Prove that a bounded nondecreasing function is differentiable Lebesgue-almost everywhere Let $g:\mathbb R\to\mathbb R$ be bounded, nondecreasing and right-continuous. We know that there is a unique finite measure $\mu_g$ on $\mathcal B(\mathbb R)$ with $$\mu_g((a,b])=g(b)-g(a)\;\;\;\text{for all }a\le b\tag1.$$ Moreover, by Lebesgue`s decomposition theroem, $$\mu_g=\mu_g^a+\mu_g^s\tag2$$ for some finite measures $\mu_g^a\ll\lambda$ and $\mu_g^s\perp\lambda$, wher $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R)$. Furthermore$^1$, $\mathbb R\setminus\mathcal D(\mu_g^a)$ is a $\lambda$-null set and $$\frac{{\rm d}\mu_g^a}{{\rm d}\lambda}={\rm D}\mu_gâ\tag5$$ and $\mathbb R\setminus\mathcal D(\mu_g^s)$ is a $\lambda$-null set and $$\mathcal D(\mu_g^s)=0\;\;\;\lambda^{\otimes d}\text{-almost everywhere}\tag6.$$

How do we conclude that $g$ is differentiable $\lambda$-almost everywhere?

From the definition we see that $${\rm D}\mu(x):=\lim_{r\to0+}\frac{\mu_g^a((x-r,x+r))}{\lambda((x-r,x+r))}\;\;\;\text{for all }x\in\mathcal D(\mu_gâ)\tag7.$$ But this looks more like this could be more useful to show symmetric differentiability, which is not enough to show differentiability.

$^1$ If $d\in\mathbb N$ and $\mu$ is a locally finite measure on $\mathcal B(\mathbb R)^{\otimes d}$, let $$\mathcal D(\mu):=\left\{x\in\mathbb R^d:\limsup_{r\to0+}\frac{\mu(B_r(x))}{\lambda^{\otimes d}(B_r(x))}<\infty\right\}$$ and $${\rm D}\mu(x):=\lim_{r\to0+}\frac{\mu(B_r(x))}{\lambda^{\otimes d}(B_r(x))}\;\;\;\text{for }x\in\mathcal D(\mu).$$ We can show that

*

*if $\mu\ll\lambda^{\otimes d}$, then $\mathbb R^d\setminus\mathcal D(\mu)$ is a $\lambda^{\otimes d}$-null set and $$\frac{{\rm d}\mu}{{\rm d}\lambda^{\otimes d}}={\rm D}\mu\tag3$$ (where ${\rm D}\mu$ is arbitrary extended to $\mathbb R^d$).

*if $\mu\perp\lambda^{\otimes d}$, then $\mathbb R^d\setminus\mathcal D(\mu)$ is a $\lambda^{\otimes d}$-null set and $$\mathcal D(\mu)=0\;\;\;\lambda^{\otimes d}\text{-almost everywhere}\tag4.$$
 A: Going in the direction of @Jonathan Hole's comment: you can prove even more and even in an easier way.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ nondecreasing, then it must
have a countable set of discontinuities (hence of Lebesgue measure
zero).

Define $\mathcal{D}:= \{x \in \mathbb{R} \rightarrow \mathbb{R} : f \text{ is discontinuous at } x\}$. Then by definition of $\mathcal{D}$, it is easy to characterize it as $\mathcal{D}= \{x \in \mathbb{R} : f(x^-):= \lim_{y \rightarrow x^-}f(y) < f(x^+):= \lim_{y \rightarrow x^+}f(y) \}$. For each $x \in \mathcal{D}$, let $I_x := (f(x^-), f(x^+)) \neq \emptyset$. Moreover, $I_{x_1} \cap I_{x_2} = \emptyset $ if $x_1, x_2 \in \mathcal{D}$ (since $f$ is monotone).
Consider now the following collection $\mathscr{G}:=\{ I_x : x \in \mathcal{D}\}$. This is a family of disjoint, nonempty, open intervals. By the axiom of choice (but you can forget this detail), there exists an injective function $\varphi: \mathcal{D} \rightarrow \mathbb{Q}$ such that $x \mapsto q_x \in I_x \cap \mathbb{Q}$. By what is known as the Cantor-Bernstein-Schroeder theorem, we get that the cardinality of $\mathcal{D}$ is $\le$ than the cardinality of the rationals. Hence $\mathcal{D}$ is at most countable, hence $\mathcal{L}(\mathcal{D})=0$.
A: Assume $\mu_g^s=0$. In that case $\mu_g=\mu_g^a$ and Radon-Nykodym tells us that $\mu_g$ admits a density wrt. $\lambda$, i.e. $g'= \frac{d\mu_g^a}{d\lambda}$ almost everywhere, since $\int_a^bg'd\lambda=g(b) - g(a)$.
If we now no longer assume $\mu_g^s=0$, then we don't know anymore if $\int_a^b\frac{d\mu_g^a}{d\lambda}d\lambda=\mu_g((a,b])$, but luckily we don't care about that:
Since $D\mu_g^s=0$ almost everywhere we have:
$$\int_a^b\frac{d\mu_g^a}{d\lambda}d\lambda = \int_a^b(\frac{d\mu_g^a}{d\lambda}+\frac{d\mu_g^s}{d\lambda})d\lambda$$
The measurable sets that are a subset of $(a,b]$, which can cause $\int_a^b\frac{d\mu_g^a}{d\lambda}d\lambda<\mu_g((a,b])$, are Lebesgue measure zero sets.
Which means that $g$ is differentiable $\lambda$-a.e. with $g'= \frac{d\mu_g^a}{d\lambda}$ $\lambda$-a.e.
