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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). – user12918723509187 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
up vote 2 down vote accepted

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 $$f_n(x) = \cases{|g(x)| & for $\sqrt{(n-1/2)\pi} \le x \le \sqrt{(n^2 -1/2) \pi}$\cr 0 & otherwise\cr}$$

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And if we take that assumption? – Matias May 25 '12 at 6:36
@Matias In that case it is true, since the functions are Riemann integrable, the limit is Riemann integrable, all of these integrals coincide with their Lebesgue counterparts and the Dominated Convergence Theorem applies to the Lebesgue integrals. I'm afraid I'm not sure how to do it without Lebesgue integration though. – user12014 May 25 '12 at 6:47
@PZZ "since the functions are Riemann integrable, the limit is Riemann integrable". But this is not true. – user17762 May 25 '12 at 6:58
@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

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