# Convergence in L1 and characteristic function

• November 16th 2009, 02:51 PM
Convergence in L1 and characteristic function
If $\mu (E_n) < \infty$ for $n \in \mathbb {N}$, and $1_{E_n} \rightarrow f$ in $L^1$, then f equals to the characteristic function of a measurable set.

Proof so far.

So I know that $\int \mid 1_{E_n} - f \mid d \mu \rightarrow 0$

And I need to find a measurable set, say M, such that $f=1_M$

I apologize for the lack of significant amount of work here, but I just don't really know how to start here, any hints?

Thank you.
• November 18th 2009, 02:21 AM
Shanks
Actually , I think , f equals to the characteristic function of a measurable set, a. e.
Hint :convergence in $L^1$ implies convergence in measure, and convergence in measure implies a subsequence convergence a. e.