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Let $f:\mathbb{R} \to \mathbb{R}$ be an bounded ($(\exists M>0)(\forall x \in \mathbb{R}) |f(x)| \leq M$) and continuous function $(\forall x \in \mathbb{R})(\forall \epsilon > 0)(\exists \delta > 0)(\forall y \in \mathbb{R})|x-y|<\delta \Longrightarrow |f(x)-f(y)|<\epsilon$ and nonconstant. Does $f$ have to be integrable?

If there is counterexample, please tell me. If there is some theorem that says so, give me a link.

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    It is necessarily *locally* integrable, but not necessarily integrable – 2017-01-13
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    if that helps, since you excluded constant functions, take $f:=\sin$ or $f:=\left|\sin\right|$. – 2017-01-13
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    @Clayton, my bad. Corrected. – 2017-01-13
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    Here is a nonintegrable (on $\mathbb{R}$) function that satisfies your conditions: $f(x)= 10 + \sin(x)$. – 2017-01-13
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    Another nontrivial example is $f(x)=\frac{x^2}{1+x^2}$. – 2017-01-13
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    To be locally integrable, continuity would suffice (not that it is necessary, but the point is that boundedness is superfluous for that result). To be integrable, a necessary but insufficient condition would be that $\liminf\limits_{x\to\infty}|f(x)|=0$. – 2017-01-13

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