L^2 convergence implies a.e. pointwise convergence? Since when?

[AxiomOfChoice]

My functional analysis professor made the following assertion the other day: If f_n \to f in the L^2 norm, then there is a subsequence f_{n_k} that converges pointwise almost everywhere to f. This is the first I've heard of that...can someone point me to a proof of this proposition? Does it have a name?

 

[micromass]

To my knowledge, it doesn't have a name. I'll write the proof down for you: (I also assume that you are working in L^p(R)).

DEFINITION: Let f_n be a measurable function. We say that f_n converges in measure to f if for each \epsilon>0

\lambda\{|f_n-f|>\epsilon\}\rightarrow 0

THEOREM: If f_n converges to f in the L²-norm, then the convergence is also in measure.

PROOF: \lambda\{|f_n-f|>\epsilon\}\leq \int_{\{|f_n-f|>\epsilon\}}{1dx}\leq\frac{1}{\epsilon^p}\int{|f_n-f|^p}\rightarrow 0

Now we need some facts about convergence almost everywhere:

THEOREM: The following are equivalent:
(1) f_n converges to f a.e.
(2) \lambda{(\limsup\{|f_n-f|>\epsilon\})}=0

I'll assume you already seen this theorem. Otherwise, I'll prove it.

LEMMA: If for all \epsilon>0 holds that \sum_{n=1}^{+\infty}{\lambda\{|f_n-f|>\epsilon\}}<+\infty, then f_n converges to f a.e.

PROOF \lambda{(\limsup\{|f_n-f|>\epsilon\})}=\lim_{n\rightarrow+\infty}\lambda\left( \bigcup_{m=n}{\{|f_m-f|>\epsilon\}}\right)

This limit is 0, since

\lambda\left( \bigcup_{m=n}{\{|f_m-f|>\epsilon\}}\right)\leq \sum_{m=n}^{+\infty}{\lambda\{|f_m-f|>\epsilon\}}\rightarrow 0


Now we can finally prove the result:

THEOREM Assume that f_n converges to f in measure, then there is a subsequence f_nk that converges to f a.e.

PROOF Choose a subsequence f_nk such that \lambda\{|f_{n_k}-f|>1/k\}\leq 1/k^2. Then

\sum_{k=1}^{+\infty}{\lambda\{|f_{n_k}-f|>\epsilon\}}<+\infty

Thus our subsequence converges a.e.


If something about the previous proof is not clear, then feel free to ask. If you rather want a reference: I got this proof from Billingsley's "Probability and measure". But the proof is a bit spread out, so you will have to do some searching...