$f_n \rightarrow f$ almost everywhere if measure is zero

Let $$(\mathcal X, \mathcal A, \mu)$$ be a measure space and $$f, f_n: X \rightarrow \mathbb R, n \in \mathbb N$$ measurble functions.

Let $$\mu(X) < \infty$$. Why does $$f_n \rightarrow f$$ a.e. iff for all $$\epsilon > 0$$ : $$\lim_{n \rightarrow \infty} \mu ( \cup_{m \geq n} \{x: |f_m(x) - f(x) | \geq \epsilon \}) = 0$$ ?

• Some further details on Robert’s answer: If we define $A \subseteq X$ as the set of all $x$ for which convergence occurs; for each $\epsilon>0$ define $R_n(\epsilon) = \cup_{m \geq n} \{x \in X : |f_m(x)-f(x)|\geq \epsilon\}$; and define $B_{\epsilon}$ as in Robert’s answer, then $$R_n(\epsilon) \searrow B_{\epsilon} \quad \mbox{(as n\rightarrow\infty)}$$ and also $$A^c = \cup_{k=1}^{\infty} B_{1/k}$$ On the other hand, a counter-example for $\mu(X)=\infty$ is $X=[0,\infty)$ and $f_n(x) = x/n$, $f(x)=0$ for all $x \in X$. (Then $f_n(x)\rightarrow 0$ for all $x \in X$.) – Michael Nov 11 at 23:58

Think of it this way. "$$f_n(x)$$ does not converge to $$f(x)$$" means that there is some $$\epsilon > 0$$ such that for every $$n$$ there in $$m$$ with $$|f_m(x) - f(x)| \ge \epsilon$$ (this is just a restatement of the $$\epsilon-n$$ definition of convergence). That says the set of $$x$$ for which $$f_n(x)$$ does not converge to $$f(x)$$ is the union of the sets $$B_\epsilon = \bigcap_{n=1}^\infty \bigcup_{m \ge n} \{x: |f_m(x) - f(x)| \ge \epsilon\}$$ Now convince yourself that
1. Instead of taking all $$\epsilon > 0$$ we can take a sequence of $$\epsilon$$ going to $$0$$.
2. $$f_n \to f$$ a.e. is equivalent to all these $$B_\epsilon$$ having measure $$0$$.
3. The measure of $$B_\epsilon$$ is the limit as $$n \to \infty$$ of the measures of $$\bigcup_{m \ge n} \{x: |f_m(x) - f(x)| \ge \epsilon\}$$.
• Perhaps some minor edits are needed: Perhaps in the first math equation you mean a union over $\epsilon$ of the form $\epsilon = 1/k$, i.e., $\cup_{k=1}^{\infty} B_{1/k}$. The third item also suggests a limit as $n\rightarrow\infty$ of an expression that does not depend on $n$. – Michael Nov 11 at 22:02