Probability laws correspond to distribution functions

If $a<b$, then $(-\mathrm{\infty },a]\subseteq (-\mathrm{\infty },b]$, so [L1] and [L2] give $F_X(a)\le F_X(b)$. Also $(-\mathrm{\infty },b]\setminus (-\mathrm{\infty },a]=(a,b],$ so the finite-measure difference formula from [L2] yields ${\mathbb{P}}_X((a,b])=F_X(b)-F_X(a)\mathrm{.}$

For fixed $x$, the sets $(-\mathrm{\infty },x+1\mathrm{/}n]$ decrease to $(-\mathrm{\infty },x]$, and ${\mathbb{P}}_X((-\mathrm{\infty },x+1])\le {\mathbb{P}}_X(\mathbb{R})=1$. Hence [L2] gives right continuity of $F_X$. Likewise $(-\mathrm{\infty },n]\uparrow \mathbb{R}$ and $(-\mathrm{\infty },-n]\downarrow \mathrm{\varnothing }$, so continuity from below and from above give ${\lim }_{n\to \mathrm{\infty }}{F}_X(n)=1,{\lim }_{n\to \mathrm{\infty }}{F}_X(-n)=0.$

Put $G(x):=F(x)-F(0)$. Then $G$ is still nondecreasing and right-continuous, so [L3] gives a unique Borel measure $\mu$ finite on compact sets with $\mu ((a,b])=G(b)-G(a)=F(b)-F(a)(a<b)\mathrm{.}$

For fixed $x$, the sets $(-n,x]$ increase to $(-\mathrm{\infty },x]$. Hence [L2] and step 1.3 give $\mu ((-\mathrm{\infty },x])={\lim }_{n\to \mathrm{\infty }}\mu ((-n,x])={\lim }_{n\to \mathrm{\infty }}(F(x)-F(-n))=F(x)\mathrm{.}$ Applying continuity from below once more to $(-\mathrm{\infty },n]\uparrow \mathbb{R}$ shows $\mu (\mathbb{R})={\lim }_{n}F(n)=1$, so $\mu$ is a probability measure. If $\nu$ is another Borel probability measure with $\nu ((-\mathrm{\infty },x])=F(x)$ for all $x$, then $\nu ((a,b])=F(b)-F(a)=\mu ((a,b])$ for every $a<b$, and [L3] gives $\nu =\mu$.

Steps 1.1 and 1.2 prove part 1, and steps 1.3 and 2.1 prove part 2.