Take a function $f \in L^1(m)$, define a new function $g = \lim_{r\to 0} \frac{1}{m(B_{r,x})} \int_{B_{r,x}}\ f(x)\ dm$ where $m$ is the Lebesgue measure and $B_{r,x}$ is the ball centred at $x$ with radius $r$. Is $g=f$ almost everywhere with respect to $m$?
I ask since this fact seems to be casually used by Rudin in the proof of Thm 7.8 in Real and Complex Analysis without reference to something earlier, and I don't know if it is supposed to be some special property of Radon-Nikodym derivatives or just a typical property of measurable functions and I missed it.
EDIT:
The accepted answer is correct, a little more detail about how to go from the defn of a Lebesgue point to this formulation, for x fixed:
$$ \lim_{r\to 0} \left(\frac{1}{m(B_{r,x})} \int_{B_{r,x}}\ f(y)\ dm - f(x)\right) = \left(\frac{1}{m(B_{r,x})} \int_{B_{r,x}}\ f(y)\ dm - \frac{1}{m(B_{r,x})} \int_{B_{r,x}} f(x)\ dm\right) \leq \frac{1}{m(B_{r,x})} \int_{B_{r,x}}\ \lvert f(y)-f(x) \rvert\ dm = 0 $$
where the last zero holds almost everywhere for $f$ in $L^1$ by Rudin Thm 7.7.
This is called a Lebesgue point. https://en.wikipedia.org/wiki/Lebesgue_point If $f$ is in $L^1$, almost every $x$ is a Lebesgue point