TITLE: Convolution of $L_1$ functions is $L_1$
QUESTION [6 upvotes]: Let $f,g\in L_1(\mathbb R, m)$ where m is the Lebesgue measure.
Prove that:
a) $f(x-t)g(t) \in L_1(\mathbb R, m)$ as a fuction of $t$ almost all $x$
b) $h\in L_1(\mathbb R, m)$ where  $h=(f \ast g)(t)=\int_{\mathbb R}f(x-t)g(t)dm(t)$
c)$||h||_{L_1}\le ||f||_{L_1}||g||_{L_1}$

REPLY [6 votes]: Hint: Use Fubini on $\mathbb{R}^2$ for the function $G(x,t)=|f(x-t)g(t)|$.
If you need more help, see p.16 here: http://people.math.gatech.edu/~heil/6338/summer08/section4c_convolve.pdf