Proof of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law

Claim 1. Let $\mathcal{L}=\{A\in\mathcal{S}:\mu(A)=\nu(A)\}$. We check that $\mathcal{L}$ is a $\lambda$-system. First, $X\in\mathcal{L}$ by hypothesis. Second, if $A,B\in\mathcal{L}$ with $A\subseteq B$, then additivity gives $\mu(B)=\mu(A)+\mu(B\setminus A)$ with all terms finite, so $\mu(B\setminus A)=\mu(B)-\mu(A)=\nu(B)-\nu(A)=\nu(B\setminus A)$ and $B\setminus A\in\mathcal{L}$. Third, if $A_1\subseteq A_2\subseteq\dots$ lie in $\mathcal{L}$ with union $A$, then writing $A$ as the disjoint union of $A_1$ and the differences $A_{m+1}\setminus A_m$ and using countable additivity, $\mu(A)$ is the limit of the nondecreasing sequence $\mu(A_m)$, and likewise for $\nu$; hence $A\in\mathcal{L}$. Since $\mathcal{P}\subseteq\mathcal{L}$, Dynkin's lemma (Dynkin's Pi-Lambda Theorem) gives $\mathcal{S}=\sigma(\mathcal{P})\subseteq\mathcal{L}$, so $\mu=\nu$.

Claim 2. Both families are $\pi$-systems: $(u,\infty)\cap(u',\infty)=(\max(u,u'),\infty)$ and $(-\infty,u]\cap(-\infty,u']=(-\infty,\min(u,u')]$. Since $(-\infty,u]$ is the complement of $(u,\infty)$, the two families generate the same $\sigma$-algebra $\mathcal{A}$. Now $\mathcal{A}$ contains the sets $(u,v]=(u,\infty)\setminus(v,\infty)$, hence the open intervals $(u,v)=\bigcup_{n\ge1}(u,v-1/n]$, hence every open subset of $\mathbb{R}$: around every point of an open set $U$ there is, by the Archimedean property, an open interval with endpoints of the form $z/n$ ($z$ an integer, $n$ a natural number) containing the point and contained in $U$, and the collection of all such intervals may be listed as a sequence (list the pairs of endpoints by increasing $n$ and, within each $n$, by increasing $\max(|z|,|z'|)$), so $U$ is a countable union of members of $\mathcal{A}$. Hence $\mathcal{A}$ contains the $\sigma$-algebra generated by the open sets, which is the Borel $\sigma$-algebra of Borel Sigma-Algebra on the Real Line; conversely each ray $(u,\infty)$ is open and each $(-\infty,u]$ is Borel, so $\mathcal{A}$ is the Borel $\sigma$-algebra.

Claim 3. The function $h$ is Borel measurable, since for every real $c$ the set $\{h>c\}$ is an interval ($\mathbb{R}$ for $c<0$; $(0,\infty)$ for $c=0$; a bounded interval with left endpoint $0$ for $0<c<1$, by monotonicity of the exponential, part of Basic Properties of the Exponential Function; empty for $c\ge1$). Let $\nu_h$ be the measure with density $h$ with respect to $\lambda$, so that $\nu_h(E)=\int\mathbf{1}_E\,h\,d\lambda$ by claim 3 of Image Measures, Measures with Densities, and Change of Variables, and let $\mu_\xi$ be the distribution of $\xi$.

For a real $x>0$, the function $h\,\mathbf{1}_{(0,x]}$ differs from the zero extension of the continuous function $s\mapsto e^{-s}$ on $[0,x]$ only at the single point $0$, an interval of Lebesgue measure $0$ by claim 4 of Existence of Lebesgue Measure on the Real Line; so by claims 2, 3, and 6 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval and the fundamental theorem of calculus at base point $0$ with the antiderivative $s\mapsto-e^{-s}$, whose derivative is $s\mapsto e^{-s}$ by Derivative of a Scaled Exponential Function, $\nu_h\bigl((0,x]\bigr)=\int_0^x e^{-s}\,ds=1-e^{-x}.$ For $0\le u<n$, additivity of $\nu_h$ on $(0,n]=(0,u]\mathbin{\cup}(u,n]$ (a disjoint union of members of the Borel $\sigma$-algebra, all values finite) gives $\nu_h((u,n])=(1-e^{-n})-(1-e^{-u})=e^{-u}-e^{-n}$. The sets $(u,n]$ increase to $(u,\infty)$ as $n\to\infty$ along the naturals; continuity of measures from below (countable additivity applied to the disjoint differences, as in the proof of claim 1) and $e^{-n}\to0$ (Basic Properties of the Exponential Function) give $\nu_h((u,\infty))=e^{-u}$ for $u\ge0$. For $u<0$, $\nu_h((u,0])=0$ since $h$ vanishes there, so $\nu_h((u,\infty))=\nu_h((0,\infty))=1$.

On the other side, $\mu_\xi((u,\infty))=P(\xi>u)$ equals $e^{-u}$ for $u\ge0$ by hypothesis, and equals $1$ for $u<0$ since $\xi\ge0$. Hence $\mu_\xi$ and $\nu_h$ agree on the $\pi$-system $\{(u,\infty):u\in\mathbb{R}\}$, which generates the Borel $\sigma$-algebra by claim 2, and they have the same total mass: $\mu_\xi(\mathbb{R})=1$ and $\nu_h(\mathbb{R})=\nu_h((-1,\infty))=1$. By claim 1, $\mu_\xi=\nu_h$.