$E_n$ is a sequence of measurable sets in finite measure space $(X,\mathcal{M},\mu)$. Then the characteristic funtions $\chi_{E_n} $ converges in $L^2(\mu)$ if and only if $\mu(\limsup E_n\setminus \liminf E_n)=0$.
I have solved the necessary part . For the essential part, i,e to prove $\mu(\limsup E_n\setminus \liminf E_n)=0$. we know that $\mu( \liminf E_n)\leq\liminf\mu( E_n)\leq\limsup\mu( E_n)\leq\mu( \limsup E_n)$, if $\mu(E_n)<\infty, \forall n$. I can only get that there is a characteristic function $\chi_E$ st $\lim_{n\rightarrow \infty}\mu(E_n\Delta E)=\lim_{n\rightarrow \infty}\int |\chi_{E_n}-\chi_E|^2=0$. how to show $\mu(\limsup E_n\setminus \liminf E_n)=0$?
Let $X = [0, 1)$ be equipped with the usual Lebesgue measure structure. Define
$$ E_{n,k} = [(k-1)/2^{n}, k/2^{n} ) \quad \text{for } 1 \leq k \leq 2^{n} $$
and re-enumerate this double sequence to form a sequence $\{E_{n}\}$. Then $\chi_{E_{n}} \to 0$ in $L^{2}$ but
$$ \limsup_{n\to\infty} E_{n} = [0, 1) \quad \text{and} \quad \liminf_{n\to\infty} E_{n} = \varnothing. $$
Therefore we have $\mu(\limsup E_{n} \setminus \liminf E_{n}) = 1$ and the claim is false.