# Can anyone explain one step in the proof of the Lebesgue theorem?

If the function f is monotone on the open interval (a, b), then it is differentiable almost everywhere on (a, b).

Proof： Assume f is increasing. Furthermore, assume (a, b) is bounded. Otherwise, express (a, b) as the union of an ascending sequence of open, bounded intervals and use the continuity of Lebesgue measure. The set of points x in (a, b) at which $\bar{D}f(x)$ > $\underline{D}$ f(x) is the union of the sets $E_{\alpha,\beta} = \{x\in E(a, b) \| \bar{D}f(x)>\alpha>\beta> Df(x)\}$ where $\alpha$ and $\beta$ are rational numbers.

Why is the following true? Should not $\alpha$ and $\beta$ be real numbers?

The set of points x in (a, b) at which $\bar{D}f(x)$ > $\underline{D}$ f(x) is the union of the sets $E_{\alpha,\beta} = \{x\in E(a, b) \| \bar{D}f(x)>\alpha>\beta> Df(x)\}$ where $\alpha$ and $\beta$ are rational numbers.

I believe you are following Royden & Fitzpatrick's Real Analysis. First let me remind you a couple of definitions pertaining to your question:

Let $f:E\to\Bbb{R}$ be a function. Then

\begin{align} \overline{D}f:&E^\circ\to\Bbb{R}& \underline{D}f:&E^\circ\to\Bbb{R}\\ &x\mapsto\limsup_{h\to0} \dfrac{f(x+h)-f(x)}{h}& &x\mapsto\liminf_{h\to0} \dfrac{f(x+h)-f(x)}{h} \end{align}

are the upper and lower derivatives of $f$ at $x$, respectively. Observe that $\forall x\in E^\circ: \overline{D}f(x)\geq\underline{D}f(x)$. $f$ is differentiable at $x\in E^\circ$ if $\infty>\overline{D}f(x)=\underline{D}f(x)>-\infty$, in which case

$$f'(x):=\overline{D}f(x)=\underline{D}f(x)$$

is the derivative of $f$ at $x$.

In Lebesgue's Theorem we are trying to show that the set $N$ where $f$ is not differentiable is negligible. By the definitions above, this set is precisely $N=\{\overline{D}f>\underline{D}f\}\subseteq E^\circ$. Even if $f$ is monotone, hence measurable, since the limit superior and inferior are not taken over a countable index we do not know yet that $N$ is measurable (hence we are using outer measure instead of the measure itself throughout).

Regardless, $\forall x\in N, \exists \alpha,\beta\in\Bbb{Q}:\overline{D}f(x)>\alpha>\beta>\underline{D}f(x)$ by the density of $\Bbb{Q}$. This means we can write $N$ as a countable union of sets:

$$N=\bigcup_{\alpha,\beta\in\Bbb{Q}\\\alpha>\beta}\{\overline{D}f>\alpha>\beta>\underline{D}f\}$$

and then we can employ the previous lemma (which is quite similar to Chebyshev's Inequality) on each one of them.

It is correct that

$$N=\bigcup_{\alpha,\beta\in\Bbb{R}\\\alpha>\beta}\{\overline{D}f>\alpha>\beta>\underline{D}f\}$$

as well, but (outer) measure can handle at best countable operations, thus in order to turn $N$ into something managable we write it as a union of countably many sets.

As a final remark observe that we use a very similar procedure to show that the sum of two measurable functions is measurable (though this is not the only way to show this).