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 L
2-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...