Motivation of Lebesgue differentiation theorem
Andrew Henderson 
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 $$\frac{1}{r}\int_{x}^{x+r}f(y)dy\overset{?}{=}\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy$$
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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}
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