= Solution
\b[True.] Suppose instead that $B=A_1\times A_2$ were a <Lebesgue measurable set> in the plane. The section consequence of <Fubini's theorem> says that there is a <Lebesgue null set> $N\subseteq\mathbb R$ such that
$$
B_x=\{y:(x,y)\in B\}
$$
is a <Lebesgue measurable set> for every $x\notin N$. This version applies to completed <Lebesgue measure>; it can be obtained on bounded rectangles and then extended by a countable exhaustion, so no finite-measure assumption on $B$ is needed.
For every $x\in A_1$ the section is exactly $A_2$, which is not measurable. Thus $A_1\subseteq N$. By <Completeness of Lebesgue measure>, every subset of $N$ is measurable, contradicting the hypothesis on $A_1$. Therefore
$$
\boxed{A_1\times A_2\text{ is not Lebesgue measurable}.}
$$
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