# 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?

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Are you using the Lebesgue integral? What is your definition of it? – Alex Becker Jun 12 '12 at 16:54
I learned $\int f d\mu = sup\{ \int s d\mu :0 \le s \le f, s \in S\}$. Is it the key for this problem? – Collins Jun 12 '12 at 17:00
$s$ is a simple function? – Alex Becker Jun 12 '12 at 17:04
Yes that was what I mean. – Collins Jun 12 '12 at 17:11

## 3 Answers

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 $0$.

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We want to have limit of x→+∞ but is that ϵ→0 come to limit of x? – Collins Jun 12 '12 at 17:20
@Collins I edited to clarify. Hope that helps. – Alex Becker Jun 12 '12 at 17:26
I got it. Thanks for your very clear and nice answer. – Collins Jun 12 '12 at 17:29

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?
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Oh, you already give me almost full solution not just hints. – Collins Jun 12 '12 at 17:22

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).

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But this is false, there are even functions in $L^p(\mathbb{R})$ that are unbounded at $\pm\infty$. – AD. Jun 12 '12 at 19:52
Example: For integers $|n|>0$ put $f(x) = |n|$ when $x\in (n,n+1/|n|^3)$, and $f(x)=0$ elsewhere; then $$\int |f(x)|dx=\sum_{|n|>0}\frac{1}{n^2}<\infty$$ so $f\in L^1$ but there is no point wise limit at $\pm\infty$. – AD. Jun 12 '12 at 19:59
Another example I saw somewhere on math.SE is $$f(x)=xe^{-x^2|\sin x|}$$ what happens at $x =2\pi n$? – AD. Jun 13 '12 at 4:52
Dear A.D.: I don't think that my argument is wrong. I think you are mixing up the variables. I call $t$ what you call $x$. Let $f$ be either one of your functions. Fix $t\in \mathbb R$. Then still $f(t)\chi_{[x,x+1]}(t)$ converges to zero for $x\rightarrow\infty$, even though your functions have large values for large $t$. But $t$ doesn't get large, $t$ stays fixed. That is what pointwise convergence means. – h.h.543 Jun 30 '12 at 12:35
You are right! I mixed up the variables, what you mean is of course $g_x = f \cdot \chi_{[x,x+1]}$, and then I agree. Thanks! (+1) – AD. Jul 1 '12 at 19:26
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