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April 7th, 2019, 02:52 PM  #1 
Newbie Joined: Apr 2019 From: Madison, Wisconsin Posts: 2 Thanks: 0 Math Focus: Measure Theory & Integration  Proving an Inequality for Locally Integrable Functions
>$\textbf{The Problem:}$ Let $f\geq 0$ be a bounded function and $E\subset\mathbb R^d$ have finite measure. Prove that there exists $R>0$ such that for all $0<r<R$ we have $$\int_{E}f(x)dx\leq 2\int_{E}\left(\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\right)dx.$$ Here $B(x,r)$ denotes the open ball of radius $r$ centered at $x$. I will use the following $\color{blue}{\text{Theorem}}$ from page $106$ in Stein and Shakarchi's Real Analysis: If $f$ is locally integrable in $\mathbb R^d$ then we have for almost every $x\in\mathbb R^d$ that $$\lim\limits_{\vert B\vert\to0}\frac{1}{\vert B\vert }\int_{B}f(y)dy=f(x),\quad x\in B,$$ where $B$ is a ball containig $x$. $\textbf{My Attempt:}$ Suppose for a contradiction that such $R>0$ does not exist. Then given $R>0$ there is some $0<r<R$ such that $$\infty>\int_{E}f(x)dx> 2\int_{E}\left(\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\right)dx.$$ In particular, this means that for almost every $x\in E$ we must have $$\infty>\frac{f(x)}{2}>\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy.$$ And since by the assumption that $f\geq0$ and bounded, we have that $$\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\geq \frac{1}{\vert B(x,p)\vert}\int_{B(x,p)}f(y)dy$$ for all $0<p<r,$ we have a contradiction of the $\color{blue}{\text{Theorem}}$. _____ Is the above proof correct? Any feedback is welcomed. Thank you for your time. 
April 7th, 2019, 05:40 PM  #2  
Senior Member Joined: Sep 2016 From: USA Posts: 685 Thanks: 461 Math Focus: Dynamical systems, analytic function theory, numerics 
The following line is wrong. Quote:
Last edited by skipjack; April 8th, 2019 at 01:20 AM.  

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functions, inequality, integrable, locally, proving, real analysis 
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