# Equivalent definition of almost uniform convergence

A sequence of measurable functions $f_1,f_2,...$ on the finite measure space $(X, \mathcal{B}, \mu)$ converges almost uniformly to the measurable function $f$ if for all $\epsilon>0$, there exists a measurable set $F_\epsilon$ such that $\mu(F_\epsilon)< \epsilon$ and $\sup_{x\in X \setminus F_\epsilon} |f_n(x) - f(x)| \to 0$ as $n \to \infty$.

Does the following equivalence hold?

$f_n \to f$ almost uniformly if and only if for all $\epsilon_1, \epsilon_2 > 0$ there exists $n(\epsilon_1, \epsilon_1) \in \mathbb{N}$ such that $$\mu\big(\big\{x: \sup_{j \geq n(\epsilon_1, \epsilon_2)}|f_j(x) - f(x)| > \epsilon_2 \big\}\big) < \epsilon_1 \tag{1}$$

If these two are indeed equivalent, then (1) seems like the more transparent definition, as it is obvious that (1) implies convergence in measure. I would wonder, then, why (1) isn't the more standard definition. Anyway, here is my attempt.

Suppose $f_n \to f$ almost uniformly and let $\epsilon_1, \epsilon_2 > 0$ be given. Take $\epsilon = \epsilon_1$ in the definition of almost uniform convergence, and get $\mu(F_\epsilon)< \epsilon$ and uniform convergence on $X \setminus F_\epsilon$. Uniform convergence on $X \setminus F_\epsilon$ implies that there exists $n=n(\epsilon_1, \epsilon_2)$ such that for all $j \geq n$ and all $x \in X \setminus F_\epsilon$, $|f_j(x) - f(x)| \leq \epsilon_2$. Hence, $F_\epsilon$ contains precisely those $x \in X$ such that $|f_j(x) - f(x)| > \epsilon_2$ for some $j \geq n$, which is to say $\sup_{j \geq n}|f_j(x) - f(x)| > \epsilon_2$. Therefore (1) holds.

For the converse, suppose (1) holds and let $\epsilon > 0$ be given. I have to find a set $F_\epsilon$ that satisfies the conditions in the definition of uniform almost sure convergence. Let $\epsilon_k = \epsilon/2^k$ and for all $k \in \mathbb{N}$ use (1) to get $$\mu\big(\big\{x: \sup_{j \geq n(\epsilon_k, 1/k)}|f_j(x) - f(x)| > 1/k \big\}\big) < \epsilon_k.$$ Then, $$\mu\big(\cup_k\big\{x: \sup_{j \geq n(\epsilon_k, 1/k)}|f_j(x) - f(x)| > 1/k \big\}\big) < \sum_k\epsilon_k = \epsilon.$$ Setting $F_\epsilon = \cup_k\{x: \sup_{j \geq n(\epsilon_k, 1/k)}|f_j(x) - f(x)| > 1/k \}$, it follows that $f_n \to f$ uniformly on $X \setminus F_\epsilon$. This is because for all $k$ and all $x \in X \setminus F_\epsilon$, $|f_j(x)-f(x)| < 1/k$ holds whenever $j \geq n(\epsilon_k, 1/k)$. This assertion is just the definition of uniform convergence. Another way of putting this would be: if $x \in X \setminus F_\epsilon$, then for all $k$ and all $j \geq n(\epsilon_k, 1/k)$, $|f_j(x)-f(x)| < 1/k$ holds. Since $n(\epsilon_k, 1/k)$ does not depend on $x$ (it is a function of just $\epsilon$ and $k$), the convergence is uniform in $x \in X \setminus F_\epsilon$.

Let $\epsilon>0.$ For $k=1,2\dots$ there exist $E_k$ and $N_k$ such that $$\mu(E_k)<\epsilon/2^k$$and $$|f_j(x)-f(x)|<1/k\quad(j>N_k,x\in X\setminus E_k).$$
Let $E=\bigcup E_k$. Then $\mu(E)<\epsilon$ and $$|f_j(x)-f(x)|<1/k\quad(j>N_k,x\in X\setminus E),$$which says precisely that $f_n\to f$ uniformly on $X\setminus E$.
• Thanks. There were several typos and silly oversights, including the $\epsilon_k$ definition. Is what I have now correct? – user435571 Mar 30 '18 at 13:38
• @user435571 More seriously: Nobody can evaluate your solution because you haven't given a solution. You need to say why $f_n\to f$ uniformly on $X\setminus F_\epsilon$ before we can say whether you're right. – David C. Ullrich Mar 30 '18 at 14:00
• @user435571 You asked whether what you had was correct - I assumed you wanted to know whether it was correct. Anyway, sure enough you're mangling the quantifiers. The sentence "$X\setminus F_\epsilon$ is precisely[etc]" is not so. In fact it makes very little sense: You can't say "$S$ is the set of all $x$ such that blah blah for all $x\in S$" - when you say "$S$ is the set of all $x$ such that [etc]", the condition [etc] should be a condition on $x$, not a condition involving $x$ and $S$. – David C. Ullrich Mar 30 '18 at 14:25