Let $1 \leq p <\infty$ and $f \in L^p(\mathbb{R})$. Prove $\lim_{x \to \infty} \int_x^{x+1} f(t) dt = 0$. 
(Jones, p. 246) Let $1 \leq p <\infty$ and $f \in L^p(\mathbb{R})$. Prove $\lim_{x \to \infty} \int_x^{x+1} f(t) dt = 0$.

This seems pretty easy to prove in the following way:
Let $g_j$ be a continuous function of compact support that approximates so that $\| g_j - f \|_{L^1} < 1/2^j$. Then, since the supp(g) is finite, for all $\epsilon$, there exists an $x$ large enough that $\int_x^{x+1} g(t) dt < \epsilon$. Since $g$ approximates $f$ in the L1-norm, we have that $\int_x^{x+1} f(t) dt < 1/2^j + \epsilon \to 0$ as $r,j \to \infty$.
Is this a good approach?
 A: Since $f\in L^p$, $$ C:= \int_{\mathbb{R}} |f|^p <\infty$$
By Holder,
$$ \int_{[c,c+1]} |f| \leq \bigg(
 \int_{[c,c+1]} |f|^p \bigg)^\frac{1}{p}  $$
Hence $$b_K:=\sum_{n=0}^K a_n \leq C,\ a_n:= \bigg(\int_{[n,n+1]} |f| \bigg)^p $$ for any $K$. Hence $\{b_K\}$ is a bounded increasing sequence. Hence it converges so that $a_{K+1}=b_{K+1}-b_K\rightarrow 0$.  
A: $$\int_{\mathbb{R}} |f|^p <\infty $$ Hence $$\lim_{b \to \infty} \int_{-\infty}^b |f|^p =\int_{-\infty}^{b} |f|^p + \sum_{i=0}^{\infty} \int_{b+ci}^{b+ci+1} |f|^p < \infty$$ $\forall c \in \mathbb{R}$.  But $$\int_{\mathbb{R}} |f|^p < \infty \Rightarrow \int_{-\infty}^{b} |f|^p< \infty$$ Therefore $$\lim_{i \to \infty} \int_{b+i}^{b+i+1} |f|^p=0$$ But by Holder$$ \lim_{i \to \infty} \int_{b+i}^{b+i+1} |f| \leq \lim_{i \to \infty} \left(  \int_{b+i}^{b+i+1} |f|^p \right)^{1/p}=0$$
A: Let $\epsilon>0$. Since $C_{0}^{\infty}\left(\mathbb{R}\right)$
is dense in $\text{L}^{p}\left(\mathbb{R}\right)$ one can find $g\in C_{0}^{\infty}\left(\mathbb{R}\right)$
such that $\left|g-f\right|_{\text{L}^{p}\left(\mathbb{R}\right)}\leq\frac{\epsilon}{2}$.
Then by Holder and properties of the $\text{L}^{p}$ norm,
$\int_{x}^{x+1}\left|f\right|\leq\left(\int_{x}^{x+1}\left|f\right|^{p}\right)^{\frac{1}{p}}\leq\left(\int_{x}^{x+1}\left|f-g\right|^{p}\right)^{\frac{1}{p}}+\left(\int_{x}^{x+1}\left|g\right|^{p}\right)^{\frac{1}{p}}.$
Appealing to the compact support of $g$, one may find $x\in\mathbb{R}$
large enough so that $\left|g\right|^{p}=0$ on $\left[x,\infty\right)$.
Therefore,
$\lim_{x\rightarrow\infty}\int_{x}^{x+1}\left|f\right|\leq\frac{\epsilon}{2}<\epsilon.$
Now send $\epsilon\downarrow0$.
