Why does $\int_{-\infty}^\infty f(x-y)g(y)dy$ exists when $\int_{-\infty}^\infty f(x)dx< \infty$ $\int_{-\infty}^\infty g(x)dx< \infty$ where $f,g$ are maps from $\mathbb{R}$ to $\mathbb{R}_+$
I was hoping to show that $f$ and $g$ are bounded as a result of the assumptions on the integrability of $f$ and $g$ and thus conclude $\int_{-\infty}^\infty f(x-y)g(y)dy<\infty$. But obviously integrability of $f$ and $g$ doesnt imply boundedness but just finiteness. I have spent quite a bit of time thinking about this. I thought about applying Holder, but since $f^2$ and $g^2$ are not necessarily integrable given an integrable $g$ and $f$,it doesnt make any sense. Any hints on how could I show this?
First note that if a measurable function $\phi:\Bbb R\to \Bbb R\cup \{\pm\infty\}$ is integrable, i.e. $\int_{\Bbb R} |\phi(x)|dx<\infty$, then $$ |\phi(x)|<\infty $$ for almost every $x\in\Bbb R$.
Define $\phi(x)=\int |f(x-y)||g(y)|dy$, which takes its values in $[0,\infty]$. By Fubini-Tonelli's theorem, it holds $$ \begin{split} \int \phi(x)dx &=\int\left[\int |f(x-y)||g(y)|dy\right]dx \\ &=\int |g(y)|\left[\int|f(x-y)|dx\right]dy =\|f\|_1\|g\|_1<\infty. \end{split} $$ This implies that $\phi(x)<\infty$ a.e., hence the integral $$(f*g)(x)=\int f(x-y)g(y)dy$$ converges absolutely for almost every $x\in \Bbb R$. This shows almost everywhere finiteness of convolution. Moreover, we can observe that since $$ |f*g|(x)\le \phi(x), $$ it holds $$ \|f*g\|_1\le \|\phi\|_1\le \|f\|_1\|g\|_1. $$ This is a special case of Young's inequality, which is saying that $$ \|f*g\|_r \le \|f\|_p\|g\|_q $$ for $\frac{1}{p}+\frac{1}{q}=1+\frac{1}{r}$, $p,q,r\in [1,\infty]$.