TITLE: Show that convolution of two measurable functions is well-defined
QUESTION [11 upvotes]: Question: 

Recall the definition of the convolution of $f$ and $g$ given by
  $$(f*g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy.$$ If  we only know that
  $f$ and $g$ are measurable, can we show that $f*g$ is well defined for
  a.e. $x$, that is, $f(x-y)g(y)$ is integrable?
(Exercise 2.5.21(c) in 'Real Analysis', by Stein and Shakarchi)

Actually, the book writes like this:

Suppose that $f$ and $g$ are measurable functions on $\mathbb{R}^d$.
(a)Prove that $f(x-y)g(y)$ is measurable on $\mathbb{R}^{2d}$.
(b)Show that if $f$ and $g$ are integrable on $\mathbb{R}^d$, then
  $f(x-y)g(y)$ is integrable on $\mathbb{R}^{2d}$.
(c)Recall the definition of the convolution of $f$ and $g$ given by
  $$(f*g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy$$ Show that $f*g$ is well
  defined for a.e. $x$, that is, $f(x-y)g(y)$ is integrable.

Can we use the assumption that $f,g$ are integrable in (c)?

REPLY [2 votes]: The assumption "$f$ and $g$ are integrable" appears in (b). I guess this is why the author (carelessly) omits it in (c). Without such assumption, e.g. $f=g=1$ on ${\bf R}^d$, the convolution might be nonsense. 
If $f, g: {\bf R}^d \to {\bf C}$ are absolutely integrable, then by the Fubini-Tonelli theorem, $f(y) g(x-y)$ is absolutely integrable on ${\bf R}^d \times {\bf R}^d$, which by further application of Fubini-Tonelli shows that $f(y) g(x-y)$ is absolutely integrable in $y$ for almost every $x$.