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!

Let $g_n(t) = |f(t)|^p\cdot \chi_{[-n,n]}(t)$. Note that $0\leqslant g_n(t)\nearrow |f(t)|^p$ pointwise a.e. as $n\to\infty$. By the Monotone Convergence Theorem,
$$
\int g_n(t)\,dt \to \int |f(t)|^p\,dt \quad\text{as $n\to\infty$}.
$$
Hence the difference
$$
\int|f(t)|^p\,dt - \int_{-n}^n|f(t)|^p\,dt  \geqslant \int_n^\infty|f(t)|^p\,dt
$$
can be made as small as desired by taking $n$ sufficiently large. In particular, the integral
$$
\int_x^\infty|f(t)|^p\,dt \geqslant \int_x^{x+1}|f(t)|^p\,dt
$$
can be made arbitrarily small by taking $x$ sufficiently large. If $p = 1$, then we are done by the triangle inequality applied to $F(x)$.
If $p > 1$, apply Hölder's inequality
\begin{align*}
|F(x)| &\leqslant \int_x^{x+1}|f(t)|\,dt \leqslant \bigg(\int_{x}^{x+1}|f(t)|^p\,dt\bigg)^{1/p}\bigg(\int_{x}^{x+1}1^q\,dt\bigg)^{1/q}=\bigg(\int_{x}^{x+1}|f(t)|^p\,dt\bigg)^{1/p},
\end{align*}
and the estimate from above.
