$\textbf{The Problem:}$ 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. If my proof is incorrect, please do not provide a solution, only hints.
Thank you for your time.
I don't understand the proof you gave. Here's a hint for another approach: It's enough to show
$$\lim_{r\to 0^+}\int_{E}\left(\frac{1}{\vert B(x,r)\vert}\int_{B(x,r)}f(y)dy\right)dx = \int_{E}f(x)dx .$$
Call the function inside the parentheses $g_r(x).$ Then $g_r(x)\to f(x)$ as $r\to 0^+$ for a.e. $x\in E.$ Apply the DCT ...