Exercise 5, § 6

a) Let $\rho$ and $\sigma$ be two positive measures on $\mathbf{R}$. Show that if $\rho([a, b]) = \sigma([a, b])$ for every semi-open interval $]a, b]$, then $\rho = \sigma$.

b) Let I be an interval of $\mathbf{R}$, with left end-point $\alpha$ and right end-point $\beta$, and let $\psi$ be an increasing finite numerical function defined on I; extend $\psi$ to $\mathbf{R}$ in any way whatsoever. In order that $\psi$ be proper for the measure $\varphi_I \cdot \mu$ ($\mu$ Lebesgue measure on $\mathbf{R}$), it is necessary and sufficient that the following two conditions hold: 1° either $\alpha$ is finite, or $\alpha = -\infty$ and $\lim_{x \to -\infty} \psi(x) = -\infty$; 2° either $\beta$ is finite, or $\beta = +\infty$ and $\lim_{x \to +\infty} \psi(x) = +\infty$.

Suppose these conditions verified. For every $y \in \mathbf{R}$, let $\theta(y)$ be the supremum of the set of $x \in I$ such that $\psi(x) \leq y$ (the supremum being equal to $\alpha$ if this set is empty). Show that $\theta$ is a finite numerical function, increasing and right-continuous in $\mathbf{R}$. If $\nu$ is the image of $\varphi_I \cdot \mu$ under $\psi$, then $\nu([a, b]) = \theta(b) - \theta(a)$ for every semi-open interval $]a, b]$ in $\mathbf{R}$.

c) Conversely, show that for every finite numerical function $\theta$, increasing and right-continuous in $\mathbf{R}$, there exists one and only one positive measure $\nu$ on $\mathbf{R}$ such that $\nu([a, b]) = \theta(b) - \theta(a)$ for every semi-open interval $]a, b]$ of $\mathbf{R}$ (the 'Stieltjes measure on $\mathbf{R}$ defined by $\theta$'; one writes $\int f \, d\theta$ instead of $\int f \, d\nu$); moreover, every positive measure on $\mathbf{R}$ may be obtained in this way. Under what condition is the measure $\nu$ diffuse? What is then the image of $\nu$ under $\theta$?