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I know the uncountable unions of measurable sets need not to be measurable. But I'm stuck at the following question:

Given two space $(X,\mathcal{M})$ and $(Y,\mathcal{N})$, if $E\in \mathcal{M}\otimes\mathcal{N}$, is $\cup_y E_y$ a measurable set, where $E_y =\{x;(x,y)\in E\}$?

I guess it's measurable, and $\cup_y E_y$ can be written as a countable union and intersection of measurable sets in $\mathcal{M}$. How to prove?

Also, I want to know if every Borel measurable set in $\mathbb{R}$ can be written as a countable union and intersection of measurable sets (intervals). It seems true because in the definition of measurable space, it only contains countable operations of the set union.

Supplement: I just noticed in the definition of Borel algebrain Wikipedia. It directly uses operations of countable union, countable intersection, and relative complement to define Borel algebra. But in the definition of my current book, the Borel algebra is defined as the sigma algebra generated by open sets.

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  • $\begingroup$ What is $\mathcal{M}\otimes\mathcal{N}?$ Sorry, just trying to remember the notation. $\endgroup$ Jun 26, 2019 at 18:49
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    $\begingroup$ $\mathcal{M}\otimes\mathcal{N}$ is the sigma algebra generated by all the rectangles $A\times B$, where $A\in\mathcal{M}$ and $B\in\mathcal{N}$ $\endgroup$
    – Jimmy Kang
    Jun 26, 2019 at 18:51

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The problem is not trivial. Let $\pi_X: X\times Y\rightarrow X$ the projection onto $X$, that is $\pi:(x,y)\mapsto x$. Your question is whether $\pi(E)$, which is $\bigcup_yE_y$ , belongs to $\mathcal{M}$ for any $E\in\mathcal{M}\otimes\mathcal{N}$.The answer is no in general. This can be studied under what is called Analytic sets and universal measurability. A good place to look at these things is Measure Theory by Cohn, chapter 8.

One of the main results of all that theory is the following:

Theorem: Suppose $(X,\mathcal{M})$ is a measurable set. Let $Y$ be a Polish space equipped with the Borel $\sigma$--algebra (noted by $\mathscr{B}(Y)$). If $E\in \mathcal{M}\otimes\mathscr{B}(Y)$, then $\pi_X(E)$ is universally measurable with respect to the measurable space $(X,\mathcal{M})$.

Some terms require some explanation:

  • An example of a Polish space is the Real line with the Borel $\sigma$--algebra.
  • A set is universally measurable with respect to $(X,\mathcal{M})$ if $E$ is in the completion of $(X,\mathcal{M})$ for any finite measure $\mu$ on $\mathcal{M}$.

Counter examples in $R^2$ things are involved. Suslin constructed such examples.

As a side note, Lebesgue himself thought that the projection of measurable sets in the product are measurable. Suslin found the mistake. Here is a nice thread https://mathoverflow.net/questions/34142/projection-of-borel-set-from-r2-to-r1 with more information.

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A reminder to myself: Elements in the $\sigma$-algebra generated by $A$ must be able to be represented by countable unions\intersections operations with complement operation of the elements in $A$. Otherwise, this $\sigma$-algebra is not the smallest $\sigma$-algebra contains $A$.

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