Motivation of Lebesgue differentiation theorem Fundamental theorem of calculus states that the derivative of the
integral is the original function, meaning that
$$
f(x)=\frac{d}{dx}\int_{a}^{x}f(y)dy.\tag{*}
$$
To motivate the statement of the Lebesgue differentiation theorem, observe
that (*) may be written in terms of symmetric differences as
$$
f(x)=\lim_{r\to 0^+}\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy.\tag{**}
$$
An $n$-dimensional version of (**) is
$$
f(x)=\lim_{r\to 0^+}\frac{1}{|B(x,r)|}\int_{B(x,r)}f(y)dy.\tag{***}
$$
where the integral is with respect $n$-dimensional Lebesgue measure. The Lebesgue differentiation theorem states that (***) holds pointwise $\mu$-a.e. for any locally integrable function $f$.
My question is how could we write (**) by using (*) ? If we define $F(x)=\int_{a}^{x}f(y)dy$. The quotient 
$$
\frac{F(x+r)-F(x)}{r}=\frac{\int_{a}^{x+r}f(y)dy-\int_{a}^{x}f(y)dy}{r}=\frac{1}{r}\int_{x}^{x+r}f(y)dy
$$
How could we say that 
$$\frac{1}{r}\int_{x}^{x+r}f(y)dy\overset{?}{=}\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy$$ 
 A: Let $g: \Bbb R \to \Bbb R$ be differentiable at a point $x \in \Bbb R$, i.e. the limit 
$$ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} $$
exists.
So it follows, that 
$$\begin{align*} \frac{g(x+h) - g(x-h)}{2h} 
&= \frac{\left[ g(x+h) - g(x) \right] + \left[ g(x) - g(x-h) \right]}{2h} \\
&= \frac 1 2 \frac{g(x+h) - g(x)}{h} + \frac 1 2 \frac{g(x + (-h)) - g(x)}{(-h)} \\
&\to \frac 1 2 g'(x) + \frac 1 2 g'(x) = g'(x) \quad \text{for } h \to 0 \; .
\end{align*}$$
For the rest, see John's answer.
A: The last equality as stated is not true. But we don't need that. 
In general we have 
$$g'(x) = \lim_{r\to 0}\frac{g(x+r) - g(x-r)}{2r}$$
for any function differentiable at $x$ (Can you show that?). Put $g(x) = \int_a^x f(s) \, ds$, by $(*)$, 
\begin{equation}
\begin{split}
f(x) &= \frac{d}{dx} \int_a^x f(s) \, ds \\
&= \lim_{r\to 0} \frac{1}{2r} \left(\int_a^{x+r} f(s) \, ds - \int_a^{x-r} f(s)\, ds\right)\\
&=\lim_{r\to 0} \frac{1}{2r} \int_{x-r}^{x+r} f(s) \, ds. 
\end{split}
\end{equation}
