# a sequence of integrables functions that converge pointwise and it's dominated

Let $f_n\colon [0,\infty) \to \mathbb{R}$ be a sequence of functions and let $g\colon [0,\infty) \to \mathbb{R}$ be such that $\left| {f_n \left( x \right)} \right| \leqslant \left| {g\left( x \right)} \right|\,$ for every $x$ and $n$. Suppose in addition that $\int\limits_0^\infty \! {f_n } \left( x \right) \, dx$ and $\int\limits_0^\infty \! {g\left( x \right) \, dx}$ exist.

It's true that if $f_n \to 0$ pointwise, then $\int\limits_0^\infty f_n\, dx \to 0$?

This is a calculus course. When we say integrable, I mean in the Riemann sense. I don't know anything about the Lebesgue integral, and I can't use it.

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Hint: Dominated Convergence Theorem (sometimes also called Lebesgue's Dominated Convergence Theorem). –  William May 25 '12 at 6:05
I think the point was to use Riemann integration. –  copper.hat May 25 '12 at 6:07
There are proof of that theorem, that not use nothing special? ( not results of measure theory, only a proof for the real numbers with integrability in the riemann sense) –  Matias May 25 '12 at 6:11
Arzela's dominated convergence theorem might apply. Although this assumes $|f_n|$ are uniformly bounded. –  copper.hat May 25 '12 at 7:14
Matias, what is the context of this problem? What tools do you have available to you? –  Antonio Vargas May 25 '12 at 8:02

No, it's not true if you only assume $\int_0^\infty g(x)\ dx$ (rather than $\int_0^\infty |g(x)|\ dx$) exists. Consider $g(x) = 2 \cos(x^2) - \sin(x^2)/x^2$ (with $g(0) = 1$), noting that $\int_0^t g(x)\ dx = \sin(t^2)/t$ so $\int_0^\infty g(x)\ dx = \lim_{t \to \infty} \sin(t^2)/t = 0$. However, $\int_{\sqrt{(n-1/2) \pi}}^{\sqrt{(n+1/2) \pi}} g(x)\ dx \approx \dfrac{2 (-1)^n}{\sqrt{n\pi}}$. So take $$f_n(x) = \cases{|g(x)| & for \sqrt{(n-1/2)\pi} \le x \le \sqrt{(n^2 -1/2) \pi}\cr 0 & otherwise\cr}$$
@Matis thats not what I meant. I meant the functions are Riemann integrable AND the limit is Riemann integrable (since it is $0$) AND all of those Riemann integrals coincide with their Lebesgue counterparts AND the DCT applies to Lebesgue integrals. All this together implies that the Riemann integrals converge to $0$. –  user12014 May 25 '12 at 7:18