# Convergence of indicator functions series

Say $$E$$ is a measurable set, and $$\{E_k\}$$ is a series of measurable sets defined by

$$E_k \subset E, m(E \setminus E_k) < \frac{1}{k}, k = 1, 2, 3, ...$$

Do their corresponding indicator functions series converge to 1, almost surely?

$$\lim_{k \to \infty} 1_{E_k} = 1, a.s. x \in E$$

If $$E_k \subset E_{k+1}$$, then it would be trivial to prove the above statement with $$A = \bigcap_{k=1}^{\infty} (E \setminus E_k)$$. However, without this condition, I'm getting suspicious about its correctness, but I cannot raise any counterexample either.

If the $$(E_k)$$ are not nested, their indicators need not converge almost everywhere. Let $$E$$ be the unit interval, consider the typewriter sequence and let $$E_k$$ be the set where the $$k$$'th function is zero.
You may need to modify the construction slightly to achieve $$m(E \setminus E_k)<\frac1k$$, but you get the idea. Note the harmonic series diverges, so this modification won't cause the typewriter to "peter out".
What you're describing is sort of a classical example of the fact that a sequence of functions may converge in $$L^P(\mu)$$ for $$1\leq p < \infty$$, but not in $$L^\infty(\mu).$$
Let $$f_k(x)=1_{E_k}(x),$$ $$f(x)=1.$$ Then $$\mu(E\backslash E_k)=\int_E f(x)-f_k(x)\, \mathrm{d}\mu < \frac{1}{k}.$$ Obviously, this shows $$f_k \to f$$ in $$L^1.$$ A similar proof applies in $$L^p.$$ for $$p<\infty.$$
However, consider if $$E=[0,1]$$ and $$f_{n,k}(x)=1_{E\backslash[\frac{k-1}{2^n},\frac{k}{2^n}]},$$ for $$k=1,...,n.$$