I'd like to know if the following statement is true:
Let $\mu$ be a probability measure and $(A_n)_{n\geq1}$ a sequence of measurable sets. Then, there exists a measurable set $A$ and a subsequence $(A_{n_k})_{k\geq1}$ such that $\chi_{A_{n_k}}$ converges to $\chi_A$ in the $L^2(\mu)$ norm.
I think a have a proof, which I present below, but I am not sure if it is correct.
Proof: First, we note that, since $\mu$ is a probability measure, the sequence $(\chi_{A_n})_{n\geq1}$ is bounded in the $L^2(\mu)$ norm. Hence, there exists a weak-convergent subsequence $(\chi_{A_{n_k}})_{k\geq1}$. Let $f$ be the weak limit.
We claim that $f = \chi_A$ for some measurable set $A$. To prove this, it is enough to show that $f$ assumes only the values $0$ or $1$ almost surely. Let $\varepsilon > 0$ and $C_\varepsilon = \{x : f(x) \geq 1+\varepsilon\}$. Then, $C_\varepsilon$ is measurable and so, by the weak convergence, $$ \int \chi_{A_{n_k}} \chi_{C_\varepsilon} d\mu \longrightarrow \int_{C_\varepsilon} f d\mu \geq (1+\varepsilon) \mu(C_\varepsilon); $$ therefore, $$\mu(C_\varepsilon) \geq \lim_k \mu(A_{n_k} \cap C_\varepsilon) = \int \chi_{A_{n_k}} \chi_{C_\varepsilon} d\mu \geq (1+\varepsilon)\mu(C_\varepsilon). $$ This is only possible if $\mu(C_\varepsilon) = 0$, that is, if $f \leq 1+\varepsilon$ almost surely. Hence, $f \leq 1$ almost surely.
A similar argument shows that $f \geq 0$ a.s. We conclude that $$ 0 \leq f \leq 1 \enspace \text{almost surely.} $$ We are going to prove that $f = \chi_A$, where $A = \limsup_{k\to+\infty} A_{n_k} = \bigcap_{\ell \geq 1} \bigcup_{k > \ell} A_{n_k}$.
Let $B_k = \bigcup_{\ell > k} A_{n_k}$. The weak convergence gives $$ \int_{B_\ell} f d\mu = \lim_{k\to+\infty} \mu(A_{n_k} \cap B_\ell) = \lim_{k\to+\infty} \mu(A_{n_k}) =\!: \delta, $$ where in the last step we used that $A_{n_k} \subseteq B_\ell$ for all big enough $k$. Now, by the weak convergence, $\int f d\mu = \delta$, so $$ \int_{B_\ell} f d\mu = \int f d\mu = \delta. $$ Being this valid for all $\ell$, we get, as $\mu$ is of probability, that $$ \int_A f d\mu = \int d\mu = \delta. \qquad (1) $$ Since $0 \leq f \leq 1$, we deduce that $f = 0$ in $A^c$.
Observe now that Faout's Lemma implies that $\mu(A) \leq \limsup_{k\to+\infty} \mu(A_{n_k}) = \delta$ EDIT: there is an error here. Combining this with (1) yields that $f = 1$ in $A$. We conclude that $f = \chi_A$.
We now use the claim to prove the initial statement. First, observe that $$ \|\chi_A - \chi_{A_{n_k}}\|_{L^2(\mu)} = \int |\chi_A - \chi_{A_{n_k}}| d\mu = \mu(A \triangle A_{n_k}) \\ = \mu(A^c \cap A_{n_k}) + \mu(A) - \mu(A \cap A_{n_k}) = \int \chi_{A^c} \cdot \chi_{A_{n_k}} d\mu + \mu(A) - \int \chi_{A} \cdot \chi_{A_{n_k}} d\mu. $$ Then, as $\chi_A = f$, the weak convergence gives that $$ \|\chi_A - \chi_{A_{n_k}}\|_{L^2(\mu)} \longrightarrow \int \chi_{A^c} \cdot \chi_A d\mu + \mu(A) - \int \chi_{A} \cdot \chi_A d\mu = 0. $$
The statement is false. Here there is a counterexample.
We consider the measure space $([0,2\pi], \mu)$, where $\mu$ is the Lebesgue measure. Let $A_n = \{x \in [0,2\pi] : \sin(nx) > 0\}$. Then, for any interval $I$, we have that $$ \lim_{n\to+\infty} \int_{[0,2\pi]} \chi_{A_n} \cdot \chi_I d\mu = \lim_{n\to+\infty} \int \frac{1}{2} \cdot \chi_I d\mu = |I|/2, $$ that is, the function $f \equiv 1/2$ is the weak limit of $(\chi_{A_n})_{n\geq1}$. This implies that $f$ is the only possible $L^2$ limit of a subsequence of $(\chi_{A_n})_{n\geq1}$. But $\|f - \chi_{A_n}\|_{L^2(\mu)} = 1/2$, so $f$ is not a $L^2$ limit of such a subsequence.