Are slices of measurable functions measurable?

[Office_Shredder]

I decided to go through @psie's measure theory notes to refresh myself since it's been a while. I got to the theorem on page 60 which I will attempt to summarize my confusion as just this statement

If X and Y are measure spaces and $f:X\times Y\to \mathbb{C}$ is measurable then the function $f_x:Y\to \mathbb{C}$ defined by $f_x(y)=f(x,y)$ for any fixed x is measurable, and similar for the other parameter.

But if X and Y are just the real numbers, and I pick some non measurable set $A$ and define $f(x,y)=0$ unless $x=0$ and $y\in A$, then $f_0$ is not a measurable function as $f_0^{-1}(\{1\})=A$, but I think $f$ is measurable as $\{0\}\times A$ is a set of measure 0 in $\mathbb{R}^2$.

What am I missing?

Edit: I guess there are two senses of a measurable function here, one where the set of measurable sets in $X\times Y$ are the product $\sigma$-algebra, and one which is the set of all measurable sets under the product measure. If it's measurable with respect to the former then my counterexample doesn't work (and in fact this part of the theorem has already been proven). Maybe that's what was meant? It seems like a not obvious choice since we literally just higher up on the page constructed the $\sigma$-algebra of measurable sets, and this doesn't correspond to the traditional definition from like $\mathbb{R}\to \mathbb{R}$ even

 

[Erland]

You are on the right track, I think.

The product $\sigma$-algebra in $\Bbb R^2$ generated by measurable rectangles (that is, cartesian products of one dimensional Lebesgue measurable sets) is not the same as the $\sigma$-algebra of Lebesgue measurable sets in $\Bbb R^2$. Your example shows precisely that: your set $\{0\}\times A$ does not lie in the product $\sigma$-algebra. The product $\sigma$-algebra is just a subalgebra of the Lebesgue $\sigma$-algebra. The product measure on the product $\sigma$-algebra is just the restriction of the Lebesgue measure to the product $\sigma$-algebra.

But we can extend the product $\sigma$-algebra to the Lebesgue $\sigma$-algebra. We then take the family of all sets in $E\subseteq\Bbb R^2$ for which there exists sets $A$ and $B$ in the product $\sigma$-algebra such that $A\subseteq E \subseteq B$ and $\mu(B\setminus A)=0$, where $\mu$ is the product measure. This family turns out to be precisely the Lebesgue measurable sets in $\Bbb R^2$, and the Lebesgue measure on $\Bbb R^2$ extends the product measure, so that $\mu(E)=\mu(A)=\mu(B)$ above.
A consequence of this is that the Lebesgue measure is complete, which means that any subset of a set of Lebesgue measure 0 is Lebesgue measurable, with Lebesgue measure 0 of course. We say that the Lebesgue measure space is the completion of the product measure space. The set in your example is then Lebesgue measurable with Lebesgue measure 0, and this shows the advantage with complete measures over non-complete measures.

 

[Office_Shredder]

I get all of that, I just don't understand what is meant by a measurable function on a product space. if you look at $\mathbb{R}\to \mathbb{R}$ you only require borel sets get pulled back to lebesgue measurable sets, and allowing lebesgue measurable sets is actually powerful.

But once you go to $\mathbb{R}^2$, is it standard to require measurable functions to pull back to something stricter than lebesgue measurable sets? I guess it gives you this nice theorem about restricting to a single variable, but is nothing lost in the process?

 

[Erland]

It's somewhat unclear to me what you mean by "pulling back".

A real or complex valued function on a product space is measurable (w.r.t. the measurable space with the product $\sigma$-algebra), if the inverse image of open sets are measurable (w.r.t this space). So, no difference here from how it is in general. As we have seen, a Lebesgue measurable function on $\Bbb R^2$ might not be measurable w.r.t the product measurable space.

Some results become a little bit more cumbersome if we use Lebesgue measure instead of the product measure om (e.g.) $\Bbb R^2$. For example, if we use Lebegue measure in the Fubini Theorem, then the function $f_x(y)=f(x,y)$ might not be measurable for all $x$, just measurable a. e., even if $f$ is a positive measurable function on $\Bbb R^2$, and hence the function $\phi(x)=\int_\infty^\infty f(x,y)dy$ is only defined a. e. This problem does not occur if we use the product measure space.

(See Rudin, W. Real and Complex Analysis, 3rd ed, McGraw-Hill, 1986, p. 168 f.)