# Sequence of indicator functions is weakly convergent.

Let $E\subseteq [0,1]$ be a measurable set.

Then we recall that $\chi_E(x)=\begin{cases} 1&\text{ if } x\in E,\\ 0 & \text{ if } x\not\in E.\end{cases}$

We are given a sequence of measurable subsets $\{E_n\}_n$ of $[0,1]$ and a measurable subset $E\subseteq [0,1]$ such that $\chi_{E_n}\rightharpoonup \chi_E$ is weakly convergent in $L^2[0,1]$.

Then prove that $\chi_{E_n}\rightarrow \chi_E$ is strongly convergent in $L^2[0,1]$.

I don't need a complete solution, even an hint is welcomed.

Thank you for your help.

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uhm... maybe try to begin with noticing that $|\chi_A-\chi_B|^2=\chi_{A\triangle B}$. This is a good point to start with I hope.. – uforoboa Aug 17 '12 at 9:34

## 1 Answer

Hint:
$$\begin{eqnarray*} \|\chi_{E_n}-\chi_{E}\|^2&=&\|\chi_{E_n}\|^2-2\langle\chi_{E_n},\chi_{E}\rangle+\|\chi_{E}\|^2\\[5pt] &=&\langle \chi_{E_n}, \chi_{[0,1]}\rangle -2\langle\chi_{E_n},\chi_{E}\rangle+\langle \chi_{E}, \chi_{[0,1]}\rangle\\ \end{eqnarray*}$$

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