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?
 A: 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.
