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I know that the $y$-section $A_x$ of a $\mu_x\otimes \mu_y$-measurable set $A$, where $\mu_x\otimes \mu_y$ is the Lebesgue extension of the product measure $\mu_x\times \mu_y$ (both measures being $\sigma$-additive complete measures defined on $\sigma$-algebras of subsets of $X$ and $Y$, respectively)$^1$, defined by $$A_y=\{x\in X:(x,y)\in A\}$$ is $\mu_y$-measurable for almost all $y$. My textbook says that (I think that the reason is that $\forall x\notin\bigcup_{y\in Y} A_y\quad A_x=\emptyset$) the integral $\int_X \mu_y(A_x)d\mu_x$ is the same as $\int_{\bigcup_{y\in Y} A_y} \mu_y(A_x)d\mu_x$.

Therefore I think that it is implicit that $\bigcup_{y\in Y} A_y$ is measurable, but I cannot see why it is. Could anybody be so kind to explain why it is measurable, provided that it is?

$^1$I want to apologise in advance if my wording may not be rigourous or formal enough, I fear I am inheriting such a style from my book, which I often find hard to understand precisely because of such a lack of formality and rigour.

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This is not true in general. Consider some $Z\subseteq X$ nonmeasurable an an arbitrary $y\in Y$ such that $\mu_Y(\{y\})=0$. Then $A=Z\times \{y\}$ is measurable, but its projection onto $X$ (i.e. the union of $y$-sections) is not.

That said, you can restrict a measure to a nonmeasurable set, but that can give rise to some funny properties. For example, if you restrict Lebesgue measure to a Bernstein set, you get a non-Radon measure, because any compact subset of a Bernstein set is countable.

Edit I thought I should expand a little on the second paragraph. $\int_Z f \, d\mu$ is usually taken to be as a shorthand $\int_X f\chi_Z \, d\mu$, and then it makes no sense if $Z$ is non-measurable, for general $f$, as $f\chi_Z$ can be non-measurable (e.g. if $f=1$).

On the other hand, if $A$ is measurable, it follows by Fubini that $\mu_Y(A_x)\cdot \chi_{\bigcup_y A_y}$ is actually measurable (trivially, since $\mu_Y(A_x)\cdot \chi_{\bigcup_y A_y}=\mu_Y(A_x)$).

We could also define $\int_Z f \,d\mu$ as a shorthand for $\int_Z f'\, d\mu_Z$ where $f'$ is the restriction of $f$ to $Z$ and $\mu_Z$ is the restriction of the outer measure $\mu^*$ to subsets of $Z$, which is a measure on the $\sigma$-algebra of intersections of the original $\sigma$-algebra with $Z$. It is an easy exercise that the two definitions coincide when $Z$ is measurable, though if it is not, there are some not so nice properties, for example, typically $\int_Z f\, d\mu+\int_{Z^c} f d\mu>\int f \,d\mu$, so this kind of thing is awkward and should be avoided in most contexts.

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    $\begingroup$ Possibly interesting for people reading your answer: math.stackexchange.com/questions/169714/… $\endgroup$ – PhoemueX Oct 23 '14 at 10:06
  • $\begingroup$ Therefore I'd find it more proper to say that $\int_X \mu_y(A_x)d\mu_x$ is equal to $\int_S \mu_y(A_x)d\mu_x$ where $S$ is any measurable set such that $\bigcup_y A_y\subset S$. I heartly thank you, Tomasz, for your answer, and Phoemuex for your comment!!! $\endgroup$ – Self-teaching worker Oct 23 '14 at 10:34

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