# Can the product of a $\sigma$-finite measure with a non-$\sigma$-finite one be a $\sigma$-finite measure?

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?

• The answer to your first question is no: let $\mu = 0$. May 12, 2022 at 23:45
• @Mason of course, you are right! I should have specified I am excluding this case. I'll edit the question accordingly.
– fmc2
May 13, 2022 at 6:26
• 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. May 13, 2022 at 8:18
• @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 :-)
– fmc2
May 13, 2022 at 17:40

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.

• 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?
– fmc2
May 14, 2022 at 10:12
• @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. May 14, 2022 at 14:39
– fmc2
May 14, 2022 at 16:27

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.