Show that F vanishes at infinity. Suppose $1 ≤ p < ∞, f ∈ L^p(R)$, and
$F(x) = \int_{x}^{x+1} f(t) dm(t)$
Prove that F vanishes at infinity.

We know that $\int_R |f|^p < \infty$, then, of course, for any $x, F(x)< \int_x^{x+1} |f|^p < \infty$. But I want to show that not only is it finite, but it goes to zero as $x$ goes to $\infty$.
I was thinking to approach it by way of contradiction. Assume that,
(WLOG assume $f \geq 0$)
$lim_{x \rightarrow \infty} F(x) > 0$ 
$\implies lim_{x \rightarrow \infty} \int_x^{x+1} f(t) dm(t) > 0$
I thought perhaps to approach this with partial sums.. and show that it contradicts that $f \in L^p(R)$. But I have had no luck with this. 
Help? Hints? I would greatly appreciate it!

For $p=1$ this is clear since $\int_x^{\infty} |f| \to 0$. For $p>1$ we have $|\int_x^{x+1} f|=|\int_x^{x+1} (1)f|\leq (\int_x^{x+1}|f|^{p})^{1/p}(\int 1^{q})^{1/q}\leq \int_x^{\infty} |f|^{p} \to 0$. [Here $q=\frac p {p-1}$].
