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.


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 $x\in I.$


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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.