Sequence of integrable functions on [0,1]

[AxiomOfChoice]

If you have a sequence of integrable functions \{f_n(x)\} on [0,1] which converges to a function f(x) pointwise for every x\in [0,1] that has the following properties:

(1) 0 \leq f_n(x) \leq f(x) for every n and every x; and

(2) \int_0^1 f_n(x)dx = 1 for every n;

does it necessarily follow that the limit function f is integrable and satisfies \int_0^1 f(x) dx = 1?

I can't think of why this would need to be true using the standard Lebesgue convergence theorems (bounded, monotone, or dominated), since none of them seem to apply. But I can't think of a counterexample to save the life of me. Can anyone help?

 

[disregardthat]

Draw an isoceles at 0.5 with width l_n = 2^{-n}, and height h_n= 2^{n+1} for each n. Let f_n(x) be the curve of the triangle where it occurs except at 0.5, and 0 elsewhere. f_n(x) is surely lebesgue integrable (but not continuous at 0.5), and:
\int^{1}_0 f_n(x) dx = \int^{\frac{1}{2}}_0 f_n(x) dx + \int^{1}_{\frac{1}{2}} f_n(x) dx =\frac{1}{2}l_nh_n=2^{-1-n+n+1}=1
which is obviously true because a continuous function need not be continuous at it's end points to be integrable.

However, this function converges pointwise to f(x) = 0. This is obvious: any point on [0,1] will surely be valued to 0 by f_n for all n > N for some N (you can find this N explicitly for each such point if you want by a simple calculation).
And since f_(1/2) = 0 for all n, f_n(x) \to 0 for all x as n \to \infty.

And surely; \int^1_0 f(x) dx \not = 1.

EDIT: Oh, I mistakingly read the first equation to be 0 \leq f(x) \leq f_n(x). Nevermind then, but I will try to find a solution to this problem as well.

 

[adriank]

Let g_n(x) = \inf_{k \ge n} f_k(x) and h_n(x) = \sup_{k \le n} f_k(x). Then g_n(x) \le f_n(x) \le h_n(x), and (g_n), (h_n) are both increasing sequences, and both converge pointwise to f. By monotone convergence, 1 = \lim \int f_n \le \lim \int h_n = \int f = \lim \int g_n \le \lim \int f_n = 1.

I think it works.edit: The statement also follows immediately and more nicely from Fatou's lemma, so check that out.

 

[AxiomOfChoice]

Let g_n(x) = \inf_{k \ge n} f_k(x) and h_n(x) = \sup_{k \le n} f_k(x). Then g_n(x) \le f_n(x) \le h_n(x), and (g_n), (h_n) are both increasing sequences, and both converge pointwise to f. By monotone convergence, 1 = \lim \int f_n \le \lim \int h_n = \int f = \lim \int g_n \le \lim \int f_n = 1.

I think it works.edit: The statement also follows immediately and more nicely from Fatou's lemma, so check that out.

Thanks for the response! I know Fatou's lemma states \int \liminf f_n \leq \liminf \int f_n, and that in this case, we have \liminf f_n = f and \liminf \int f_n = 1, which gives \int f \leq 1 < \infty, implying f is integrable. Does the other inequality follow from monotonicity of the integral: 1 = \int f_n \leq \int f?

 

[adriank]

It sure does.