure, while the Lebesgue measure on $[0,1]^2$ is not just the product measure, but completion of it.

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I think the following argument works:

Let $(I\times I,\mathcal L, \lambda_2)$ be the completion of $(I\times I,\mathscr B([0,1])\times \mathscr B([0,1]), \lambda\times \lambda ),$ set $f=1_A$ and choose a $\mathscr B([0,1])\times \mathscr B([0,1])$- measurable function $g$ such that $g=f\ \text{a.e.}\ \lambda_2.$ Note that the sections $g_x$ are $\mathscr B([0,1])$-measurable.

Then, $h=f-g=0\ $a.e.-$\lambda_2$ and therefore, there is a $B\in \mathscr B([0,1])\times \mathscr B([0,1])$ such that $\left \{ h\neq 0 \right \}\subseteq B$ and $\lambda_2(B)=\lambda \times \lambda (B)=0.$

We have now $0=\lambda \times \lambda (B)=\int_I \lambda (B_x)d\lambda$ so $\lambda(B_x)=0$ for almost all $x\in I.$ But then, $\lambda (\left \{ h\neq 0 \right \}_x)\le \lambda (B_x)=0\Rightarrow A_x=f_x=g_x$ for almost every $x\in I.$

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This is just a more detailed and extended version of the accepted answer. I just want to make sure everything is understood properly.

Let $\lambda_d$ denote the $d$-dimensional Lebesgue measure and $\mathcal{L}_d$ the set of Lebesgue-measurable subsets of $\mathbb{R}^d$. Moreover, let $\mathcal{L}_d\times\mathcal{L}_{d'}$ denote the product $\sigma$-algebra, i.e., the smallest $\sigma$-algebra containing all cartesian products $X\times Y$ for $X\in\mathcal{L}_d$ and $Y\in\mathcal{L}_{d'}$. (That is, $\mathcal{L}_d\times\mathcal{L}_{d'}$ is not just the cartesian product of the two sets.)

Note that $\mathcal{L}_2$ is the completion of the product $\sigma$-algebra $\mathcal{L}_1\times\mathcal{L}_1$, i.e., every set $X\in\mathcal{L}_2$ can be expressed as $X=Y\Delta Z$ for $Y\in\mathcal{L}_1\times\mathcal{L}_1$ and $Z$ a null-set.

We will be using the standard Fubini's theorem for product measure spaces that claims that for $X\in\mathcal{L}_d\times\mathcal{L}_{d'}$, we have $X_x\in\mathcal{L}_{d'}$ for every $x\in\mathbb{R}^d$.

Let $A$ be given and decompose $A=B\Delta C$ for $B\in\mathcal{L}_1\times\mathcal{L}_1$ and $C$ a null-set.

Claim. There exists $D\in\mathcal{L}_1\times\mathcal{L}_1$ such that $C\subseteq D$ and $\lambda_2(D)=0$.

Proof of claim. Since $C$ is null-set, for every $\varepsilon>0$ there exists a countable collection $O_1^\varepsilon, O_2^\varepsilon,\ldots$ of open boxes such that $\sum_{n=1}^\infty |O_n^\varepsilon|\leq\varepsilon$ and $C\subseteq\bigcup_{n=1}^\infty O_n^\varepsilon$. Then we can set $$D=\bigcap_{m=1}^\infty\bigcup_{n=1}^\infty O_n^{1/m},$$ which is an intersection and union of open boxes, hence Borel, hence from $\mathcal{L}_1\times\mathcal{L}_1$.

So by Fubini we have $$0=\int_{[0,1]^2}1_D(x,y)dxy=\int_{[0,1}\left(\int_{[0,1]}1_D(x,y)dy\right)dx=\int_{[0,1]}\lambda_1(D_x)dx.$$

Since $\int f=0$ implies that $f\equiv 0$ almost everywhere, it follows that $\lambda_1(D_x)=0$ for almost every $x\in[0,1]$. Moreover, since $C\subseteq D$, we also have that $\lambda_1(C_x)=0$ for almost every $x\in[0,1]$.

Finally, for almost every $x\in[0,1]$ we have $A_x=B_x\Delta C_x$ with $B_x\in\mathcal{L}_1$ (by Fubini) and $\lambda_1(C_x)=0$, it follows that $A_x$ is Lebesgue measurable (since $\mathcal{L}_1$ is itself a complete measure space).

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  • $\begingroup$ I made a couple of small changes to your post. They are supposed to be helpful, but please check that I haven't made any mistakes. $\endgroup$ – user940 Aug 13 '17 at 14:10
  • $\begingroup$ I think you can just use the outer regularity of Lebesgue measure to say there are open sets $U_n\supseteq C$ s.t. $\lambda_2(U_n)<1/n.$ Then $U=\cap_n U_n$ is Borel, contains $C$ and satisfies $\lambda_2(U)=0.$ $\endgroup$ – Matematleta Aug 13 '17 at 14:15

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