Exercise 1.6.6 (Lebesgue differentiation theorem, second formulation)

Question

Show that Theorem 1.6.11 follows from Theorem 1.6.12.

Elu Thingol · Accepted Answer

We have already demonstrated that $F$ is continuous. We now demonstrate the almost everywhere differentiability. We have
\begin{align*}
F'(x) &= \lim_{y	o x, y\in[a,b]/\{x\}}\frac{F(y) - F(x)}{y-x}
= \lim_{y	o x, y\in[a,b]/\{x\}} \frac{1}{y-x} \cdot \left( \int_{[-\infty,y]}{f\,\mathrm{d}m} - \int_{[-\infty,x]}{f\,\mathrm{d}m} ight)
\intertext{In case that $y$ approaches $x$ from left, we have by change of variables:}
& =\lim_{y	o x, y\in[a,b]/\{x\}} \frac{1}{x-y}\int_{[y,x]}{f\,\mathrm{d}m}
= \lim_{h	o 0, h
eq 0} \frac{1}{h}\int_{[x-h,x]}{f\,\mathrm{d}m}
= f(x)
\end{align*}
The case when $y$ approaches $x$ from right follows similarly, and the mixed case follows by combining both.

Exercise 1.6.6 (Lebesgue differentiation theorem, second formulation)

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Exercise 1.6.6 (Lebesgue differentiation theorem, second formulation)

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