If $f$ is in $L^{p}$, prove that $\lim \int_{x}^{x+1} f(t) dt = 0$ 
If $f$ is in $L^p$, prove that $\lim_{x \to \infty} \int_{x}^{x+1} f(t) dt = 0$

It is easy to think that integration must be vanish as $x \to \infty $ but I cannot write them with math. 
Suppose not. Then there exist $P$ such that if $p\geq P$ then $\int_{p}^{p+1} f(t)dt >0$. So with $p_1=P$ and $p_k=p_{k+1}-1$, 
$\lim_{n \to \infty}\sum_{k=1}^{n}\int_{p_k}^{p_{k+1}}f(t)dt$ should be $>\infty$(I think)
Please help. Thank you!
 A: Thanks to zhw for catching an oversight.
Write $\mathbb{R} = \bigcup_n [n,n+1)$ and then
$$\int_\mathbb{R} |f|^p = \sum_n \int_{[n,n+1)} |f|^p,$$ and
hence $$\int_{[n,n+1)} |f|^p \stackrel{n\to\infty}{\longrightarrow} 0.$$
If $p=1$, we are finished, otherwise suppose $p>1$.
Since $$\left(\int_{[a,b)} |f|\right)^p = \left(\int_{[a,b)} |f|1\right)^p \le \int_a^b |f|^p  \left( \int_a^b 1^{p \over p-1} \right)^{p-1},$$ we have
$$\int_{[n,n+1)} |f| \le \int_{[n,n+1)} |f|^p .$$
A: Use Holder to see 
$$(1)\,\,\,\,|\int_x^{x+1}f\,| \le \int_x^{x+1}|f| = \int_x^{x+1}|f|\cdot 1 \le (\int_x^{x+1}|f|^p)^{1/p}\cdot (\int_x^{x+1}1)^{1/q}=(\int_x^{x+1}|f|^p)^{1/p}.$$
Because $f\in L^p, \int_x^{x+1}|f|^p \to 0$ by the dominated convergence theorem. That shows $(1) \to 0$ as desired.
A: You haven't got the correct negation of the limit hypothesis. Remember that the opposite of 

$f(x) \to b$ as $x \to \infty$ if for every $\varepsilon>0$ there is an $n$ such that $\lvert f(x)-b \rvert < \varepsilon$ for all $x>n$

is

$f(x) \not\to b$ as $x \to a$ if there is an $\varepsilon>0$ such that for every $n$ there is an $x>n$ for which $\lvert f(x)-b \rvert > \varepsilon$.

In other words, for some "bad $\varepsilon$", I can find "bad $x$" as large as I like, which don't do the thing I want.
(Remark: this is rather difficult to think about. It's easier to think to yourself "I can make a sequence $(x_n)$ so that $f(x_n)$ either doesn't converge, or converges to something other than $b$.")
In this case, suppose the limit is not zero. Then there is $\varepsilon>0$ such that
$$ \left\lvert \int_x^{x+1} f \right\rvert > \varepsilon \tag{1} $$
for $x$ arbitrarily large. Hence there is an infinite increasing sequence $(x_n)_{n=0}^{\infty}$ for which $x_{n+1}-x_n>1$ (so the intervals of integration are disjoint), and (1) holds. But then
$$ \infty > \frac{1}{\varepsilon^p}\int_{-\infty}^{\infty} \lvert f \rvert^p \geqslant \sum_{n} \int_{x_n}^{x_{n+1}} \left\lvert \frac{f}{\varepsilon} \right\rvert^p  $$
Now, Hölder's inequality says
$$ \int_A \lvert f \rvert \leqslant (\mu(A))^q \int_A \lvert f \rvert^p, $$
so the next inequality is
$$ \sum_{n} \int_{x_n}^{x_{n+1}} \left\lvert \frac{f}{\varepsilon} \right\rvert^p \geqslant 1^{-q}\sum_{n} \frac{1}{\varepsilon} \int_{x_n}^{x_{n+1}} \lvert f \rvert \geqslant\sum_{n} \frac{1}{\varepsilon} \left\lvert \int_{x_n}^{x_{n+1}} f \right\rvert > \sum_n 1 = \infty, $$
which gives a contradiction.
(This generalises to any measurable dissection of $\mathbb{R}^n$ in the same way, for example. Or, indeed, any sufficiently non-degenerate measure space with infinite measure.)
