$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ >[!idea] Note for Experts >The main takeaway is that monotone non-decreasing right-continuous functions $g:[0,T]\to \RR_+$ are in bijection with finite positive measures $\theta$ on $[0,T]$. This is really deep, because it implies that right-continuous functions have measure-theoretic regularity, not topological regularity, which lends credence to the ubiquity of right-continuous functions in [[Stochastic Calculus Homepage|stochastic calculus]]. Now that we've shown how to construct a _uniform_ distribution over the real line, we can try to make some modifications. Specifically, instead of letting $\mu((a,b]) = b - a$ for all $a,b$ we can let $\mu((a,b]) = F(b) - F(a)$ for some function $F$. This $F$ is then required to satisfy two properties: 1. For any $a < b$, we know $\mu((0, a])\leq \mu((0,b])$. Thus $F(b)\geq F(a)$, and $F$ must be ==**nondecreasing**==. 2. If $x_n \downarrow x$, then the intervals $(0, x_n] \downarrow (0,x]$. $F(x_1) - F(x_0) < \infty$, so by continuity of measures we must have $F(x_n) \downarrow F(x)$. Thus $F$ must be ==**right-continuous**==. >[!idea] >Actually, all [[Radon measures]] on $\RR$ must take this form! The function $F$ is just $F((0, x])$ or $-F((-x, 0])$. > >This will be clear by the uniqueness portion of Caratheodory. > [!definition] Stieltjes Measure Function > A ==**Stieltjes measure function**== is a function $F: \overline{\RR} \to \RR$ that is nondecreasing and right-continuous. > [!example] Uniform on the interval > Take $F(x) = 0$ for $x \leq 0$, $F(x) = x$ for $0\leq x \leq 1$, and $F(x) = 1$ for $x\geq 1$. This is a Stieltjes measure function that gives the uniform distribution on $[0,1]$. > [!example] Dirac Measure > Take $F(x) = 0$ for $x < 0$, and $F(x) = 1$ for $x \geq 0$. This is a Stieltjes measure function that gives the Dirac measure at $0$. In particular, this example shows that Stieltjes measure functions do not hae to be left-continuous. >[!example] Distributions >Later, we will study random variables and their distributions; these are of course not generally uniform over any interval; as such, they are completely characterized by these Stieltjes measure functions. And of course, the main theorem is that one can extend any Stieltjes measure function to a measure on the Borel sets. > [!theorem] Lebesgue-Stieltjes Measure > Let $F$ be a Stieltjes measure function. Then, there exists a unique measure $\mu$ on the Borel sets such that $\mu((a,b]) = F(b) - F(a)$ for all $a,b$. A la [[Standard Measure Construction]], we simply define our pre-measure. > [!definition] Lebesgue-Stieljes Pre-measure on $\RR$ > Let $\LL_0$ be the algebra on $\RR$ generated by the intervals of the form $(a,b]$. Then, the ==**Lebesgue-Stieljes pre-measure**== $\lambda_0$ is defined on $\LL_0$ by $\lambda_0((a,b]) = F(b) - F(a)$. To keep things reasonable, $\lambda_0((a,\infty)) = F(\infty) - F(a)$, and $\lambda_0((-\infty, a]) = F(a) - F(-\infty)$. >[!proof]- Not deep > $\lambda_0$ is a well-defined pre-measure for identical reasons to the Lebesgue Measure. The results of the [[Caratheodory Extension Theorem]] then imply that there exists a unique measure $\mu$ on the Borel sets such that $\mu((a,b]) = F(b) - F(a)$ for all $a,b$. # Alternative route We can also bootstrap a measure on top of the [[Lebesgue Measure]] via the [[pushforward measure]]. Our function $F$ doesn't point the right way for this to work, so we need to take a pseudo-inverse of it first: >[!claim] >Let $F:\RR\to \RR$ be non-constant, right-continuous, and non-decreasing. Let $F(\pm \infty) = \lim_{x\to \pm \infty} F(x)$. Let $I = (F(-\infty), F(\infty))$. Then, let $f$ be a sort of inverse of $F$, with ties broken downwards: >$ > f(x) = \inf\left\{y\in \RR: x\leq F(y)\right\} >$ >Then: > - For $x\in I$ and $y\in \RR$, $f(x)\leq y$ iff $x\leq F(y)$; in particular $f$ is Borel measurable. > - $f$ is left-continuous and non-decreasing. >[!proof]- Not deep > If $x\leq F(y)$, then clearly $f(x)\leq y$, it doesn't get more obvious than this. > > If $f(x)\leq y$, then by definition of infimum, there exists a sequence $y_i\downarrow y$ such that $x\leq F(y_i)$ for all $i$. By definition of right continuity, it follows that $x\leq F(y)$. > > This means $f$ is Borel measurable; the pre-image of $(-\infty, y]$ is $(-\infty, F(y)]$. > > $f$ is non-decreasing because if $x < x'$, then $\{y\in \RR: x\leq F(y)\}$ is a strict superset of $\{y\in \RR: x'\leq F(y)\}$. It is left-continuous: suppose $x_i\uparrow x$. Then, suppose all $f(x_i)\leq y$, yet $f(x) > y$. This is absurd, because this means $x_i\leq F(y)$ for all $x_i$, yet $x\not\leq F(y)$. Now, the induced measure is $ f_\sharp\mu((a,b]) = \mu\left(\{x: a<f(x)\leq b\}\right) = \mu\left((F(a), F(b)]\right) = F(b) - F(a) $ which is exactly what we want.