To show $\lim_{x \to \infty} \int_{x}^{x+1}f(t)dt=0$ if $\lim_{x\to\infty} f(x)=0$

$f:\mathbb{R}\to \mathbb{R}$ is Riemann integrable on any bounded interval and $\lim_{x\to\infty} f(x)=0$.

Define $g(x)=\int_{x}^{x+1}f(t)dt$, we need to show $\lim_{x\to\infty} g(x)=0$.

Please give me hint, I want to try myself. Thank you, I was trying to apply fundamental theorem of calculus, like taking the derivative of $g$ which is $f(x+1)-f(x)$ but then don't know what to do.

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The proof I just gave holds if $f$ is continious – Amr Jan 13 '13 at 10:06
@Amr you should have edited it instead of deleting the whole answer.anyway thank you for response. – Un Chien Andalou Jan 13 '13 at 10:08
I will edit it now – Amr Jan 13 '13 at 10:08
fixed it ${}{}{}{}{}{}{}{}$ – Amr Jan 13 '13 at 10:12
@Amr I don't see what's wrong, i.e. where you require $f$ to be continuous ... – Hagen von Eitzen Jan 13 '13 at 10:13

Hint:

Choose a large number $M$ such that $\forall x>M [|f(x)|<\epsilon]$. Now it follows thatfor all $x>M$:

$$\left|\int_x^{x+1} f(x) \right|\leq \int_x^{x+1} |f(x)|dx \leq \int_x^{x+1} \epsilon \,dx=\epsilon$$

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I guess you could use as well the following (part of the integral MVT):

$$|g(x)|\leq (x+1-x)\cdot \max_{t\in [x,x+1]}|f(t)|\xrightarrow[x\to\infty\Longrightarrow t\to\infty]{}0$$

since $\,t\to\infty\,$ as $\,x\to\infty\,$ and $\,f(x)\xrightarrow[x\to\infty]{}0\,$

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This was my first solution. Integral MVT holds for continuous functions. It is not mentioned in the question that $f$ is continuous . – Amr Jan 21 '13 at 19:38
I shall delete the MVT reference to avoid misunderstandings, but the above's true for functions as given in the OP. Put $$M_x:=\max_{t\in[x,x+1]}|f(t)|\Longrightarrow M_x\xrightarrow[x\to\infty]{}0$$ and from here $$|g(x)|\leq\int\limits_x^{x+1}|f(t)|\,dt\leq M_x\xrightarrow [x\to\infty]{} 0$$ – DonAntonio Jan 21 '13 at 20:18