The limit of integral Let $1 \le p < \infty$ and assume $f \in L^p(\mathbb{R})$.
I'm trying to prove the limit of integral
$$\lim_{x \to \infty} \int^{x+1}_x f(t)dt =0.$$
Can I use Riesz Theorem for Banach spaces?
 A: Not a full solution I just provide some hints:


*

*Using Hölder's inequality (or Jensen), we get 
$$\left|\int_{[x,x+1]}f(t)dt\right|^p\leq \int_{[x,x+1]}|f(x)|^pdx,$$
hence we just have to deal with the case $f\in L^1$. 

*Write $\int_{[x,x+1]}f(t)dt=\int_{(-\infty,x+1]}f(t)dt-\int_{(-\infty,x)}f(t)dt$. What is the limit, when $x\to +\infty$, of each term?

A: I would argue that $f(t) \chi_{[x,x+1]}(t)$ is a sequence of functions that converge pointwise to $0$ for $x\rightarrow\infty$. Then use theorem of dominated convergence with upper bound $f$ (assume $f$ is non-negative real-valued function first, then extend to all functions).
A: Note that we need only prove this for real-valued $f\geq 0$, since 
$$\left|\int_x^{x+1}f(x)dx\right|\leq \int_x^{x+1}|f(x)|dx.$$
Let $\epsilon>0$, and choose some simple function $s\leq f$ such that $\int fd\mu<\int sd\mu+\epsilon/2$. Since $s$ is simple and integrable, we have that 
$$s(x)=\sum\limits_{n=0}^k \alpha_n\chi_{A_n}(x)$$
for some collection $\{A_n\}$ of measurable sets, and each $A_n$ has finite measure except $A_0$ (for which $\alpha_0=0$). Thus $A=\bigcup\limits_{n=1}^k A_n$ has finite measure, so we have some interval $I$ such that $\mu(A\setminus I)<\epsilon/2\max\{|\alpha_1|,\ldots,|\alpha_n|\}$. Thus for sufficiently large $x$ we have $(x,x+1)\cap I=\emptyset$ so
$$\int_x^{x+1} f(x)dx<\int_x^{x+1}s(x)dx+\epsilon/2< \max\{|\alpha_1|,\ldots,|\alpha_n|\}\cdot \epsilon/2\max\{|\alpha_1|,\ldots,|\alpha_n|\}+\epsilon/2=\epsilon$$
i.e. we have $\lim\limits_{x\to \infty} \left|\int_x^{x+1}f(x)dx\right|<\epsilon$. Since this is true for all $\epsilon>0$, the limit must be $0$.
