How does one prove that partial functions of measurable functions of 2 var. are nearly all measurable?

Question

How does one prove that if $F: X \times Y \to Z$ is measurable for a product measure, then the partial function $F(x,.) : Y \to Z$ (defined by $y \mapsto F(x,y) )$ is measurable for almost all $x$ in $X$? I avoid being more precise for a start so that I have the best chances of getting a usable answer for my purpose. (Some not mentioned hypotheses are probably needed.)

Motivation: In the book Operator Theory in Function Spaces by Kehe Zhu integral operators are introduced with $Tf(x) = \int_X K(x,y)f(y)d\mu(y)$ where $K$ and $f$ are in $L^2$ (and therefore measurable) and to make sure that the integral exists it is necessary to have the partial functions also in $L^2$ which will be easy once they are measurable (for almost all $x$)

Answer 1

First, we can prove the following lemma:

If $E$ is a measurable subset of $X \times Y$, then for any $x \in X$, the slice $E_x := \{ y\in Y: (x,y)\in E\}$ is a measurable subset of $Y$.

To prove this, let $\mathcal E$ be the set of all subsets $E \subset X \times Y$ such that $E_x$ is a measurable subset of $Y$ for all $x \in X$.

Notice that:

• $\mathcal E$ is a sigma algebra: it contains $X \times Y$, and it is closed under taking complements and countable unions.
• $\mathcal E$ contains all measurable rectangles in $X \times Y$.

But the product sigma algebra $\mathcal P$ on $X \times Y$ is by definition the smallest sigma algebra on $X \times Y$ that contains all measurable rectangles. So every set in $\mathcal P$ is also a set in $\mathcal E$, and our lemma is proved.

We now return to your original question:

Let $F : X \times Y \to Z$ be a measurable function. For some $x \in X$, let $F_x : Y \to Z$ be the slice function, $F_x(y) = F(x, y)$. Then $F_x$ is measurable.

To prove this, consider an open subset $U \subset Z$. We must show that $F_x^{-1}(U)$ is a measurable subset of $Y$.

Since $F$ is measurable, $E := F^{-1}(U)$ is a measurable subset of $X \times Y$. By our lemma, the slice $E_x$ is a measurable subset of $Y$. But $E_x$ is precisely $F_x^{-1}(U)$, so we're done.

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Edit: As per the discussion in the comments below, the OP requires a stronger version of the lemma:

Let $\mu$ and $\nu$ be complete measures on $X$ and $Y$. Let $\mu \times \nu$ be the product measure on $X \times Y$, and let $\mathcal P$ be the corresponding sigma algebra. Let $\mathcal P^\star$ be the completion of $\mathcal P$ with respect to $\mu \times \nu$. Let $E$ be a subset of $X \times Y$, measurable with respect to $\mathcal P^\star$ (but not necessarily measurable with respect to $\mathcal P$). Then, for almost all $x \in X$, the slice $E_x$ is a measurable subset of $Y$.

To prove this, we write $E = A \cup B$, where:

• $A$ and $B$ are disjoint.
• $A$ is measurable with respect to $\mathcal P$.
• $B$ is contained inside some set $C$, where $C$ is measurable with respect to $\mathcal P$ and has measure zero.

My previous lemma shows that $A_x$ is measurable for all $x$. However, it still remains to show that $B_x$ is measurable, at least for almost all $x$. So we have more work to do...

Let's start from the following equation:

$$\int_X \nu(C_x) d\mu(x) = (\mu \times \nu) (C) = 0.$$

[The equation $(\mu \times \nu) (W) = \int_X \nu(W_x) d\mu(x)$ holds for all $\mathcal P$-measurable sets $W$, since this equation is literally the definition of the product measure $\mu \times \nu$ on the sigma algebra $\mathcal P$.]

Since $\int_X \nu(C_x) d\mu(x) = 0$, it must be the case that $\nu(C_x) = 0$, for almost all $x \in X$.

Now $B \subset C$, so $B_x \subset C_x$. And by assumption, $Y$ is complete. Thus $B_x$ is measurable and has measure zero, for almost all $x \in X$. And we're done.

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Another edit: We still need to prove our final result.

If $F : X \times Y \to Z$ is measurable with respect to $\mathcal P^\star$, then $F_x$ is measurable, for almost all $x \in X$.

As pointed out by the OP in the comments, this is actually quite subtle, and with the results we've established so far, we're going to have to make place some restrictions on $Z$. Let's assume that $Z$ is second countable, i.e. the topology for $Z$ has a countable base $\mathcal B = \{ U_n \}_{n \in \mathbb N}$. For example, $Z$ could be $\mathbb C$: in this case, $\mathcal B$ could consist of the open balls with rational centres and rational radii.

For each $n \in \mathbb N$, let $Q_n$ be a null set such that $F_x^{-1}(U_n)$ is measurable for all $x \notin Q_n$. (The existence of such a null set was established by our previous lemma.)

Let $Q = \bigcup_{n \in \mathbb N} Q_n$, which, too, is null, by countable additivity.

Consider an arbitrary open set $U \subset Z$, which can be expressed as $U = \bigcup_{n \in S} U_n$ for some collection of indices $S \subset \mathbb N$.

Observe that $F_x^{-1}(U) = \bigcup_{n \in S} F_x^{-1}(U_n)$. If $x \notin Q$, then $F_x^{-1}(U_n)$ is measurable for all $n \in \mathbb N$, and in particular, is measurable for all $n \in S$. Thus $F_x^{-1}(U)$, being the union of at most countably many measurable sets, is measurable, provided that $x \notin Q$.

Thus $F_x$ is a measurable function for all $x \notin Q$, i.e. $F_x$ is measurable for almost all $x$.