Any Sequence of a Measurable Sets Has a Convergent Subsequence?

Question

For each $n\geq 1$, let $A_n$ be a measurable (Borel) set in the unit interval $I=[0, 1]$.

Does there necessarily exist a subsequence $(A_{n_k})_{k=1}^\infty$ of $(A_n)_{n=1}^\infty$ such that the characteristic functions $\chi_{A_{n_k}}$ converge pointwise to $\chi_A$ for some measurable subset $A$ of $I$?

Answer 1

Let $A_n$ be the union of the intervals $[j/2^n, (j+1)/2^n)$ for even values of $j$, $0\le j<2^n$. Define $$r_n=2\chi_{A_n}-1,$$so $r_n$ is just the $n$th Rademacher function.

If we had $A_{n_j}\to A$ almost everywhere then $r_{n_j}\to f=2\chi_A-1$ almost everywhere. Dominated convergence then shows that $||r_{n_j}-f||_1\to0$, which is impossible because $||r_n-r_m||_1=1/2$ for $n\ne m$.