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Given $(\Omega, \mathbb F, \mu)$ a measure space, $(A_n)$ a sequence of measurable sets, $f: (\Omega, \mathbb F) \to (\mathbb R, \mathbb B(\mathbb R))$ an integrable function such that: $\displaystyle \lim_{n\to\infty}\int_\Omega |1_{A_n} - f|d\mu = 0$, where $1_{A_n}$ is indicator function.

1) Show that: $|f| \leq 2, \mu$-a.e.

2) Show that: $\exists A \in \mathbb F: f = 1_A, \mu$-a.e.

3) Show that: if $\displaystyle \sum_{n \geq 0}\mu(A_n \Delta A) < \infty$, then $1_{A_n} \to 1_A, \mu$-a.e.

I did the last question as follows:

As $\displaystyle \sum_{n \geq 0}\mu(A_n \Delta A) < \infty$, applying Borel-Cantelli lemma, we obtain: $\mu (\limsup A_n \Delta A) = 0$. Thus, $$x \in (\limsup A_n \Delta A)^c = \liminf (A_n \Delta A)^c$$ $$\Leftrightarrow \exists n \in \mathbb N: \forall k \geq n, x \in (A_n \Delta A)^c$$ $$\Leftrightarrow \exists n \in \mathbb N: \forall k \geq n, x \in A_n \cap A \text{ or } x \in \Omega \backslash (A_n \cup A)$$ As both cases imply $1_{A_n} = 1_A$, we obtain that $1_{A_n}$ will eventually equal to $1_A$, $\mu$-a.e.

Is my proof correct?

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Yes, your proof is correct.

Alternatively, we can notice that the assumption is $\sum_{n\geqslant 0}\lVert \mathbf 1_{A_n}-\mathbf 1_A\rVert_1\lt \infty$. We can use the fact that if $\sum_{n\geqslant 1}\lVert f_n\rVert_1\lt \infty$, then $\sum_{n\geqslant 1}\lvert f_n\rvert\lt \infty$ almost everywhere hence $f_n\to 0$ almost everywhere.

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