A set of measure zero in a product space intersects almost every fiber in a set of measure zero

Question

Let $C\subset A\times B$ be a set of content zero. Let $A'\subset A$ be the set of all $x\in A$ such that $\{y\in B: (x,y)\in C\}$ is not of content zero. Show that $A'$ is a set of measure zero.

I couldn't finish this problem, after a few steps I'm not sure what could I do next.

This much I know: $C$ has content zero, then $\partial C$ has content zero ($\partial C$ denotes the boundary of $C$) which means that the characteristic function $\chi_c$ is integrable in $A\times B$.

Then I applied Fubini's theorem to have $\displaystyle\int_{A\times B}\chi_C=\displaystyle\int_A\left(L\displaystyle\int_B \chi_C\; dy\right)dx=\displaystyle\int_A\left(U\displaystyle\int_B\chi_C\;dy\right)dx$ where $L$ and $U$ denote that those are the lower and upper integral.

Now follows $\displaystyle\int_A\left(L\displaystyle\int_B \chi_C\; dy-U\displaystyle\int_B\chi_C\;dy\right)dx=0$ which means that $\displaystyle\int_B \chi_C\;dy$ is integrable.

Since $C$ is a set of content zero, is a set of measure zero. I know that if a bounded set $C$ is of measure zero and $\displaystyle\int_B\chi_c$ exists then $\displaystyle\int_B\chi_c=0$. Here I'm not sure how to go on, I'm considering taking a partition $P$ of $B$ and knowing that $\sup\{L(\chi_C,P): \text{P partition of B}\}=0$ is possible to take $P$ such that $U(f,P)=\sum_{S\in P} M_S(\chi_C)v(S)\leq\sum_{S\in P} v(S)<\epsilon$.

But it doesn't seem okay... Any thoughts about it?.

Answer 1

I assume that $A = B = \mathbb R$.

We have that $A' = \bigcup_{\epsilon > 0} A_\epsilon$ where $A_\epsilon = \{x : \mu(\{ y \in B : (x, y) \in C \}) > \epsilon \}$. Denote by $\bar \mu$ the outer Lebesgue measure. By properties of the upper integral and by Fubini, $$\begin{align*} \bar \mu(A_\epsilon) &= \overline{\int_{A}} \chi_{A_\epsilon} \\ & \leq \epsilon^{-1} \overline{\int_A} \chi_{A_\epsilon} \int_B \chi_C(a, b) \\ & \leq \epsilon^{-1} \overline{\int_A} \int_B \chi_C(a, b) \\ &= \epsilon^{-1} \int_A \int_B \chi_C(a, b) \\ &= \epsilon^{-1} \int_C \chi_C(a, b) \\ & = 0 \end{align*}$$ so that $\mu(A_{\epsilon}) = 0$. Because $A'$ is the countable union of $A_{1/n}$, we have $\mu(A') = 0$.