Let $f$ be a measurable function from $\mathbb{R}^n/\{0\}\to \mathbb{R}$,under which condition the integral below near the origin goes to zero?
$$\lim_{\epsilon\to 0}\int_{B_\epsilon / \{0\}} f(x)dx = 0$$
I try to show $\frac{1}{|x|^s}$ is not integrable near to origin (inside $B_1(0)$) when $s\ge n$ , and here uses the fact to split the ball into two parts and the small $\epsilon$ part is negligible. For general $f(x)$ under which condition the integral above is negligible?
I try to gives an ansewer using Lebesgue differentiation theorem ,which state for Locally integrable function,with $x$ a.e. we have: $$\lim _{{B\rightarrow x}}{\frac {1}{|B|}}\int _{{B}}f\,{\mathrm {d}}\lambda $$
the question then is why 0 it one of the desired point?
If $f$ is integrable on $B_r \setminus \{0\}$ for some $r>0$ then this is true. Take any sequence $\epsilon_n$ decreasing to $0$ and apply Monotone Convergence Theorem to the sequence $|f|\chi_{B_{\epsilon_k\setminus B_{\epsilon_{k+n}}}}$. [This sequence increases to $|f|\chi_{B_{\epsilon_k}}$].