1. function pointwise convergence

$\{f_n\}$ converges to $f$ pointwise on $E$ if the following equation holds, $$f(x) = \lim_{n\to\infty}f_n(x) , $$ for every $x\in E$.

2. Strong Law of Large Numbers

A sequence of random variables $\{X_n\}$ converges almost surely to a random variable $X$ if for almost every $\omega \in \Omega$, $$\lim_{n\to\infty}{X_n(\omega)}=X(\omega).$$

3. The relation of above

I found out that these two are amazingly similar. When the almost is removed, they become exactly the same.

The first definition is from Principles of Mathematical Analysis Definition 7.1. The second definition is from What is the difference between the weak and strong law of large numbers?.


No it is not equal.

We say that $f_n \to f$ $\mu$-almost everywhere if $\mu\{\omega: f_n(\omega) \not\to f(\omega)\} = 0$.

It is obvious that pointwise convergence implies convergence almost surely, since if $f_n \to f$ pointwise, then $\mu \{\omega: f_n(\omega) \not\to f(\omega)\} = \mu(\emptyset) = 0$.

For a counterexample of the other direction, we work with the lebesgue measure $\mu = \lambda$. Consider your favorite pointwise convergence limit $f_n \to f$ on the reals and change the value of $f$ in your favorite point and denote this adaption of $f$ with $g$. Then $f_n \to g$ almost surely (since singeltons have Lebesgue measure zero) but not pointwise because we destroyed the convergence in the point where we changed $f$.

  • $\begingroup$ So $f_n \to f$ almost everywhere means there can be a measure zero exceptional set, but $f_n \to f$ pointwise means there is no single exceptional point. $\endgroup$
    – dawen
    Jan 20 '20 at 15:35
  • 1
    $\begingroup$ Yes, the convergence is everywhere except on a set of measure zero. $\endgroup$ Jan 20 '20 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.