Let $L^1_{\text{loc}}$ denote locally compact and $(Arf)$ denote the average of a function $f$. I already have the proof for the following Theorem state as
$\text{Theorem: If} ~~f\in L^1_{loc}~~\text{then}~~ \lim_{r\to 0}(Ar f )(x) = f (x)$ for $m$ — a.e. $x\in \mathbb{R}^n$.
Now, I want to construct a parallel proof for the following statement:
$$\text{If}~ f(x)~\text{ is continuous on}~ \mathbb{R,}~\text{ prove that}~\lim_{r\to 0}(Arf)(x) = f(x)$$ for all $x$.
My questions are; what modifications do I have to make in the above Theorem so as to write an independent proof to my statement? What is the relationship between $\mathbb{R}$ and $L^1_{\text{loc}}$?
If $f$ is continuuous at $x_0$ then for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-f(x_0)|\le\varepsilon$ for all $|x-x_0|\le\delta$. Hence, for $0<r<\delta$, $$|Arf(x_0)-f(x_0)|=\left\vert \frac1{\text{meas}(B(x_0,r))}\int_{B(x_0,r)}(f(x)-f(x_0))\,dx \right\vert \\\le \frac1{\text{meas}(B(x_0,r))}\int_{B(x_0,r)}|f(x)-f(x_0)|\,dx \\ \le \varepsilon \frac1{\text{meas}(B(x_0,r))}\int_{B(x_0,r)}1\,dx=\varepsilon.$$