Prove that for $A \subseteq [0,1]^2$, with Lebesgue measure $m_2(A)=1$, then $m(s_x(A)) = 1$ for almost all $x \in [0,1]$

Question

I can picture in my mind why this works, but I have problems proving it rigorously.

Problem:

For some Lebesgue-measurable subset $A \subseteq [0,1]^2$ with $m_2(A)=1$, prove that for almost every $x\in[0,1]$, we have $m(s_x(A)) = 1$ where $s_x(A) = \{y \in [0,1] : (x,y) \in A\}$

First, since $m_2$ is a measure, it is true that

$$A \subseteq [0,1]^2 \Rightarrow m_2(A)\leq m_2([0,1]^2)=1$$

Therefore, the integral that gives the measure of $A$ is bounded above by $1$. That is,

$$\int_{[0,1]} \left[ \int_{[0,1]} \chi_A(x,y) m(dy)\right] m(dx) \leq 1$$

Therefore, to have the case with the equality, the function given by the inside integral, that is: $\int_{[0,1]}\chi_A(x,y)m(dy)$ has to be $1$ almost everywhere since it is bounded above by $1$, and (this is the part I have trouble being rigorous at) you can't "compensate" values strictly less than $1$ with other values strictly greater than $1$. Therefore, it must be true that this function equals $1$ for almost all $x \in [0,1]$

Thank you for the help!

Answer 1

It is much easier to work with $A^{c}$. $m_2(A^{c})=0$ and this gives you $m(s_x(A^{c}))=0$ almost everywhere: if a nonnegative measurable function integrates to $0$ then it is $0$ almost everywhere. Now verify that $s_x(A^{c})=(s_x(A))^{c}$ and that gives you $m(s_x(A^{c}))=1$ almost everywhere.

Answer 2

You are on the right track. Assume by contradiction that there is a set of positive measure $B$ such that $m(s_x(A)) < 1$ for all $x \in B$. Fubini now implies \begin{align*} m_2(A) &= \int_0^1 m(s_x(A)) \, \mathrm{d}x\\ &= \int_B m(s_x(A)) \, \mathrm{d}x + \int_{A \setminus B} m(s_x(A)) \, \mathrm{d}x\\ &< \int_B 1 \, \mathrm{d}x + \int_{A \setminus B} 1 \, \mathrm{d}x\\ &= m(B) + m(A \setminus B)\\ &= 1, \end{align*} a contradiction.