If $f(x)g(y)$ is a measurable function, and $f$, $g \in L^{1}(dm)$, does this imply $g(y - x) \in L^{1}(dm)$? Question rephrased
Suppose we are working in $(\mathbb{R}, \Sigma(m^{*}) \times \Sigma(m^{*}), m \times m)$ where $m$ is Lebesgue measure.  Note that our $\sigma$-algebra is not necessarily complete.
Suppose $f$ and $g$ are in $L^{1}(dm)$.  I am trying to understand my analysis notes on convolution, and in one part of the notes, a fact is used that for the functions $f(x)$ and $g(y)$, if the product $f(x)g(y -x)$ is measurable, then not only is $g( y -x )$ measurable, but also in $L^{1}(dm)$.  Why?
 A: Not if nonmeasurable sets exist. Let $h$ be a $0-1$ valued nonmeasurable function. The product of $f = h+1$ and $g = 1/f$ is constant, but $f$ is not measurable.
Your second question must take place on some type of group, not an arbitrary measure space.
A: The answer to your two new questions is yes. If you consider $S_f=\{f^{-1}(E)|E\in\Sigma\}$, then if $g(x)=f(-x)$ and $h(x)=f(x+s)$ then we actually have $S_f,S_g,S_h$ are all measurable iff any one of them is. This is because measurable sets remain measurable when rotated or translated, and in fact the measure stays the same under these actions (assuming we are in a place where those operations can be carried out).
In addition to Umberto's example, we have the trivial example where one of the functions is the zero function. There is a bit of ambiguity about if you mean composition or multiplication, as $fg$ can mean either depending on the context, however the zero function is a counterexample for both meanings, and, if composition is intended, then any constant function is.
