This is a problem that appears in Stein and Shakarchi's Real Analysis text. Below are some results relevant to this problem that appear elsewhere in the text, stated exactly as they're written there.
Corollary 3.7 : Suppose $f$ is a measurable function on $\mathbb{R}^d$. Then the function $f$ defined by $\tilde{f}(x,y) = f(x)$ is measurable on $\mathbb{R}^{d_1} \times \mathbb{R}^{d_2}$.
Proposition 3.9 : If $f$ is a measurable function on $\mathbb{R}^d$, then the function $\tilde{f}(x,y) = f(x-y)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d$.
Property 5 (ii) : If $f$ and $g$ are measurable, then $f + g$ and $fg$ are measurable if both $f$ and $g$ are finite-valued.
Remark on Property 5 (ii) : Property 5 (ii) also holds when $f$ and $g$ are finite-valued almost everywhere.
By Corollary 3.7, both $\tilde{f}(x,y) = f(x)$ and $\tilde{g}(x,y) = g(y)$ are measurable on $\mathbb{R}^d \times \mathbb{R}^d = \mathbb{R}^{2d}$. By Proposition 3.9, the function $\tilde{\tilde{{f}}}(x,y) = f(x-y)$ is measurable on $\mathbb{R}^{2d}$.
We have that $f(x-y)$ and $g(y)$ are each measurable on $\mathbb{R}^{2d}$. It remains to show that their product is measurable on $\mathbb{R}^{2d}$, and this is where my question comes in.
By Property 5 (ii) and the remark that follows, the product of two measurable functions is not, in general, measurable ; we need to know that $f(x-y)$ and $g(y)$ are each either finite-valued or finite-valued almost everywhere. However, in the statement of the problem, the only information that we're given is $f$ and $g$ are measurable on $\mathbb{R}^{2d}$. In general, measurable functions are not finite-valued or finite-valued almost everywhere.
Am I missing something ? Is there an underlying assumption about $f$ and $g$ that I'm not paying attention to ? How can I show that $f$ and $g$ are each either finite-valued or finite-valued almost everywhere to complete the proof ?
Thanks !
To address the comment, we could define $(-\infty)(\infty)=-\infty$, which I'm sure is the usual convention. Also, $0\times \infty = 0\times (-\infty)=0$ and so on. The only issues with arithmetic are things like $\frac{1}{\infty}$, $\frac{1}{0}$ and $\infty-\infty$. As long as we avoid these things, everything else works "nicely".
Property 5(ii) can be generalized as follows:
The proof of course will have to be different. For instance we could proceed directly. Recall the following equivalences for measurability of a function $f:\Bbb{R}^d\to [-\infty,\infty]$:
So, for the case of the product, we have: \begin{align} \{fg=\infty\} &= \bigg(\{f=\infty\}\cap\{0< g\leq \infty\}\bigg) \cup\bigg(\{f=-\infty\}\cap \{-\infty \leq g < 0\}\bigg)\\ & \cup \bigg(\{g=\infty\}\cap\{0< f\leq \infty\}\bigg) \cup\bigg(\{g=-\infty\}\cap \{-\infty \leq f < 0\}\bigg) \end{align} Each of the eight sets on the right is measurable because $f$ and $g$ are. These are all basic set operations hence $\{fg=\infty\}$ is also measurable. Similar reasoning holds for $\{fg=-\infty\}$. To finish the proof, we need to show that for every Borel $U\subset \Bbb{R}$, the set $(fg)^{-1}(U)$ is measurable; but this follows from the finite version.
Indeed, let $\phi=f$ whenever $f$ is finite, and $0$ elsewhere. Define $\gamma$ in terms of $g$ similarly. Then, $\phi, \gamma$ are measurable because $f$ and $g$ are. Then, for every Borel set $U\subset \Bbb{R}$, we have \begin{align} (fg)^{-1}(U) &= (fg)^{-1}(U\setminus \{0\}) \cup (fg)^{-1}(0)\\ &= (\phi\gamma)^{-1}(U\setminus \{0\})\cup f^{-1}(0)\cup g^{-1}(0) \end{align} This is a union of measurable sets hence measurable. This completes the proof that $fg$ is measurable.
Essentially it's just a bit of more case work, but apart from that the result is true. Thus, $(x,y)\mapsto f(x-y)g(y)$ is also measurable.