Suppose $f$ and $g$ are in $L^1(\mathbb{R}^n,m)$ where $m$ is a Lebesgue measure. and suppose that $f=g$ almost everywhere with respect to $m$.
Define $C_f$ and $C_g$ to be the set of points $x$ for which $\lim_\limits{r\to 0 } A_rf(x)$ and $\lim_\limits{r\to 0 } A_rg(x)$ exist respectively. Can $C_f$ be described in term of $C_g$? here $(A_rf)(x)=\frac{1}{m(B(r,x))}\int_{B(r,x)}f(y)\;dm(y)$ and $m$ is a Lebesgue measure.
could you please explain to me why $C_f$ and $C_g$ must be the same sets. Thanks in advance
I don't entirely understand your question. $\newcommand{\RR}{\mathbb{R}}$
For all $r$ and $x$, $(A_rf)(x) = (A_rg)(x)$, since $$\frac{1}{m(B_r(x))}\int_{B_r(x)} f(y)\,dm(y) = \frac{1}{m(B_r(x))}\int_{B_r(x)} g(y)\,dm(y),$$ since $f=g$ $m$-a.e. Thus if $Af : (0,\infty)\times \RR^n \to \RR$ and $Ag : (0,\infty)\times \RR^n \to \RR$ are the functions $Af(r,x)=(A_rf)(x)$ and $Ag(r,x)=(A_rg)(x)$, then as functions, $Af=Ag$. Thus the set of $x$ for which $$\lim_{r\to 0}Af(r,x)$$ exists and the set of $x$ for which $$\lim_{r\to 0} Ag(r,x)$$ exists are the same, by definition of equality.