# prove that f is a characteristic function

$( X, \mu)$ is a complete measure space and $E_{n}$ are measurable sets such that $\mu (E_{n}) < \infty$ for all $n$. Let $\chi_{E_{n}}$ converge to $f$ in $L_{1}$. Prove that $f$ is the characteristic function of a measurable set. (almost everywhere)

So far, I proved that $f = lim \chi _{E_{n_{k}}}$ for some sub sequence $E_{n_{k}}$ of $E_{n}$. But I don't know where to go from there. Is the limit of characteristic functions a characteristic function? because i was not able to prove that.

thank you

-
It is false that $f=\lim_{k\to\infty} \chi_{E_{n_k}}$ for some subsequence $n_k$; it is only true a.e. (Corollary 2.32 in Folland) – Zev Chonoles Nov 19 '11 at 6:11

If $a=\lim a_n$, where $a_n=0$ or $a_n=1$ for all $n$, then $a=0$ or $a=1$. Once you have a subsequence $(E_{n_k})$ such that $\lim\limits_{k\to\infty} \chi_{E_{n_k}}(x)=f(x)$ for almost every $x$, this can be applied at each $x$ for which the convergence holds. A function whose only possible values are $0$ and $1$ (a.e.) is a characteristic function (a.e.).
Alternatively, suppose that it is not true that $f(x)=0$ or $f(x)=1$ a.e. Then the set $$\bigcup_nf^{-1}\left[\left(-\infty,-\frac{1}{n}\right)\cup\left(\frac{1}{n},1-\frac{1}{n}\right)\cup\left(1+\frac{1}{n},+\infty\right)\right]$$ has positive measure, so one of the sets in the union has positive measure, which implies that there is a set $E$ of positive measure and a $c>0$ such that $|f(x)|>c$ and $|f(x)-1|>c$ for all $x\in E$. This implies that the $L^1$ distance between $f$ and any characteristic function is at least $c\cdot\mu(E)$.
Where does the assumption that $\mu$ be complete enter the argument? It seems to me that this isn't used anywhere. – t.b. Nov 19 '11 at 6:33
@t.b. I don't believe it's necessary - Folland states this problem on p.63 (#36) without the assumption that $\mu$ is complete, and I also answered it (hopefully correctly) on my homework earlier in the semester without needing to assume it :) – Zev Chonoles Nov 19 '11 at 6:42