By Transcendental


2016-10-30 20:46:04 8 Comments

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!

2 comments

@rumpf 2016-11-01 22:48:32

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.

@Transcendental 2016-11-01 23:04:00

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?

@drumpf 2016-11-01 23:12:41

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.

@Nate Eldredge 2016-11-02 05:11:37

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 $J$.

@Transcendental 2016-11-02 18:26:26

@Nate: drumpf’s improved argument does appear to check out. What do you think?

@Nate Eldredge 2016-11-02 18:31:48

@Transcendental: Yes, I think so too. +1 from me. Maybe a simpler way of describing $f$ is that it is just $f(x,y) = 1$ iff $x=y \in A$ and $0$ otherwise.

@Transcendental 2016-11-02 22:02:22

@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).

@PhoemueX 2016-10-31 16:59:59

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 $$ \pi_1 (M) = \{x \,:\, \exists y : (x,y) \in M\} = \{x \,:\, F(x) > 0\}. $$

@Christian Remling 2016-10-31 18:50:14

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/…

@Transcendental 2016-10-31 21:19:18

@PhoemueX: Thanks! I must shamefully admit that the fact you quoted wasn’t known to me.

@Transcendental 2016-10-31 21:38:27

@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.

@Transcendental 2016-11-01 07:04:36

@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!

@PhoemueX 2016-11-01 07:32:15

@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 :(

@Transcendental 2016-11-01 10:46:42

@PhoemueX: Please don’t delete your answer as I (and many other people) considered it very helpful. Even if you aren’t able to find a fix in the end, somebody else might. :)

@PhoemueX 2016-11-01 13:22:41

@Transcendental: Ok, sure :) I just thought it would draw more attention to the question, since at the moment it seems to be answered. I will just edit it to show more clearly that it is not a complete answer.

@Christian Remling 2016-11-01 19:11:59

To make one more unhelpful remark, the fact that $M$ is Borel in $\mathbb R^2$ is actually not that relevant perhaps since one could have taken $\mathcal P(Y)$ as the $\sigma$-algebra on $Y$.

@Transcendental 2016-11-01 21:27:34

@Christian: It’s clear that $ M $ is a member of the $ \sigma $-algebra $ \mathcal{B}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R}) $, but any act of whittling down each vertical cross-section of $ M $ to make it finite might threaten this membership.

@Nate Eldredge 2016-11-02 15:18:59

As discussed here, the Lusin-Novikov theorem says that your example cannot satisfy the first property.

@PhoemueX 2016-11-02 15:23:48

@NateEldredge: Nice, that looks fascinating. I always wanted to learn more about descriptive set theory :)

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