[proofplan] We first record the basic properties of the distribution function $F$ that follow from the fact that $\mu$ is a Borel probability measure: monotonicity, right-continuity, and the limits at $\pm\infty$. These properties imply that the quantile set $\{y\in\mathbb R:F(y)\ge p\}$ is nonempty and bounded below for every $p\in(0,1)$, so $Q(p)$ is a real number. The main point is the inverse-threshold equivalence $Q(p)\le x$ if and only if $p\le F(x)$; right-continuity is used in the direction from $Q(p)\le x$ to $p\le F(x)$. This equivalence gives measurability of $Q$ and then converts the event $\{X\le x\}$ into the event $\{U\le F(x)\}$, whose probability is $F(x)$ by uniformity. [/proofplan]

[step:Verify the distribution function properties needed for the quantile argument]For $x_1,x_2\in\mathbb R$ with $x_1\le x_2$, we have $(-\infty,x_1]\subseteq(-\infty,x_2]$, hence monotonicity of the measure $\mu$ gives \begin{align*} F(x_1)\le F(x_2). \end{align*} Thus $F$ is nondecreasing. We next prove right-continuity. Fix $x\in\mathbb R$, and let $(x_n)_{n\ge 1}$ be any sequence in $\mathbb R$ such that $x_n\downarrow x$. Define Borel sets $A_n:=(-\infty,x_n]$ for $n\in\mathbb N$. Then $(A_n)_{n\ge 1}$ is decreasing and \begin{align*} \bigcap_{n=1}^{\infty}A_n=(-\infty,x]. \end{align*} Since $\mu(A_1)\le 1<\infty$, continuity from above for finite measures gives \begin{align*} \lim_{n\to\infty}\mu(A_n)=\mu\left(\bigcap_{n=1}^{\infty}A_n\right). \end{align*} Therefore \begin{align*} \lim_{n\to\infty}F(x_n)=F(x). \end{align*} This proves right-continuity of $F$. Finally, define $B_n:=(-\infty,-n]$ and $C_n:=(-\infty,n]$ for $n\in\mathbb N$. Then $B_n\downarrow\varnothing$, so continuity from above gives \begin{align*} \lim_{n\to\infty}F(-n)=0. \end{align*} Also $C_n\uparrow\mathbb R$, so continuity from below gives \begin{align*} \lim_{n\to\infty}F(n)=1. \end{align*} Since $F$ is nondecreasing, these sequential limits imply \begin{align*} \lim_{x\to-\infty}F(x)=0 \end{align*} and \begin{align*} \lim_{x\to\infty}F(x)=1. \end{align*}[/step]

[guided]We need exactly the structural facts about $F$ that make an inverse-threshold construction work. First, if $x_1\le x_2$, then the interval $(-\infty,x_1]$ is contained in $(-\infty,x_2]$. Because $\mu$ is a measure, it is monotone on Borel sets, and therefore \begin{align*} F(x_1)=\mu((-\infty,x_1])\le \mu((-\infty,x_2])=F(x_2). \end{align*} So $F$ is nondecreasing. The next point is right-continuity. Fix $x\in\mathbb R$, and let $(x_n)_{n\ge 1}$ be a decreasing sequence with $x_n\downarrow x$. Define \begin{align*} A_n:=(-\infty,x_n] \end{align*} for each $n\in\mathbb N$. These sets are Borel sets because they are closed intervals of the form $(-\infty,a]$. Since $x_{n+1}\le x_n$, the sets decrease: \begin{align*} A_{n+1}\subseteq A_n. \end{align*} Their intersection is precisely $(-\infty,x]$: a real number belongs to every $(-\infty,x_n]$ exactly when it is at most the limit $x$. Since $\mu(A_1)\le \mu(\mathbb R)=1<\infty$, continuity from above for finite measures applies and gives \begin{align*} \lim_{n\to\infty}\mu(A_n)=\mu\left(\bigcap_{n=1}^{\infty}A_n\right)=\mu((-\infty,x]). \end{align*} By the definition of $F$, this is \begin{align*} \lim_{n\to\infty}F(x_n)=F(x). \end{align*} This is the right-continuity of $F$. Finally, we verify the endpoint limits. Define $B_n:=(-\infty,-n]$ for $n\in\mathbb N$. Then $B_n\downarrow\varnothing$, and continuity from above gives \begin{align*} \lim_{n\to\infty}F(-n)=\lim_{n\to\infty}\mu(B_n)=\mu(\varnothing)=0. \end{align*} Define also $C_n:=(-\infty,n]$. Then $C_n\uparrow\mathbb R$, and continuity from below gives \begin{align*} \lim_{n\to\infty}F(n)=\lim_{n\to\infty}\mu(C_n)=\mu(\mathbb R)=1. \end{align*} Because $F$ is nondecreasing, the integer limits force the full real limits: \begin{align*} \lim_{x\to-\infty}F(x)=0 \end{align*} and \begin{align*} \lim_{x\to\infty}F(x)=1. \end{align*} These facts will be used to ensure that each quantile is finite and to prove the threshold equivalence.[/guided]

[step:Show that every quantile value is finite] Fix $p\in(0,1)$, and define \begin{align*} A_p:=\{y\in\mathbb R:F(y)\ge p\}. \end{align*} Since $\lim_{y\to\infty}F(y)=1$ and $p<1$, there exists $b\in\mathbb R$ such that $F(b)\ge p$, so $A_p\neq\varnothing$. Since $\lim_{y\to-\infty}F(y)=0$ and $p>0$, there exists $a\in\mathbb R$ such that $F(a)<p$. For every $y\le a$, monotonicity of $F$ gives $F(y)\le F(a)<p$, hence $y\notin A_p$. Therefore $A_p\subseteq(a,\infty)$. Thus $A_p$ is nonempty and bounded below, so its infimum is a real number. Consequently $Q(p)\in\mathbb R$ for every $p\in(0,1)$. [/step]

