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I spent some time searching MathOverflow for a problem that would resemble the one given below, but it turned out to be a rather futile endeavor. I was led to this problem in my attempts to construct a counterexample refuting a result that had already been published in a peer-reviewed article.

Problem. Find measure spaces $ (X,\mathcal{S},\mu) $ and $ (Y,\mathcal{T},\nu) $, at least one of which is not $ \sigma $-finite, and an $ (\mathcal{S} \otimes \mathcal{T}) $-measurable function $ f: X \times Y \to \mathbb{R}_{\geq 0} $ with the following properties:

  • The function $ f(x,\bullet): Y \to \mathbb{R}_{\geq 0} $ belongs to $ {L^{1}}(Y,\mathcal{T},\nu) $ for every $ x \in X $.
  • The function $ \left\{ \begin{matrix} X & \to & \mathbb{R}_{\geq 0} \\ x & \mapsto & \displaystyle \int_{Y} f(x,\bullet) ~ \mathrm{d}{\nu} \end{matrix} \right\} $ is not $ \mathcal{S} $-measurable.

Does anyone know if this problem can even be solved? Thank you very much for your time!

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    $\begingroup$ Doesn't seem like the measure $\mu$ is even used here. $\endgroup$
    – Daniel McLaury
    Commented Oct 30, 2016 at 21:56
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    $\begingroup$ A short note to anyone studying this problem: In accordance with Daniel’s comment, one may let $ \mu $ be the trivial measure on $ (X,\mathcal{S}) $, which makes $ (X,\mathcal{S},\mu) $ a $ \sigma $-finite measure space. Therefore, in order for a counterexample to exist, it is necessary that $ (Y,\mathcal{T},\nu) $ be non-$ \sigma $-finite, otherwise the second function in my post would be $ \mathcal{S} $-measurable (as can be seen by analyzing any standard proof of the Fubini-Tonelli Theorem). $\endgroup$
    – Transcendental
    Commented Oct 31, 2016 at 16:49
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    $\begingroup$ Another short note: PhoemueX has provided a nice counterexample below that satisfies the second property, but it is unclear whether or not it also satisfies the first property. A brief analysis of PhoemueX’s argument leads one to the following question: Does there exist a Borel subset $ M $ of $ \mathbb{R}^{2} $ such that (i) each vertical cross-section of $ M $ is finite, i.e., $ \{ y \in \mathbb{R} \mid (x,y) \in M \} $ is finite for each $ x \in \mathbb{R} $, and (ii) the projection of $ M $ onto the first coordinate is a non-Borel analytic subset of $ \mathbb{R} $? $\endgroup$
    – Transcendental
    Commented Nov 1, 2016 at 7:01

2 Answers 2

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Let $\mathcal{B}$ denote the class of Borel subsets of $[0,1]$ and let $A \subseteq [0, 1]$ be a non Borel set. Let $f$ be the characteristic function of the graph of a bijection from $A$ to $[0, 1]$. Then $f$ is $\mathcal{B} \otimes \mathcal{P}([0, 1])$-measurable (check) and the map $x \mapsto \int f(x, y) d\mu(y)$ is non-zero precisely on $A$ where $\mu$ is counting measure.

Edit: I was asked to provide more details so here they are.

Suppose $W \subseteq \mathbb{R}^2$ is such that every horizontal section $W^y = \{x : (x, y) \in W\}$ is closed. Then $W \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R})$. To see this, define, for each interval $J$ with rational endpoints, $Y_J = \{y : W^y \cap J = \phi\}$. As each $W^y$ is closed, we have $$ W = \mathbb{R}^2 \Big\backslash \bigcup \{J \times Y_J : J \text{ is an interval with rational endpoints}\} \in \mathcal{B} \otimes \mathcal{P}(\mathbb{R}). $$ It follows that the graph of every partial bijection on $\mathbb{R}$ is in $\mathcal{B} \otimes \mathcal{P}(\mathbb{R})$.

Now choose any non Borel set $A \subseteq \mathbb{R}$ and an injection $i: A \to \mathbb{R}$. Define $f: \mathbb{R}^2 \to \mathbb{R}$ to be the characteristic function of the graph of $i$. Let $\mu$ be the counting measure on $\mathbb{R}$. The map $x \mapsto \int f(x, y) d\mu(y)$ is precisely the characteristic function of $A$ and hence is non-Borel.

