Prove $\lim_{r \to 0} \frac{1}{2r} \int_{x-r}^{x+r} f(z) dz = f(x)$ From 2.27 here

Let $f$ be Lebesgue integrable on $[0,1] \to \mathbb R$.
$$\lambda\left(\lim_{r \to 0} \frac{1}{2r} \int_{x-r}^{x+r} f(z) dz = f(x)\right) = 1$$

The proof might actually involve things beyond my level but here's my attempt anyway:
$$\lim_{r \to 0} \frac{1}{2r} \int_{x-r}^{x+r} f(z) dz$$
$$\lim_{r \to 0} \frac{1}{2r} [\int_{0}^{x+r} f(z) dz - \int_{0}^{x-r} f(z) dz]$$
$$\stackrel{LHR}{=} \lim_{r \to 0} \frac{1}{2} [f(x+r) - f(x-r)(-1)]$$
$$\stackrel{?}{=} \frac{1}{2} [f(x+0) + f(x-0)] \tag{*}$$
$$= f(x)$$
$(*)$ Guess based on Using Fubini's Theorem in Stochastic Calculus, $f$ is not continuous, but the justification is that $f$ is measurable. $(*)$ does not work for finitely many points and hence the $\lambda$-almost everywhere.
If it's completely wrong to approach this in a Riemann/basic calculus/non-measure theory way, which parts and why please?
 A: This is a non-trivial result in measure theory called Lebesgue' Theorem. There is a lot of difference between the case where f is continuous and the one where f is only integrable.
A: As Kavi Rama Murthy said, this is a nontrivial theorem in measure theory called the Lebesgue differentiation theorem. The statement of the theorem is as follows: 
If $f : \mathbb{R}^n \to \mathbb{R}$  is integrable, then $\frac{1}{|B(x, r)|}\int_{B(x,  r)}f(y)\,d y \to f(x)$ as $r \to 0$ for almost every $x \in \mathbb{R}^n$. Everything here is with respect to the Lebesgue measure on $\mathbb{R}^n$, and $|B(x,  r)|$ denotes the Lebesgue measure of the ball at $x$ of radius $r$. Proving this for a general integrable function $f$ requires more sophisticated techniques than for a continuous function $f$. 
The quantity $\frac{1}{|B(x,r )|}\int_{B(x,r )}f(y)\,d y$ is the average  of $f$  over the ball $B(x, r)$. What this theorem is saying is that as we take smaller and smaller balls centered at a point $x$, the average of $f$ on these balls approaches the value of $f$ at the center of these balls. For continuous $f$, this is relatively intuitive. In full generality, the theorem says that this holds at "most"  points.
