# Proving pointwise convergence almost everywhere

I would like to prove the following statement:

Let $$(X, F, \mu)$$ be a measure space, not necessarily finite. Suppose that for every $$\epsilon > 0$$ there exists a natural number $$N$$ such that $$\mu(\bigcup_{n = N}^{\infty} \{x \in X : |(f_n(x) - f(x)| > \epsilon \}) < \epsilon$$.

Then $$f_n \to f$$ pointwise almost everywhere.

My idea was to prove the statement by contradiction, i.e. assuming the negation of pointwise convergence almost everywhere:

Suppose there exists an $$\epsilon'$$ such that for all $$N$$ there exists an $$n \geq N$$ with $$|f_n(x) - f(x)| > \epsilon$$ for almost every $$x \in X$$.

By assumption we know that for any $$\epsilon$$, in particular for $$\epsilon'$$, we can find an $$N_0$$ such that for all $$n \geq N_0$$ we have $$|f_n(x) - f(x)| > \epsilon'$$ for almost every $$x \in X$$.

I thought this would imply that

$$\mu(\bigcup_{n = N_0}^{\infty} \{x \in X : |(f_n(x) - f(x)| > \epsilon' \}) = \mu(X)$$,

but even if that were the case, I do not see a way to reach a contradiction as I cannot assume that $$\mu(X) > \epsilon$$, I've probably made a mistake, but I cannot see where.

Is there any way I could complete this proof, or is there a better way?

Thank you in advance.

• Recall that we only care about small $\epsilon$, so you certainly can assert that $\epsilon < \mu(X)$ for all relevant cases. You only need a contradiction for some $\epsilon$. Regardless, you might find the direct route more profitable. Let $A$ be the set of points where there is not pointwise convergence, and $B_N$ be that set in the hypothesis. What is the relation between these sets? What does the size of the sets say about the size of $A$? Oct 6, 2018 at 21:26

Since OP asks for an alternative way, I give a constructive approach. I prefer this since we see the truth along the proof. Moreover, the skill of converting an uncountable union into a countable one is often re-used.

For any fixed $$\epsilon' > 0$$,

\begin{align} & \{x \in X : f_n(x) \not\to f_n(x) \} \\ =& \{x \in X : \exists \epsilon > 0, \forall N \in \Bbb{N}, \exists n \ge N, |f_n(x) - f(x)| > \epsilon \} \\ =& \bigcup_{\epsilon > 0} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon \}\\ =& \bigcup_{k \in \Bbb{N}} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}. \end{align}

Show that the last equality is true:

• $$\subseteq$$: take $$k$$ large enough so that $$\epsilon > \epsilon'/2^k$$
• $$\supseteq$$: for any $$k \in \Bbb{N}$$, choose $$\epsilon$$ sufficiently small so that $$\epsilon'/2^k > \epsilon$$

For each $$k \in \Bbb{N}$$, invoke the given condition (with $$\epsilon = \epsilon'/2^k$$) to find $$N_k \in \Bbb{N}$$ so that

$$\mu\left(\bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}\right) < \epsilon'/2^k.$$

It's not hard to check that $$\bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \subseteq \bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \},$$ from which we get our desired conclusion

\begin{align} & \mu\left(\bigcup_{\epsilon > 0} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon \}\right) \\ =& \mu\left(\bigcup_{k \in \Bbb{N}} \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \}\right) \\ \le& \sum_{k \in \Bbb{N}} \mu\left( \bigcap_{N\in\Bbb{N}} \bigcup_{n \ge N}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \right) \\ \le& \sum_{k \in \Bbb{N}} \mu\left( \bigcup_{n = N_k}^{\infty} \{x \in X : |f_n(x) - f(x)| > \epsilon'/2^k \} \right) \\ \le& \sum_{k \in \Bbb{N}} \epsilon'/2^k = \epsilon'. \end{align}

Since $$\epsilon' > 0$$ is arbitrary, we conclude that $$\mu(\{f_n \not\to f\}) = 0$$, i.e. $$f_n \to f$$ a.e.