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April 7th, 2019, 01:52 PM   #1
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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.
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April 7th, 2019, 04:40 PM   #2
SDK
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The following line is wrong.


Quote:
Originally Posted by Gaby Alfonso View Post

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
$$\infty>\frac{f(x)}{2}>\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy.$$
This is not true. You have taken an inequality for integrals and tried to use it to imply a pointwise inequality which in general is not valid even on a set of full measure.

Last edited by skipjack; April 8th, 2019 at 12:20 AM.
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