I am trying to prove or disprove this statement :
For $\phi$ a finite nonnegative measure on $[0,1]$ and $\left(\mathbf A_n\right)_{n\in \mathbb N}$ a sequence of measurable subsets of $[0,1]$, there is a subsequence $\left(n_k\right)_{k \in \mathbb N}$ and a subset $\mathbf A \subset [0,1]$ such that
$$\lim_{k \to \infty}\int_{[0,1]} \left|1_{\mathbf A_{n_k}}(x) - 1_{\mathbf A}(x)\right|\mathrm d \phi(x) = 0.$$
I am struggling with proving this statement, can any one help me to do it?
Hint. Let $\phi = \operatorname{Leb}$ be the Lebesgue measure restricted to $[0, 1]$ and consider
$$\mathbf{A}_{n} = \{x \in [0,1] : \lfloor 2^n x\rfloor \equiv 0 \text{ (mod 2)} \}.$$
If $\mathbf{1}_{\mathbf{A}_{n_k}} \to \mathbf{1}_{\mathbf{A}}$ in $L^1$ for some subsequence $(n_k)$ and a measurable set $\mathbf{A}$, then for any $f \in C_{b}([0,1])$ we have
$$ \int f(x) \mathbf{1}_{\mathbf{A}_{n_k}} \, \mathrm{d}\phi \xrightarrow[k\to\infty]{} \int f(x) \mathbf{1}_{\mathbf{A}} \, \mathrm{d}\phi = \int_{\mathbf{A}} f(x) \, \mathrm{d}x. $$
On the other hand, we can easily check that
$$ \int f(x) \mathbf{1}_{\mathbf{A}_{n}} \, \mathrm{d}\phi \xrightarrow[n\to\infty]{} \frac{1}{2} \int_{[0, 1]} f(x) \, \mathrm{d}x. $$
Now argue that there is no such $\mathbf{A}$ satisfying
$$ \frac{1}{2} \int_{[0, 1]} f(x) \, \mathrm{d}x = \int_{\mathbf{A}} f(x) \, \mathrm{d}x, \qquad \forall f \in C_b([0,1]). $$