convergence in $L^1$ for product of functions

Question

If $f_n$ converges to $f$ in $L^1$ and $g_n$ converges to $g$ in $L^1$. Does it necessarily mean that $f_ng_n$ converges to $fg$ in $L^1$ for finite measure spaces.

Answer 1

As Did already pointed out, $f \cdot g$ and $f_n \cdot g_n$ are in general not integrable, therefore we need stronger assumptions. Using the ($L^1$-)triangle inequality and Hölder's inequality, it's not difficult to prove the following result:

Let $f_n \to f$ in $L^2$ and $g_n \to g$ in $L^2$. Then $f_n \cdot g_n \to f \cdot g$ in $L^1$.

Answer 2

Of course not. Consider the Lebesgue measure on [0,1] and the functions $f_{n}=g _{n}=\frac{1}{x^{\frac{1}{2}+\frac{1}{n}}}$ for $n>1$