6-1

Question: 
(a)Let E be a measurable subset of \mathbf{R^2} such that for almost every x \in \mathbf{R^1} , the set \left \{ y|(x,y) \in E \right \} has \mathbf{R^1}-measure zero. Show that E has measure zero and that for almost every y \in \mathbf{R^1}, the set \left \{ x|(x,y) \in E \right \} has measure zero.
(b) Let f(x,y) be nonnegative and measurable in \mathbf{R^2}. Suppose that for almost every x \in \mathbf{R^1}, f(x, y) is finite for almost every y. Show that for almost every y \in \mathbf{R^1}, f(x, y) is finite for almost every x. 
Solution:
(a) Consider the indicator function \chi_{E} on \mathbf{R^2}. This is a nonnegative measurable function, so by Fubini’s theorem we have 
\iint_{\mathbb{R}^{2}} \chi_{E}=\int_{\mathbb{R}}\left(\int_{\mathbb{R}} \chi_{E}(x, y) d y\right) d x=\int_{\mathbb{R}}\left(\int_{\mathbb{R}} \chi_{E}(x, y) d x\right) d y .
For almost every x \in R, we have
\int_{\mathbb{R}} \chi_{E}(x, y) d y=|\{y:(x, y) \in E\}|=0.
so all the integrals in the first line above are 0. Since \iint_{\mathbb{R}^{2}} \chi_{E} = | E|, we have | E|=0. 
Moreover, since 
\int_{\mathbb{R}}\left(\int_{\mathbb{R}} \chi_{E}(x, y) d x\right) d y=\int_{\mathbb{R}}|\{x:(x, y) \in E\}| d y
and since |\{x:(x, y) \in E\}| is a non-negative function of y whose integral over y is zero,it follows that |\{x:(x, y) \in E\}| = 0 for almost every y.
(b) Let E be the set of points on which f is not finite. This is a measurable subset of \mathbf{R^2}, and E_{x} = \{y:(x, y) \in E\} has measure zero for almost every x \in R . so by part (a), we could obtain E_{y} = \{x:(x, y) \in E\} has measure 0 for almost every y \in R , which is equivalent to what we want to prove.