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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*).2012-05-25
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    I think the point was to use Riemann integration.2012-05-25
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    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)2012-05-25
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    Arzela's dominated convergence theorem might apply. Although this assumes $|f_n|$ are uniformly bounded.2012-05-25
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    Matias, what is the context of this problem? What tools do you have available to you?2012-05-25
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    @Matias Could you add the assumption that $\lvert g(x) \rvert$ is integrable as well?2012-05-25

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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?2012-05-25
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    @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.2012-05-25
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    @PZZ "since the functions are Riemann integrable, the limit is Riemann integrable". But this is not true.2012-05-25
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    @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 $0$.2012-05-25