Problem: Show that $$f \in L^1(\Bbb R) \Longleftrightarrow g \in L^1 (\Bbb R \times \Bbb R)$$ where $g(x,y)=f(x+y)f(x).$
Here is my solution, but there is one part that I believe I am convincing. In order to apply Tonelli we have to have $g$ to be measurable and nonnegative. In my answer I said $g$ is measurable because it is product of two measurable sets. This is true on a space but I am not sure it is true on a product spaces Any help how to prove $g$ is measurable? Thanks
$"\Rightarrow"$ Let $f \in L^1(\Bbb R)$ which means that $f$ is integrable. First $g$ is a measurable function as a product of two measurable functions. By Tonelli theorem we have \begin{align*} \int_{\Bbb R}\int_{\Bbb R} |g(x,y)| dy dx &= \int_{\Bbb R}\int_{\Bbb R} |f(x+y) f(x)| dydx\\ &= \int_{\Bbb R} |f(x)|\Big[ \int_{\Bbb R} f(x+y)|dy \Big] dx\\ &=\int_{\Bbb R} |f(x)| \int_{\Bbb R} |f(y)|dy dx\\ &= \int_{\Bbb R}|f(x)|\cdot \|f\|_{L^1}\\ &= \|f\|_{L^1} \cdot \|f\|_{L^1}<\infty \end{align*} therefore $g \in L^1(\Bbb R \times \Bbb R)$. \
$"\Leftarrow"$ Consider now that $g \in L^1(\Bbb R \times \Bbb R)$ which means that $g$ is integrable on $\Bbb R \times \Bbb R$. By Fubini theorem we have \begin{align*} \int_{\Bbb R}\int_{\Bbb R} |g(x,y)|dydx & = \int_{\Bbb R}\int_{\Bbb R} |f(x) f(x+y)| dy dx\\ &= \int_{\Bbb R} |f(x)| \Big[\int_{\Bbb R} |f(x+y)dy\Big] dx \end{align*} therefore $\int_{\Bbb R} |f(x+y)|dy $ is finite for a.e. $x$. Since Lebesgue measure is translation invariant we have $$\int_{\Bbb R} |f(x+y)| dy = \int_{\Bbb R} |f(x)| dy \text{ so } f \in L^1$$