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!
 A: Let $q$ be the conjugated exponent of $p$.
Hölder inequality on $[x,x+1]$ and Lebesgue dominated convergence theorem imply
\begin{align}
|F(x)| &\le \int_{x}^{x+1} |f| \\
&\le \left(\int_x^{x+1} |f|^p\right)^{1/p}\left(\int_x^{x+1} 1\right)^{1/q} \\
&=\left(\int_{\langle -\infty, x+1]}|f|^p - \int_{\langle -\infty, x]}|f|^p\right)^{1/p} \\&\xrightarrow{x\to\infty} \left(\|f\|_p^p - \|f\|_p^p\right)^{1/p} \\
&= 0
\end{align}
so $\lim_{x\to\infty} F(x) = 0$.
A: $$
\|f\|_p^p
\ge\sum_{k=0}^\infty\int_k^{k+1}|f(x)|^p\,\mathrm{d}x\\
$$
Since $f\in L^p(\mathbb{R})$, the sum converges; therefore, the terms must tend to $0$:
$$
\begin{align}
\lim_{k\to\infty}\int_k^{k+1}|f(x)|\,\mathrm{d}x
&\le\lim_{k\to\infty}\left(\int_k^{k+1}|f(x)|^p\,\mathrm{d}x\right)^{1/p}\\[6pt]
&=0
\end{align}
$$
Thus,
$$
\begin{align}
\lim_{x\to\infty}|F(x)|
&\le\lim_{x\to\infty}\int_{\lfloor x\rfloor}^{\lfloor x\rfloor+1}|f(x)|\,\mathrm{d}x+\lim_{x\to\infty}\int_{\lfloor x+1\rfloor}^{\lfloor x\rfloor+2}|f(x)|\,\mathrm{d}x\\[6pt]
&=0
\end{align}
$$
A: 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.
A: 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 $q=\frac p {p-1}$].
