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If a function $f:[0,\infty)\to\mathbb{R}$ is non-negative, integrable, and uniformly continuous, show that $\underset{x\to\infty}{\text{lim}}f(x)=0$.

My attempt: Let $\epsilon>0$. Since $f$ is integrable, $\int_0^{\infty}f(x)dx<\infty\implies$ \begin{align} 0&=\underset{n\to\infty}{\text{lim}}\int_0^{n+1}f(x)dx-\underset{n\to\infty}{\text{lim}}\int_0^{n}f(x)dx\\ &=\underset{n\to\infty}{\text{lim}}\int_{n}^{n+1}f(x)dx \end{align} So $\exists N$ such that $n\geq N\implies\int_n^{n+1}f(x)<\epsilon$. Pick $x_0$ large enough so $x_0\in [N,N+1]$. Then $x\geq x_0\implies x\in [n,n+1]$ where $n\geq N$. So for $x\geq x_0$, $0\leq f(x)\leq \int_n^{n+1}f(x)dx<\epsilon$. Hence $f(x)\to 0$ as $x\to\infty$.

Does this look correct? Why is uniform continuity assumed?

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    $\begingroup$ Why is $f(x)\leq \int_{n}^{n+1}f(x)\,dx$ if $x\in[n,n+1]$? $\endgroup$ – Thomas Andrews Jul 1 '17 at 21:49
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    $\begingroup$ Note that $f(x)=\max\{0,\max\{\,1-|x-n|^2:n\in\Bbb N\,\}\}$ is non-negative, integrable, and continuous. Yet $\lim_{x\to\infty}f(x)$ does not exist. Do you see where your proof fails for this $f$? $\endgroup$ – Hagen von Eitzen Jul 1 '17 at 21:52
  • $\begingroup$ Possible duplicate. $\endgroup$ – hamam_Abdallah Jul 1 '17 at 22:02
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    $\begingroup$ No, $f$ could be a tall thin triangular spike centered over the midpoint of the interval of almost arbitrary height. $\endgroup$ – zhw. Jul 1 '17 at 22:13
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    $\begingroup$ $\int_{n}^{n+1} f(x)\,dx$ is the "average" of $f(x)$ on $[n,n+1]$. In particular, the only way for $\int_{n}^{n+1}f(x)\,dx\geq f(x)$ for all $x\in[n,n+1]$ is for $f$ to be constant on that interval. @TomChalmer $\endgroup$ – Thomas Andrews Jul 1 '17 at 22:39
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Suppose $\limsup\limits_{x\to\infty}f(x)\ge\epsilon$.

By the uniform continuity, there is a $\delta$ so that if $|x-y|\le\delta$, then $|f(x)-f(y)|\le\frac\epsilon3$

Since $\limsup\limits_{x\to\infty}f(x)\ge\epsilon$, for any $L$, there is an $x\ge L$ so that $f(x)\ge\frac{2\epsilon}3$. Then $$ x-\delta\le t\le x+\delta\implies f(t)\ge\frac\epsilon3 $$ Therefore, $$ \int_{x-\delta}^{x+\delta}f(t)\,\mathrm{d}t\ge2\delta\cdot\frac\epsilon3 $$ Since there is a sequence such $x$'s out to infinity, $$ \int_0^\infty f(t)\,\mathrm{d}t=\infty $$

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