Assume that $f\in L^1(\mathbb{R}^n)$. Let $B(x,r)$ be a ball centered at $x$ with radius $r$. By Lebesgue Differentiation Theorem, $$ \lim_{r\to 0} \dfrac{1}{m(B(x,r))}\int_{B(x,r)}f(y)dy=f(x)$$ for $x\in\mathbb{R}^n$ almost everywhere. Assume that $x_0\in\mathbb{R}^n$ is a point that such that $f(x_0)$ is finite and such that the equality above holds. Define
$$F(r):=\dfrac{1}{m(B(x_0,r))}\int_{B(x_0,r)}f(y)dy.$$ Is $F(r)$ a continuous function of $r$ on $(0,+\infty)$? I think so but I’m not sure how to prove it. I tried to prove it by writing the definitions but feel completely stuck. Here’s some of my attempts.
By Lebesgue Differentiation Theorem, $$\lim_{r\to 0^+} F(r)=f(x_0)<\infty.$$ And $$\lim_{r\to +\infty} F(r)= \lim_{r\to +\infty} \dfrac{1}{m(B(x_0,r))}\int_{B(x_0,r)}f(y)dy\le \lim_{r\to +\infty} \dfrac{1}{m(B(x_0,r))}\int_{\mathbb{R}^n} f(y)dy = \lim_{r\to +\infty} \dfrac{1}{m(B(x_0,r))}\cdot \Vert f\Vert_1=0.$$
Could you give me some ideas? Thanks!
It is a ratio of two continuous functions. Note that $m(B(x,r))=r^{n} m(B(0,1))$. To show that the numerator is continuous use the following facts:
For any integrable function $f$ , $\int_E f(x)dx \to 0$ as $m(E) \to 0$
$|\int_E f(x)dx - \int_F f(x)dx | \leq \int_{E\Delta F} |f(x)| dx$.
$m(B(x,r+h)\Delta B(x,r))=|(r+h)^{n} -r^{n}| m(B(0,1)) \to 0$ as $ h \to 0$.
Notation: $A \Delta B=(A \setminus B) \cup (B \setminus A)$