5
$\begingroup$

EDIT: To avoid the situation described in Mason's comment, all the measures considered below are assumed to be non-zero.

Given two measurable spaces $(X,\mathcal{A})$ and $(Y,\mathcal{B})$, assume there is a $\sigma$-finite measure $\pi\neq 0$ be a $\sigma$-finite measure on $(X\times Y,\mathcal{A}\times\mathcal{B})$ such that $$ \pi(A\times B)=\mu(A)\nu(B) $$ for all measurable sets $A\in\mathcal{A}$ and $B \in\mathcal{B}$, where $\mu$ is a $\sigma$-finite measure on $(X,\mathcal{A})$ and $\nu$ is a measure on $(Y,\mathcal{B})$. Can we conclude that $\nu$ is $\sigma$-finite too?

Stated differently, let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite measure space (i.e., the measure $\mu\neq 0$ is $\sigma$-finite) and $(Y,\mathcal{B},\nu)$ be a non-$\sigma$-finite measure space (i.e., the measure $\nu\neq 0$ is not $\sigma$-finite). Is there a product measure $\pi$ on $(X\times Y,\mathcal{A}\times\mathcal{B}) $ which is $\sigma$-finite?

$\endgroup$
4
  • 1
    $\begingroup$ The answer to your first question is no: let $\mu = 0$. $\endgroup$
    – Mason
    May 12, 2022 at 23:45
  • $\begingroup$ @Mason of course, you are right! I should have specified I am excluding this case. I'll edit the question accordingly. $\endgroup$
    – fmc2
    May 13, 2022 at 6:26
  • 1
    $\begingroup$ I think the construction in math.stackexchange.com/questions/70888/… shows that if $\nu$ is semifinite then it must be $\sigma$-finite. You'd think the non-semifinite case would be easier, but I can't come up with a proof. $\endgroup$ May 13, 2022 at 8:18
  • $\begingroup$ @NateEldredge I gave a quick at the construction you pointed out but I did not see your point but this is mainly due to the fact that I am not very comfortable swimming in the sea of measure theory. I will keep thinking about it, and, perhaps, I will find the courage to consult Fremlin's book :-) $\endgroup$
    – fmc2
    May 13, 2022 at 17:40

2 Answers 2

4
$\begingroup$

Partial answer: if $\nu$ is semifinite then it is $\sigma$-finite.

Following the setup of the first version of your statement, suppose that $\pi$ is $\sigma$-finite and satisfies $\pi(A \times B) = \mu(A) \nu(B)$ and that $\nu$ is semifinite. (We do not actually need the assumption that $\mu$ is $\sigma$-finite.) We show $\nu$ is $\sigma$-finite.

We follow the idea from Uniqueness of product measure (non $\sigma$-finite case). Since $\pi$ is $\sigma$-finite, there is a sequence of sets $E_n \subset X \times Y$ with $\pi(E_n) < \infty$ and $X \times Y = \bigcup_n E_n$. For each $n$, let $$a_n = \sup\{ \pi((X \times B) \cap E_n) : \nu(B) < \infty\}.$$ Note that $a_n \le \pi(E_n) < \infty$. Now by definition of sup, there is a sequence of sets $B_{n,k} \subset Y$ with $\nu(B_{n,k}) < \infty$ and $\pi((X \times B_{n,k}) \cap E_n) \uparrow a_n$. Set $C = Y \setminus \bigcup_{n,k} B_{n,k}$. I claim $\nu(C) =0$, which would complete the proof.

By semifiniteness, it is enough to show that for every measurable $C' \subset C$ with $\nu(C') < \infty$, we have $\nu(C') =0$. Now since such $C'$ is disjoint from all the $B_{n,k}$, we have for every $k$ that $$\pi((X \times B_{n,k}) \cap E_n) + \pi((X \times C') \cap E_n) = \pi((X \times (B_{n,k} \cup C')) \cap E_n) \le a_n$$ since $\nu(B_{n,k} \cup C') < \infty$. But by definition of the $B_{n,k}$, we have $\sup_k \pi((X \times B_{n,k}) \cap E_n) = a_n$, so we conclude $\pi((X \times C') \cap E_n) = 0$. Since $X \times Y$ is the countable union of the $E_n$, it follows that $0 = \pi(X \times C') = \mu(X) \nu(C')$. Since $\mu(X) > 0$ by assumption, we have $\nu(C') = 0$ as desired.

I don't know what happens if $\nu$ is not semifinite. You would think that if $\nu$ is not semifinite then it would make it even harder for $\pi$ to be $\sigma$-finite, but I don't see how to prove that at the moment.

