convergence in measure does not imply weak convergence

Question

Suppose $\sup_n\|f_n\|_1<\infty$ and $f_n\rightarrow f$ a.e.. However it is not necessary that $f_n\rightarrow f$ weakly in $L^1$.

Can someone raise an example?

Thanks in advance.

Answer 1

Try $f_n = n \, \chi_{(0,1/n)}$. The key to weak convergence in $L^1$ is uniform integrability. This does not hold for the above $f_n$.