# Lebesgue integral measurable in the variable not of integration?

Let $X$, respectively $Y$, be a space endowed with the $\sigma$-additive complete measure $\mu_x$, respectively $\mu_y$.

If $f: X\times Y\to\mathbb{R}$ is $\mu_x\otimes\mu_y$-measurable$^1$ and Lebesgue summable in one variable, is its integral measurable in the other variable? I mean if, for example, $$\int_{Y}f(x,y)d\mu_y$$is $$x\mapsto\int_{Y}f(x,y)d\mu_y$$always a measurable function? If it is, how is it proved? I heartily thank any answerer.

If I correctly understand this argument, for which I thank PhoemueX, once proved for $f$ as a non-negative function, the more general case that my question covers would follow.

$^1$ Measurable according to the Lebesgue extension $\mu_x\otimes\mu_y$ (a complete measure) of the measure $\mu_x\times\mu_y$ defined by $(\mu_x\times\mu_y)(A\times B)=\mu_x(A)\mu_y(B)$ for the measurable sets $A\subset X$ and $B\subset Y$.

• What is $A$? If $A$ is a measurable set, then $f \cdot 1_A$ is measurable with respect to the product sigma algebra. Then my answer here (math.stackexchange.com/questions/1055496/…) shows the needed measurability. – PhoemueX Mar 5 '16 at 20:09
• @PhoemueX Thank you for your comment. Yes, in fact what I need precisely is a proof of the statement in this comment (this question of mine assumes that the integral exists finite, but a proof including the case where it diverges would be even more interesting). – Self-teaching worker Mar 6 '16 at 9:19