3
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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 $A$? $\endgroup$
    – user24142
    Oct 6, 2018 at 21:26

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .