By Lebesgue's dominated convergence is integrable, since is a vector space is integrable for all and so is also integrable and and so applying Lebesgue again you get the result.
Hi all,..
i try to solve this problem, but i don't know what is the connection between convergence in measure and the integral. can someone help me. . .
Let be a sequence of function and in measure.
Let be an integrable function such that .
Show that
thks
But hmmm where do you use the assumption that fn converges in measure to f ?
It's not an almost everywhere convergence, so we can't say that |fn-f| goes to 0, almost everywhere...
Are there more precisions about the measure we're working with ? If it's a finite measure, then it's possible to get a proof by splitting the space into {|fn-f|<e} and {|fn-f|>e}, for some given e.
Yeah, I read what I wanted. Anyway, if the measure is finite the result holds, I think, because Lebesgue dominated convergence is still valid in this setting, and I assume this is the case since the nature of the problem clearly suggests that particular theorem (although it could be one of those tricky questions).
Hello,
Here is a solution (not from me)
Let's fix
Let
Because we need n in it...
We know that .
We have , except over (this deals with inclusion of sets), which is a set which measure goes to 0 as n goes to infinity.
By taking the limit, we can say that , except over a set of measure 0.
Thus it is true almost everywhere.
Since this is true for any , we can take and consider that
But the measure of this intersection is the measure of
Hence almost everywhere.
And finally, we can use Lebesgue's dominated convergence theorem.