Assuming countable choice, finite-on-compacts Borel measures on $\mathbb{R}$ correspond to nondecreasing right-continuous functions modulo constants

If $0\le x<y$, then $(0,x]\subseteq (0,y]$, so $F_{\mu }(x)\le F_{\mu }(y)$. If $x<y\le 0$, then $(y,0]\subseteq (x,0]$, so $-\mu ((x,0])\le -\mu ((y,0])$, again giving $F_{\mu }(x)\le F_{\mu }(y)$. If $x<0\le y$, then $F_{\mu }(y)-F_{\mu }(x)=\mu ((0,y])+\mu ((x,0])=\mu ((x,y])\ge 0$. [given, algebra]

$0={\mu }_F((a,b])-{\mu }_G((a,b])=(F(b)-G(b))-(F(a)-G(a))\mathrm{.}$

So $F(b)-G(b)=F(a)-G(a)$ for all $a<b$, and therefore $F-G$ is constant. [algebra]

In the three sign cases:

$\mu ((0,b])-\mu ((0,a])=\mu ((a,b])(0\le a<b),$

$-\mu ((b,0])+\mu ((a,0])=\mu ((a,b])(a<b\le 0),$

and

$\mu ((0,b])+\mu ((a,0])=\mu ((a,b])(a<0\le b)\mathrm{.}$

So the displayed interval formula always holds. [step 1.1, given, algebra]

For all large $n$ one has $x+h_n<0$ when $x<0$, while for $x\ge 0$ no sign change occurs. In either case, step 2.1 gives

$F_{\mu }(x+h_n)-F_{\mu }(x)=\mu ((x,x+h_n])\mathrm{.}$

The sets $(x,x+h_n]$ decrease to $\mathrm{\varnothing }$, and the first one has finite measure because it is contained in a compact interval. Therefore [L1] gives $\mu ((x,x+h_n])\to 0$, so $F_{\mu }(x+h_n)\to F_{\mu }(x)$. [step 2.1, L1]

Step 2.1 says that ${\mu }_{F_{\mu }}$ and $\mu$ agree on every half-open interval, so [L4] gives ${\mu }_{F_{\mu }}=\mu$. [step 2.1, step 3.1, L3, L4]

The interval values of ${\mu }_F$ and ${\mu }_G$ therefore agree by [L3], and [L4] yields ${\mu }_F={\mu }_G$. Together with steps 1.1, 1.2, 1.3, 2.1, and 3.1 this proves the theorem. [step 1.1, step 1.2, step 1.3, step 2.1, step 3.1, L3, L4] ∎