Monotone convergence theorem on a function of two variables (integrating over one variable)

Question

Consider a measurable space $\big( E \times F, \mathcal{E} \otimes \mathcal{F} \big)$, an $(\mathcal{E} \otimes \mathcal{F})$-measurable and positive function $f$, and a measure $\nu$ on $\big( F, \mathcal{F} \big)$.

Consider the function $\nu f$ defined as

\begin{align*} \nu f \colon E & \longrightarrow \mathbb{R}_+ \\ x & \longmapsto \nu f (x) = \int_F \nu(dy) f(x,y) \end{align*}

Under what conditions on $\nu$, is $\nu f$ $\mathcal{E}$-measurable and positive?

My attempt

I would proceed by showing first the measurability of indicator functions in $(\mathcal{E} \otimes \mathcal{F})$, then measurability of simple functions, and finally measurability of positive measurable functions by monotone convergence theorem. This would allow to make no requirements on $\nu$. Here it is.

Consider a simple function $\mathbb{I}_{A \times B}$, where $A \in \mathcal{E}$ and $B \in \mathcal{F}$.

Then, $\nu \mathbb{I}_{A \times B} = \nu \mathbb{I}_A \mathbb{I}_B = \nu(B) \mathbb{I}_A $ is $\mathcal{E}$-measurable as it is the product of a constant $\nu(B)$ and $\mathbb{I}_A$. Of course, $A \in \mathcal{E}$ and $B \in \mathcal{F}$.

So, for any measurable rectangle $A \times B \in \mathcal{E} \otimes \mathcal{F}$, the function $\nu \mathbb{I}_{A \times B}$ is $\mathcal{E}$- measurable and positive.

Now, consider the simple function $f = \sum_{n=1}^{N}a_n \mathbb{I}_{A_n \times B_n}$, with $a_n$ postive $\forall \ n \geq 1$. It is $(\mathcal{E} \otimes \mathcal{F})$-measurable and positive as a linear combination of measuarble indicator functions.

Then, $\nu \big( \sum_{n=1}^{N}a_n \mathbb{I}_{A_n \times B_n} \big) = \sum_{n=1}^{N} \big( a_n \ \nu \mathbb{I}_{A_n \times B_n} \big) $ is $\mathcal{E}$-measurable as it is a linear combination of $\mathcal{E}$-measurable and positive functions from the previous step.

Finally, consider an arbitrary $(\mathcal{E} \otimes \mathcal{F})$-measurable and positive function $f$. Then, $\exists \big( f_n \big)_{n=1}^{\infty} \subset (\mathcal{E} \otimes \mathcal{F})_+$, a sequence of $(\mathcal{E} \otimes \mathcal{F})$-measurable and positive simple functions, such that $f_n \nearrow f$. Then, one can apply monotone convergence theorem with respect to $\nu$ and claim that

$\nu f = \lim_{n \rightarrow \infty} \nu f_n$ is $\mathcal{E}$-measurable and positive as the limit of a sequence of $\mathcal{E}$-measurable and positive functions $\nu f_n$.

The issue

My textbook proves the measurability of $\nu f$ under the assumption that $\nu$ is $\Sigma$-finite. So, this means that my proof above is not correct, i.e. $\nu f$ is not necessarily measurable. I do not know what is wrong in my proof: I suspect issues can arise when applying the monotone convergence theorem, but I cannot imagine how, because the limit of $\mathcal{E}$-measurable functions is supposed to be $\mathcal{E}$-measurable itself. Any help would be much appreciated. I can provide a potential counterexample of a non-$\Sigma$-finite measure $\nu$ such that $\nu f$ is not $\mathcal{E}$-measurable even if $f$ is $\mathcal{E} \otimes \mathcal{F}$-measurable an positive.

Answer 1

I think you have confused yourself by conflating two different notions of "simple function".

Here I'll use the term "simple function" in its usual sense, for a function of the form $f(x,y) = \sum_{n=1}^N a_n 1_{C_n}$ where $C_n$ is an arbitrary $\mathcal{E} \otimes \mathcal{F}$-measurable set. Note that $C_n$ does not have to be of the form $A_n \times B_n$.

I'll use the term "rectangle function" for a simple function with the special property that each $C_n$ is of the form $C_n = A_n \times B_n$. This is what you were calling a "simple function" in your proof, and I think that is the source of the confusion.

It is true that each measurable function $f$ is approximated by a monotone increasing sequence of simple functions. However, your argument did not prove that $\nu f(x)$ is measurable for an arbitrary simple function $f$. This step is where you would need to use the hypothesis of $\sigma$-finiteness.

You did prove that $\nu f(x)$ is measurable when $f$ is a rectangle function. But it is not true that every measurable function is approximated by a monotone increasing sequence of rectangle functions.

For a counterexample, let $E = F = [0,1]$, and consider the "diagonal" set $D = \{ (x,x) : x \in [0,1]\} \subset E \times F$. Let $f = \mathbb{1}_D$. Now let $g = \sum_{n=1}^N a_n \mathbb{1}_{A_n \times B_n}$ be a positive rectangle function in which, without loss of generality, every $a_n > 0$ and every $A_n, B_n$ are nonempty. If $g \le f$ then necessarily $A_n \times B_n \subset D$ for every $n$. But this is only possible if $A_n = B_n = \{x\}$ for some $x$, i.e. they must be singletons. Therefore $g$ is only nonzero at a finite number of points (has finite support).

Hence if $f_n$ is a sequence of positive rectangle functions with $f_n \le f$, then each $f_n$ has finite support. This means their pointwise limit, if it exists, has countable support, and so it cannot equal $f$.