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    $\begingroup$ Hi rumpf. Thank you for your response. Unfortunately, I don’t consider myself to be smart enough to check your claim in the third line. Could you elaborate further? $\endgroup$
    – Transcendental
    Commented Nov 1, 2016 at 23:04
  • $\begingroup$ For each rational interval $J$, let $Y_J$ be the set of points where the horizontal section of the graph of the bijection is disjoint with $J$. Now consider the union of $J \times Y_J$'s. $\endgroup$
    – drumpf
    Commented Nov 1, 2016 at 23:12
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    $\begingroup$ I am getting confused by these definitions; I think it would help if more things had names. Let's call $\phi : A \to [0,1]$ the desired bijection. It looks to me like you want $Y_J = \phi(A \cap J^c)$? That doesn't seem to make sense. In particular if $y = \phi(x)$ then $(x,y)$ is not in $J \times Y_J$ for any $J$. $\endgroup$
    – Nate Eldredge
    Commented Nov 2, 2016 at 5:11
  • $\begingroup$ @Nate: drumpf’s improved argument does appear to check out. What do you think? $\endgroup$
    – Transcendental
    Commented Nov 2, 2016 at 18:26
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    $\begingroup$ @Nate: It also appears that his argument works if we let $ X $ be any uncountable second-countable $ T_{1} $ topological space whose Borel $ \sigma $-algebra $ \mathcal{S} $ isn’t all of $ \mathcal{P}(X) $ (i.e., $ X $ has a non-Borel subset). Any uncountable standard Borel space automatically satisfies this property (assuming the Axiom of Choice, of course). $\endgroup$
    – Transcendental
    Commented Nov 2, 2016 at 22:02
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EDIT: The following only provides a partial answer, since it is not clear at all that the first property of the question ($f(x, \cdot) \in L^1(Y, \mathcal{T}, \nu)$) is fulfilled for the given example.

Let $(X, \mathcal{S})$ and $(Y, \mathcal{T})$ both be the real line with the Borel sigma algebra. Note that the product sigma algebra is again the Borel sigma algebra (but on $\Bbb{R}^2$).

It is well-known (see Projection of Borel set from $R^2$ to $R^1$) that not every projection of a Borel set is a Borel set. Hence, let $M \subset \Bbb{R}^2$ be a Borel set such that the projection $\pi_1 (M)$ is not Borel measurable.

Let $\mu$ be the counting measure on the real line. If $$ F(x) := \int_{\Bbb{R}} 1_M (x,y) d \mu(y) $$ was measurable, then so would be the set $$ \pi_1 (M) = \{x \,:\, \exists y : (x,y) \in M\} = \{x \,:\, F(x) > 0\}. $$

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    $\begingroup$ Wikipedia claims that Fubini always works for the "maximal product measure." I wonder what that means if the iterated integrals aren't even defined. en.wikipedia.org/wiki/… $\endgroup$
    – Christian Remling
    Commented Oct 31, 2016 at 18:50
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    $\begingroup$ @Christian: Fubini’s Theorem states that if $ (X,\mathcal{S},\mu) $ and $ (Y,\mathcal{T},\nu) $ are measure spaces, and if $ f \in {L^{1}}(X \times Y,\mathcal{S} \otimes \mathcal{T},(\mu \otimes \nu)_{\text{max}}) $, where $ (\mu \otimes \nu)_{\text{max}} $ denotes the maximal product measure, then the iterated integrals exist and are equal to $ \displaystyle \int_{X \times Y} f ~ \mathrm{d}{(\mu \otimes \nu)_{\text{max}}} $. I suspect that the function $ \mathbf{1}_{M} $ defined by PhoemueX above isn’t integrable on $ \mathbb{R}^{2} $ with respect to the maximal product measure. $\endgroup$
    – Transcendental
    Commented Oct 31, 2016 at 21:38
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    $\begingroup$ @PhoemueX: Your counterexample clearly satisfies the second property, but can you show that it also satisfies the first property? In other words, can you prove that $ \{ y \in \mathbb{R} \mid (x,y) \in M \} $ is finite for each $ x \in \mathbb{R} $? Thanks! $\endgroup$
    – Transcendental
    Commented Nov 1, 2016 at 7:04
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    $\begingroup$ @Transcendental: Yes, I realized this problem just this morning. With the current construction, I don't think this is true. I will delete my answer if I see no way to fix it until this evening :( $\endgroup$
    – PhoemueX
    Commented Nov 1, 2016 at 7:32
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    $\begingroup$ As discussed here, the Lusin-Novikov theorem says that your example cannot satisfy the first property. $\endgroup$
    – Nate Eldredge
    Commented Nov 2, 2016 at 15:18

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