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Possible Duplicate:
If $\int_0^\infty fdx$ exists, does $\lim_{x\to\infty}f(x)=0$?

Let $ \int_{-\infty}^\infty |f| < \infty$. Then $$ \lim_{x \to \infty} f(x) =0 \;?$$ If this is true, then how can I prove this?

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    Basically the same question as this one: [If $\int_0^\infty fdx$ exists, does $\lim_{x\to\infty}f(x)=0$?](http://math.stackexchange.com/questions/197450/if-int-0-infty-fdx-exists-does-lim-x-to-inftyfx-0)2012-10-13
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    Consider $$f(x) = \begin{cases}1 & \text{if }x \in \mathbb{Q} \\0 & \text{otherwise} \end{cases}$$ for a counterexample.2012-10-13
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    @Marvis Thank you for your comment. But how about the case of $f \in L^1(\Bbb R)$ ?2012-10-13
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    @Ann $f(x)=\sum_{n=0}^\infty 2^{n} \chi_{[n,n+4^{-n}]}$.2012-10-13
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    Marvis's function *is* in $L^1$!2012-10-13
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    If you let $f$ be uniformly continuous,then $\lim_{x\to\infty}f(x)=0$2012-10-13

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Let $$f(x)=\begin{cases} n,&\text{if }n\le x\le n+\frac1{n^3}\text{ for some }n\in\Bbb Z^+\\ 0,&\text{otherwise}\;; \end{cases}$$

then $$\int_{-\infty}^\infty f(x)\,dx=\sum_{n\ge 1}\frac{n}{n^3}=\frac{\pi^2}6\;,$$

but $\limsup\limits_{x\to\infty}f(x)=\infty$. You can replace the steps with ‘tents’ to make $f$ continuous without qualitatively affecting the example; you can even round off the corners to make it arbitrarily differentiable.

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    An easy way to make it $C^\infty$ is to use the same idea, but scale Gaussian functions instead.2012-10-13
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    E.g. $f(x)=\sum\limits_{k=0}^\infty e^{-4^k(x-k)^2}$?2012-10-13
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    Noting that $\liminf\limits_{x\to\infty}f(x)=0$ shows that the limit does not exist. However, if the limit exists...2012-10-13