[step:Prove the inverse-threshold equivalence using right-continuity]We claim that for every $p\in(0,1)$ and every $x\in\mathbb R$, \begin{align*} Q(p)\le x\quad\iff\quad p\le F(x). \end{align*} Fix $p\in(0,1)$ and $x\in\mathbb R$. If $p\le F(x)$, then $x\in A_p$, where \begin{align*} A_p:=\{y\in\mathbb R:F(y)\ge p\}. \end{align*} Since $Q(p)=\inf A_p$, it follows that $Q(p)\le x$. Conversely, assume $Q(p)\le x$. For each $n\in\mathbb N$, the number $x+\frac{1}{n}$ is strictly larger than $\inf A_p$. By the defining property of the infimum, there exists $y_n\in A_p$ such that \begin{align*} y_n<x+\frac{1}{n}. \end{align*} Since $y_n\in A_p$, we have $F(y_n)\ge p$. Also $y_n<x+\frac{1}{n}$, so monotonicity gives \begin{align*} p\le F(y_n)\le F\left(x+\frac{1}{n}\right). \end{align*} Taking the limit as $n\to\infty$ and using right-continuity of $F$ at $x$, because $x+\frac{1}{n}\downarrow x$, gives \begin{align*} p\le F(x). \end{align*} This proves the equivalence.[/step]

[guided]The equivalence \begin{align*} Q(p)\le x\quad\iff\quad p\le F(x) \end{align*} is the core of the theorem. It says that the event that the quantile at level $p$ lies below $x$ is exactly the same as the condition that the level $p$ lies below the distribution value at $x$. Fix $p\in(0,1)$ and $x\in\mathbb R$, and define \begin{align*} A_p:=\{y\in\mathbb R:F(y)\ge p\}. \end{align*} By definition, $Q(p)=\inf A_p$. First suppose $p\le F(x)$. Then $x$ itself belongs to $A_p$. Since the infimum of a set is less than or equal to every element of the set, we get \begin{align*} Q(p)=\inf A_p\le x. \end{align*} Now suppose $Q(p)\le x$. The subtle point is that $Q(p)$ may not itself belong to $A_p$ before we use right-continuity, so we cannot immediately write $F(Q(p))\ge p$. Instead, we approximate the infimum from above. For each $n\in\mathbb N$, the number $x+\frac{1}{n}$ is larger than $Q(p)=\inf A_p$. Therefore the defining property of the infimum gives an element $y_n\in A_p$ satisfying \begin{align*} y_n<x+\frac{1}{n}. \end{align*} Because $y_n\in A_p$, we have \begin{align*} F(y_n)\ge p. \end{align*} Because $F$ is nondecreasing and $y_n<x+\frac{1}{n}$, we also have \begin{align*} F(y_n)\le F\left(x+\frac{1}{n}\right). \end{align*} Combining these inequalities yields \begin{align*} p\le F\left(x+\frac{1}{n}\right) \end{align*} for every $n\in\mathbb N$. The sequence $x+\frac{1}{n}$ decreases to $x$, so right-continuity of $F$ at $x$ gives \begin{align*} \lim_{n\to\infty}F\left(x+\frac{1}{n}\right)=F(x). \end{align*} Passing to the limit in the inequality $p\le F(x+\frac{1}{n})$ gives \begin{align*} p\le F(x). \end{align*} This proves the reverse implication and hence the equivalence.[/guided]

[step:Deduce measurability of the quantile map and of the transformed random variable] For each $x\in\mathbb R$, the inverse-threshold equivalence gives \begin{align*} Q^{-1}((-\infty,x])=\{p\in(0,1):p\le F(x)\}. \end{align*} Since $F(x)\in[0,1]$, this set is either $\varnothing$, or $(0,F(x)]$, or all of $(0,1)$. In every case it belongs to $\mathcal B((0,1))$. Therefore $Q:(0,1)\to\mathbb R$ is Borel measurable, because the half-lines $(-\infty,x]$ generate $\mathcal B(\mathbb R)$. Since $U:(\Omega,\mathcal F)\to((0,1),\mathcal B((0,1)))$ is measurable and $Q:(0,1)\to\mathbb R$ is Borel measurable, the composition \begin{align*} X=Q\circ U:\Omega\to\mathbb R \end{align*} is $\mathcal F/\mathcal B(\mathbb R)$-measurable. [/step]

[step:Compute the distribution function of the transformed random variable] Fix $x\in\mathbb R$. Using the definition $X=Q\circ U$ and the inverse-threshold equivalence with $p=U(\omega)$, we obtain \begin{align*} \{\omega\in\Omega:X(\omega)\le x\}=\{\omega\in\Omega:U(\omega)\le F(x)\}. \end{align*} Since $F(x)\in[0,1]$ and $U\sim\operatorname{Unif}(0,1)$, the defining distribution function of the uniform law gives \begin{align*} \mathbb P(\{\omega\in\Omega:U(\omega)\le F(x)\})=F(x). \end{align*} Therefore \begin{align*} \mathbb P(\{\omega\in\Omega:X(\omega)\le x\})=F(x). \end{align*} Since $x\in\mathbb R$ was arbitrary, $X$ has distribution function $F$. [/step]