Can anyone give me a hint with this measure theory problem?

Question

Let (X, M, u) be a measure space (X a set, M a sigma-algebra on X and u a measure defined on M). Suppose f_n is a sequence of measurable functions defined on X and with values on [0, oo]. If lim Integral f_n du = Integral f du < oo (integrals over X), then, for ever set E of M, it's true that lim Int_E f_n du = Int_E f du, the integrals being taken over E. This conclusion may, however, fail if lim Integral f_n du = Integral f du = ool.

I could prove the theorem for the finite case, but as for the infinite one, couldn't find yet an example of a measurable set E for which lim Integral_E f_n du = Integral_E f du does not hold (either because the limit on the left side doesn't exist or exists but is larger than the right hand side. By Fatou's Lemma, it can't be smaller).

Oh ! I forgot to say that lim f_n = f This is another assumption

Answer 1

I think the issue (c.f. thefreevariable's response) is that the convergence doesn't have to be uniform. so for example, let X = (0, ∞) and let f_n = 1/nx. then f_n converges to the constant function f = 0. but if you integrate f_n over, say, the inverval (0,1), you'll get ∞ for all n, whereas you will of course get 0 if you integrate f.

Answer 2

Try f_n(x)=infinity for n Then lim f_n(x)=f=0 for all x,
so Integral f du=0,
but Integral f_n du=infinity for each n,
so lim Integral f_n du=lim infinity=infinity
thus for this example
Integral f du < lim Integral f_n du.

Answer 3

Honestly I would not have guessed that the conclusion fails for the infinite case. But I haven't actually taken measure theory yet, so a couple of the concepts are a little sketchy to me. I'll be interested to see if anyone does find a counterexample.

I'm just thinking back to real analysis where you can always pull the limit out of the integral if f_n converges uniformly.