In Rudin's proof of Theorem 8.14, which states that convolutions of Lebesgue integrable functions over the real line are Lebesgue integrable, he first proves the result for Borel measurable functions, as opposed to the more general Lebesgue measurable functions. I fail to see why this is necessary -- the proof for the measurability of the product function
$$f(x-y)g(y)$$
over the plane can be shown without that assumption, I think; just note that compositions of measurable functions with continuous functions are measurable, and so are products of pairs of measurable functions. Am I missing something?
You say we should just note that "compositions of measurable functions with continuous functions are measurable". If we're applying that to $f(x-y)$ we must be talking about $f\circ\phi$ where $f$ is measurable and $\phi$ is continuous (as opposed to $\phi\circ f$). But in general $f$ measurable and $\phi$ continuous does not imply that $f\circ \phi$ is measurable! Of course here $f(x-y)$ is measurable, but not for the reason you say it is.
An explicit example of the phenomenon Daniel Fischer mentions:
Say $C\subset[0,1]$ is the middle-thirds Cantor set and $K\subset[0,1]$ is a "fat Cantor set": $K$ is homeomorphic to $C$ but $m(K)>0$. Assume $0,1\in K$. Then there is a homeomorphism $\phi:[0,1]\to[0,1]$ with $\phi(K)=C$.
Now since $m(K)>0$ there is a non-measurable set $E\subset K$. Let $F=\phi(E)$. Then $F$ is measurable since $F\subset C$ and $m(C)=0$. So $\chi_F$ is measurable, while $\chi_F\circ\phi=\chi_E$ is not measurable.