Show that convolution of two measurable functions is well-defined

Solution 1:

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 $x$.

Solution 2:

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.