Show that $\int\vert \int f(y)g(x-y)\,dy \rvert \,dx \le \iint \vert f(y)g(x-y)\rvert\, dy\, dx$

Question

I'm reading a theorem as following about the convolution, but didn't understand the first step:

Why is this inequality true? i.e. why is $$\int\left| \int f(y)g(x-y)\,dy \right| \,dx \le \it \vert f(y)g(x-y)\rvert\, dy\, dx$$ Is this a property that we need to prove?

Answer 1

It follows from two properties of integrals:

1. $u \leq v \implies \int u \leq \int v$
2. $\left| \int u \right| \leq \int \left| u \right|$

Answer 2

You always have that $|\int h| \le \int |h|$, so $|\int f(y)g(x-y)dy| \le \int |f(y)g(x-y)|dy$ a.e. in $x$, and integrating respects this monotonicity.