Question: Is it true that if two functions are not $ L^1$ functions, then their product is not a $ L^1$ function? How can we prove it?

Answer: 
I don't think so. Let $\theta(x)=\cases{1 \quad x>0\\0\quad x\leq0}$

we have $\int dx\, \theta(\pm x)=\infty$, but $\theta(x)\theta(-x)=0$ which is clearly $L^1$
No.

Take $f(x)=\frac{1}{x}1\_{[1,+\infty)}$ and $g(x)=\frac{\sin{x}}{x}1\_{[1,+\infty)}$

$f,g$ are not in $L^1$ but $fg$ is.