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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.

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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".

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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 $k=1,...,n.$

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