# Sequence of integrable functions on [0,1]

1. Aug 9, 2010

### 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?

Last edited: Aug 9, 2010
2. Aug 9, 2010

### 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.

Last edited: Aug 9, 2010
3. Aug 10, 2010

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.

Last edited: Aug 10, 2010
4. Aug 10, 2010

### AxiomOfChoice

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: $1 = \int f_n \leq \int f$?

Last edited: Aug 10, 2010
5. Aug 10, 2010

It sure does.