# 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

1. 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 $$\mathbb R^d$$).
2. if $$\mu\perp\lambda^{\otimes d}$$, then $$\mathbb R^d\setminus\mathcal D(\mu)$$ is a $$\lambda^{\otimes d}$$-null set and $$\mathcal D(\mu)=0\;\;\;\lambda^{\otimes d}\text{-almost everywhere}\tag4.$$
• Rudin introduces the concept of sets shrinking nicely to $x$. That is needed to go from symmetric derivative to actual derivative. Jun 21 at 8:29
• As a remark, this is true without the boundedness and right-continuity assumption Jun 22 at 14:46
• @JonathanHole Well, but in this case the proof attempt that I provided isn't available (we need boundedness and right-continuity to ensure that $\mu_g$ exists). I'm particularly intereted in how we can finish this proof. Jun 23 at 9:15