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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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    @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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    @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