Let $\lambda$ be the lebesgue measure on the Borel sets. Suppose $\lambda(A) < \infty$
Is the following true? $\lambda(B_n) \uparrow \lambda (A), B_n \subseteq A \implies I_{B_n} \to I_A$ a.e.
I think it should be true. I cannot apply monotone convergence theorem because that would be the other implication.
I tried rewriting $\lambda(B_n) = \int I_{B_n}d\lambda$ and do something with that but that didn't work either.
Any hints?
Take $B_n= [n,2n]$ and $A=(-\infty,\infty)$ to get a counterxample in the infinite case.
If you assume $A$ to be finite, construct a counterexample as follows. Let $A=[0,1]$
Step 1: Divide $A$ into halves. $B_1$ will be $[0,1/2]$, then $B_2$ will be $[1/2,1]$. Notice how each has measure 1/2.
Step 2: Divide $A$ into quarters. $B_3$,$B_4$,$B_5$ and $B_6$ will each have measure 3/4 by picking exactly three of these quarters.
Step 3: Divide $A$ into eighths. The next 8 $B_n$'s will be formed by choosing exactly seven of these eighths. These each have measure 7/8.
and so on... The measure of $B_n$ approaches 1. The indicators do not converge, since every point will be 0 infinitely often.