Randomizing across CDF jumps gives a uniform variable

By [F1] and [F2], $F$ is nondecreasing. For each real $c$, $\{x:F(x)>c\}$ is an upper set, hence is empty, all of $\mathbb{R}$, or a ray with or without its finite endpoint; each such ray is Borel by [F11]. Thus every strict superlevel set is Borel, and [F10] makes $F$ Borel-measurable. For each $x$, the sets $(-\mathrm{\infty },x+1\mathrm{/}n]$ decrease to $(-\mathrm{\infty },x]$ and have finite measure, so [F4] gives right continuity. The sets $(-\mathrm{\infty },-n]$ decrease to $\mathrm{\varnothing }$ and $(-\mathrm{\infty },n]$ increase to $\mathbb{R}$, so [F3] and [F4] give the limits $0$ and $1$ at the two infinities. Also $(-\mathrm{\infty },x-1\mathrm{/}n]$ increases to $(-\mathrm{\infty },x)$, so [F3] gives $F(x-)=\mu ((-\mathrm{\infty },x))$; [F5] then gives $F(x)-F(x-)=\mu (\{x\})$.

For every $x$, step 1.1 gives $F(x-)={\lim }_{n}F(x-1\mathrm{/}n)$. Each translate $x\mapsto F(x-1\mathrm{/}n)$ is measurable because translation is continuous and [F8] preserves measurability under composition; [F9] makes their increasing limit $F(x-)$ measurable. Thus $\mathrm{\Delta }F=F-F(\cdot -)$ is measurable by [F9]. Since $Y$ is measurable, $F(Y-)$ and $\mathrm{\Delta }F(Y)$ are measurable. Fix a bijection $e:\mathbb{Q}\to \mathbb{N}$ supplied by [F12]. Each rational open rectangle is encoded by four natural numbers, and iterating the injection $(a,b)\mapsto 2^a(2b+1)$ shows there are only countably many such rectangles. Density of $\mathbb{Q}$ makes them a base of ${\mathbb{R}}^2$. Every Euclidean open set is therefore a countable union of rectangles in $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$. Conversely the coordinate projections are continuous, and [F8] makes the inverse image of each Borel set under either projection Borel; finite intersections then show every Borel rectangle is Euclidean Borel. Thus the product Borel sigma-algebra equals $\mathcal{B}({\mathbb{R}}^2)$. The pair $(U,\mathrm{\Delta }F(Y))$ is measurable into the product sigma-algebra because preimages of generating rectangles are intersections of coordinate preimages, hence it is measurable into $\mathcal{B}({\mathbb{R}}^2)$. The map $(u,d)\mapsto ud$ is continuous, as $\mathrm{\mid }ud-u_0{d}_0\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }\cdot \mathrm{\mid }d-d_0\mathrm{\mid }+\mathrm{\mid }{d}_0\mathrm{\mid }\cdot \mathrm{\mid }u-u_0\mathrm{\mid }$ and $\mathrm{\mid }u\mathrm{\mid }$ is bounded near $u_0$; hence [F8] makes it Borel-measurable. Composing and then summing with $F(Y-)$ proves $V$ is measurable.

Fix $0<u<1$ and set $A_u:=\{x:F(x)\ge u\}$. The upper-tail limit in step 1.1 makes $A_u$ nonempty, and the lower-tail limit makes it bounded below. Applying [F6] to $-A_u$ gives $q:=-\sup (-A_u)=\inf A_u$. For every $n$, the infimum property gives $F(q+1\mathrm{/}n)\ge u$, so right continuity from step 1.1 yields $F(q)\ge u$. Since $q-1\mathrm{/}n<q$ and $q$ is a lower bound for $A_u$, $F(q-1\mathrm{/}n)<u$; the sets $(-\mathrm{\infty },q-1\mathrm{/}n]$ increase to $(-\mathrm{\infty },q)$, so [F3] and step 1.1 give $F(q-)={\lim }_{n}F(q-1\mathrm{/}n)\le u$. Put $a:=F(q-)$ and $b:=F(q)$. If $Y<q$, then $F(Y)\le a\le u$, so $V\le u$; this event has probability $a$. If $b>a$, then $P(Y=q)=b-a$ and, conditional on $Y=q$, $V=a+U(b-a)$ is uniform on $[a,b]$, contributing $(u-a)$ to $P(V\le u)$. If $b=a$, the inequalities force $a=b=u$ and $P(Y=q)=0$, so this contribution is zero.

For $Y>q$, monotonicity gives $F(Y-)\ge F(q)=b\ge u$. If $F(Y-)>u$, then $V>u$. If $F(Y-)=u$ and $\mathrm{\Delta }F(Y)>0$, the event $V\le u$ requires $U=0$, which has probability zero by [A1]. In the remaining case $F(Y-)=F(Y)=u$, put $L:=\{y>q:F(y-)=F(y)=u\}$. By steps 1.1 and 2.1, both $F$ and $F(\cdot -)$ are Borel-measurable, so $L=\{y>q\}\cap \{F(y-)=u\}\cap \{F(y)=u\}$ is Borel and therefore $\mu$-measurable. If $L$ is empty it contributes no mass. Otherwise let $r=\sup L$, allowing $r=+\mathrm{\infty }$ when $L$ is unbounded. For every $z$ with $q<z<r$, some $y\in L$ has $z<y$, so monotonicity gives $u=F(q)\le F(z)\le F(y-)=u$ and hence $F(z)=u$. If $r<\mathrm{\infty }$, define $c_n=q+(r-q)n\mathrm{/}(n+1)$; if $r=+\mathrm{\infty }$, define $c_n=q+n$. In either case $c_n\uparrow r$ (with the limit interpreted as $+\mathrm{\infty }$ in the second case), and [F5] gives $\mu ((q,c_n])=F(c_n)-F(q)=0$. By [F3], $\mu ((q,r))=0$. If finite $r$ belongs to $L$, its mass is $\mathrm{\Delta }F(r)=0$, so $\mu (L)=0$. Thus $Y>q$ contributes no probability, and step 2.2 yields $P(V\le u)=a+(u-a)=u$.

For $u_n=1\mathrm{/}n$ with $n\ge 2$, the events $\{V\le u_n\}$ decrease to $\{V\le 0\}$; [F4] and the formula from step 3.1 give $P(V\le 0)=0$. For $v_n=1-1\mathrm{/}n$ with $n\ge 2$, the events $\{V\le v_n\}$ increase to $\{V<1\}$; [F3] and step 3.1 give $P(V<1)=1$. Finally $0\le V\le 1$ by its definition, so $P(V\le 1)=1$. Thus $P(V\le u)=u$ for all $u\in [0,1]$, which is the uniform law.