# dominated convergence theorem for convergence in measure

• Dec 5th 2009, 04:06 AM
ramdayal9
dominated convergence theorem for convergence in measure
hi,
Ive just read this on wikipedia:
"(X, M, μ) - measure space. If μ is σ-finite, Lebesgue's dominated convergence theorem also holds if almost everywhere convergence is replaced by convergence in measure."
How could i go about proving this?

I know that if a sequence f_n --> f in measure, then there is a subsequence which converges to f a.e. I can apply the DCT on this subsequence, but how would i show the it works for the whole sequence? Also, how would i use the fact that μ is σ-finite?
Thank you
• Oct 1st 2014, 11:56 AM
Juju
Re: dominated convergence theorem for convergence in measure
I think you have the right idea. The point is that every subsequence has in turn a subsequence that converges to f a.e.
• Oct 1st 2014, 12:11 PM
Juju
Re: dominated convergence theorem for convergence in measure
The measure $\displaystyle \mu$ is $\displaystyle \sigma$-finite iff every subsequence has in turn a subsequence that converges a.e.