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?.

  • $\begingroup$ two questions just to be sure: $A\times B\subset \mathbb{R}^{2}$ right? The definition of "zero content" in two dimensions is that for all $\epsilon$ there exist a finite cover of closed rectangles with area sum less than $\epsilon$ , and in 1 dimension(e.g.$ {y∈B:(x,y)∈C})$ its the same, just with closed rectangles? $\endgroup$ – Conformal Jul 22 '14 at 18:13
  • $\begingroup$ Sorry the last two words should be "closed intervals" $\endgroup$ – Conformal Jul 22 '14 at 18:19

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