Let $(X,\mu)$ be a measure space and suppose that $f,g\in L^1(\mu)$, i.e. $$\|f\|_1=\int_X|f|d\mu<\infty\quad\text{and}\quad\|g\|_1=\int_X|g|d\mu<\infty.$$ How to show that $fg\in L^1(\mu)$?
Attempt: I know how to prove that $$\|fg\|_1 \leq \|f\|_p\|g\|_q$$ for any $1<p,q<\infty$ such that $\frac{1}{p}+\frac{1}{q}=1$ and for $p=1,q=\infty$ and $p=\infty,q=1$. But can we use this to prove that $\|fg\|<\infty$? So we have, for example $$\|fg\|_1\leq \|f\|_1\|g\|_\infty,$$ and $\|f\|_1<\infty$, but I don't think we can say $\|g\|_\infty<\infty$ so this doesn't show anything.
You cannot show $fg\in L^1(\mu)$ because it is not true in general. Let $X = (0, \pi/2)$ and let $\mu$ be the Lebesgue measure on $X$. If $f(x) = \sqrt{\sec(x)}$, then $f\in L^1(X,\mu)$ but $f^2\notin L^1(X,\mu)$.