Show that convolution of two measurable functions is well-defined 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)?

Hint: do that first when $f, g \ge 0$. Recall that if the integrand function has a sign, you can safely change the order of integration in a double integral. This is sometimes known as Tonelli's theorem. 
Tonelli's theorem is easier than the closely related Fubini's theorem, which regards integrand functions which possibly change sign. In the latter, you need to check that the integrand function is absolutely integrable with respect to both variables before you can do anything. 
