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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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    Yes that was what I mean.2012-06-12

3 Answers 3

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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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    I got it. Thanks for your very clear and nice answer.2012-06-12
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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.2012-06-12
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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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    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)2012-07-01