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 $$\mathcal D(\mu)=0\;\;\;\lambda^{\otimes d}\text{-almost everywhere}\tag4.$$
  • $\begingroup$ Rudin introduces the concept of sets shrinking nicely to $x$. That is needed to go from symmetric derivative to actual derivative. $\endgroup$ Jun 21 at 8:29
  • $\begingroup$ As a remark, this is true without the boundedness and right-continuity assumption $\endgroup$ Jun 22 at 14:46
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Jun 23 at 9:15


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