$\endgroup$
3
  • $\begingroup$ Thank you very much elaborating. I have some perhaps silly questions. Assuming k ranges in a countable set, the set C is measurable, right? Then, what happens if the only measurable C' containing C has infinite measure? $\endgroup$
    – fmc2
    May 14, 2022 at 10:12
  • $\begingroup$ @F.M.C.: Yes, $n,k$ range over positive integers, so $C$ is measurable. The set $C'$ is to be contained in $C$, not the other way around. The definition of semifinite is that every measurable set $C$ of infinite measure contains a measurable set $C'$ of finite nonzero measure. $\endgroup$ May 14, 2022 at 14:39
  • $\begingroup$ Thank you very much again! I should have paid more attention in reading your answer :-) I will keep thinking about the non-semifinite case, and accept your answer. $\endgroup$
    – fmc2
    May 14, 2022 at 16:27
1
$\begingroup$

This is an old question. In case anyone is still interested though, here is an example showing that, perhaps somewhat surprisingly, the semifinite assumption is necessary in @NateEldredge ’s answer:

Let $(X, \mathcal{A}) = (Y, \mathcal{B}) = (S^1, \mathcal{B}(S^1))$. Then $(X \times Y, A \times B) = (S^1 \times S^1, \mathcal{B}(S^1 \times S^1))$. We define $\pi$ as follows: Fix a countable dense subset $\{\theta_n\}$ of $S^1$. Let $R_{\theta_n}: S^1 \rightarrow S^1$ be the rotation by $\theta_n$. Let $\Delta \subseteq S^1 \times S^1$ be the diagonal set. Let $p_1: S^1 \times S^1 \rightarrow S^1$ be the projection onto the first coordinate. Let $\chi$ be the normalized Haar measure on $S^1$. Then, for any $E \in \mathcal{B}(S^1 \times S^1)$, we define,

$$\pi(E) = \sum_n \chi(p_1(E \cap [(\mathrm{Id} \times R_{\theta_n})(\Delta)]))$$

Note that $p_1$, when restricted to $(\mathrm{Id} \times R_{\theta_n})(\Delta)$, is a homeomorphism, so $p_1(E \cap [(\mathrm{Id} \times R_{\theta_n})(\Delta)])$ is indeed Borel and $\pi$ is well-defined. $\pi$ is $\sigma$-finite. Indeed, $\cup_n (\mathrm{Id} \times R_{\theta_n})(\Delta)$ is co-null and each $(\mathrm{Id} \times R_{\theta_n})(\Delta)$ has finite measure - in fact measure $1$.

We now show that $\pi$ is a product measure of a finite measure and a non-semifinite measure. Indeed, let $\mu = \chi$ be, again, the normalized Haar measure on $S^1$. Let $\nu$ be the measure defined on $(S^1, \mathcal{B}(S^1))$ by,

$$\nu(E) = \begin{cases} 0&, \mathrm{ if }\chi(E) = 0\\ +\infty&, \mathrm{ otherwise}\end{cases}$$

Clearly, $\nu$ is not semifinite. For any $A, B \in \mathcal{B}(S^1)$, if either is null (under $\chi$), then $p_1((A \times B) \cap [(\mathrm{Id} \times R_{\theta_n})(\Delta)]) = A \cap R_{-\theta_n}(B)$ is null for all $n$, whence $\pi(A \times B) = 0 = \mu(A)\nu(B)$. Now, assume $A, B$ both have positive $\chi$ measure. By Iosif Pinelis‘s answer here (https://mathoverflow.net/a/447929/504602), there exists $\theta \in S^1$ s.t. $p_1((A \times B) \cap [(\mathrm{Id} \times R_\theta)(\Delta)]) = A \cap R_{-\theta}(B)$ has positive measure, say $\chi(A \cap R_{-\theta}(B)) = \epsilon > 0$. We observe that $\chi(A \cap R_{-\theta}(B))$ is continuous in $\theta$, so, as $\{\theta_n\}$ is dense in $S^1$, there must exist an infinite subset $\{\theta_{n_k}\}$ s.t. $\chi(A \cap R_{-\theta_{n_k}}(B)) \geq \frac{\epsilon}{2}$. Thus,

$$\begin{split}\pi(A \times B) &= \sum_n \chi(p_1((A \times B) \cap [(\mathrm{Id} \times R_{\theta_n})(\Delta)]))\\ &\geq \sum_k \chi(p_1((A \times B) \cap [(\mathrm{Id} \times R_{\theta_{n_k}})(\Delta)]))\\ &= \sum_k \chi(A \cap R_{-\theta_{n_k}}(B))\\ &\geq \sum_k \frac{\epsilon}{2}\\ &= \infty\\ &= \mu(A)\nu(B)\end{split}$$

Thus, $\pi$ is a product measure of $\mu$ and $\nu$, as required.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .