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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 $\epsilon > 0$ : $\lim_{n \rightarrow \infty} \mu ( \cup_{m \geq n} \{x: |f_m(x) - f(x) | \geq \epsilon \}) = 0$ ?

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  • $\begingroup$ 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 $x \in X$.) $\endgroup$ – Michael Nov 11 at 23:58
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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\}$.
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  • $\begingroup$ 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 $n$. $\endgroup$ – Michael Nov 11 at 22:02

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