Why does L^1 convergence follow from uniform convergence for finite measure spaces? I think I have an idea of how to show it, but I'd like to discuss it.
err what is this in relation to specifically what topic?
Real analysis, measure theory, etc.
is L a random variable??
L^1 is the space of functions that satisfy the following condition: \(\int |f| < \infty\)
ahh sorry this is beyond my scope...apologies
@Princer_Jones perhaps might be able to help u
So, let's start with a finite measure space \((X, \mathcal{M}, \mu)\). In particular this means that \(\mu(X) < \infty\). We want to show that for any sequence of functions \(\{f_n \}_1^\infty\subset L^1(\mu)\) which converges uniformly \(f_n \to f\) to some function \(f\). That we also have \(L^1(\mu)\) convergence and by that I mean \(f \in L^1(\mu)\) and \(\int f_n \to \int f \ \)
My idea is to consider \(f_n - f\) and apply the dominated convergence theorem. Since we have uniform convergence for every \(\varepsilon > 0\) we can certainly find some \(N\) such that for all \(n > N\) we have \(|f_n - f| < \epsilon\). Then, can't we apply the dominated convergence theorem since the constant function \(g(x) = \epsilon\) is surly \(L^1\) for finite measure spaces.
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Of course the argument fails (as it should) for non-finite measure spaces since a constant function would have an infinite integral.
yes this seems the correct way,any sequence {f_n(z)} will be uniform convergent ,if for every given positive numbet (epsilon)>0 ,and for all Z in D and there should be an \[N \in IN~exist~which~is~independent~of~Z~such~that\\|f_n(Z)-f_m(Z)|<\epsilon,\forall n,m \ge N\]
it is cauchys criteria of uniform convergence i think
and this definition also holds when {f_n(x)} is a sequence of real function in a space like L^1
is this a million dollar problem ? :P
No, it is just a problem in Folland Real Analysis: http://www.amazon.ca/Real-Analysis-Modern-Techniques-Applications/dp/0471317160
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