[proofplan] We first identify the left limit $F(a-)$ with the measure of the open half-line $(-\infty,a)$ by approximating that half-line from below with closed half-lines. The atom identity then follows by decomposing $(-\infty,a]$ into $(-\infty,a)$ and the singleton $\{a\}$. Finally, for any $p$ between the left limit and the value at $a$, all points strictly below $a$ have distribution value below $p$, while $a$ itself has distribution value at least $p$; the generalized inverse therefore has infimum exactly $a$. [/proofplan] [step:Identify the left limit with the measure below $a$] For each positive integer $n$, define the Borel set \begin{align*} A_n:=(-\infty,a-\frac{1}{n}]. \end{align*} The sequence $(A_n)_{n\ge1}$ is increasing and satisfies \begin{align*} \bigcup_{n=1}^{\infty}A_n=(-\infty,a). \end{align*} By countable additivity of $\mu$ applied to the increasing union, \begin{align*} \mu((-\infty,a))=\lim_{n\to\infty}\mu(A_n). \end{align*} Since $F(x)=\mu((-\infty,x])$ for every $x\in\mathbb R$, this gives \begin{align*} \mu((-\infty,a))=\lim_{n\to\infty}F(a-\frac{1}{n}). \end{align*} Because $a-\frac{1}{n}\uparrow a$ and $F(a-)$ is the left limit of $F$ at $a$, \begin{align*} F(a-)=\mu((-\infty,a)). \end{align*} [guided] We want to convert the analytic object $F(a-)$ into a measure of a set. The natural set is $(-\infty,a)$, because it is obtained by approaching $a$ from the left. To make this precise, for each positive integer $n$ define \begin{align*} A_n:=(-\infty,a-\frac{1}{n}]. \end{align*} Each $A_n$ is a Borel subset of $\mathbb R$, and the sequence is increasing: if $n\le m$, then $a-\frac{1}{n}\le a-\frac{1}{m}$, hence $A_n\subset A_m$. The union is exactly the open half-line below $a$: \begin{align*} \bigcup_{n=1}^{\infty}A_n=(-\infty,a). \end{align*} Indeed, every point of every $A_n$ is strictly less than $a$. Conversely, if $x<a$, then $a-x>0$, so for all sufficiently large $n$ we have $\frac{1}{n}<a-x$, equivalently $x<a-\frac{1}{n}$, and hence $x\in A_n$. By countable additivity of the measure $\mu$ on the increasing union, we obtain \begin{align*} \mu((-\infty,a))=\lim_{n\to\infty}\mu(A_n). \end{align*} Using the definition of the distribution function, \begin{align*} \mu(A_n)=\mu((-\infty,a-\frac{1}{n}])=F(a-\frac{1}{n}) \end{align*} for every positive integer $n$. Therefore \begin{align*} \mu((-\infty,a))=\lim_{n\to\infty}F(a-\frac{1}{n}). \end{align*} Since the sequence $a-\frac{1}{n}$ increases to $a$, the last limit is precisely the left limit $F(a-)$. Hence \begin{align*} F(a-)=\mu((-\infty,a)). \end{align*} [/guided] [/step] [step:Subtract the mass below $a$ from the mass up to $a$] The Borel sets $(-\infty,a)$ and $\{a\}$ are disjoint, and \begin{align*} (-\infty,a]=(-\infty,a)\cup\{a\}. \end{align*} By finite additivity of $\mu$, \begin{align*} F(a)=\mu((-\infty,a])=\mu((-\infty,a))+\mu(\{a\}). \end{align*} Using $F(a-)=\mu((-\infty,a))$ from the previous step and rearranging gives \begin{align*} \mu(\{a\})=F(a)-F(a-). \end{align*} [/step] [step:Locate the generalized inverse on the atom interval] Fix $p\in(F(a-),F(a)]$. Define the threshold set \begin{align*} S_p:=\{x\in\mathbb R:F(x)\ge p\}. \end{align*} Since $p\le F(a)$, we have $a\in S_p$, so $S_p$ is nonempty and \begin{align*} \inf S_p\le a. \end{align*} If $x<a$, then $(-\infty,x]\subset(-\infty,a)$, and therefore \begin{align*} F(x)=\mu((-\infty,x])\le\mu((-\infty,a))=F(a-)<p. \end{align*} Thus no point $x<a$ belongs to $S_p$, so $S_p\subset [a,\infty)$ and \begin{align*} \inf S_p\ge a. \end{align*} Combining the two inequalities gives \begin{align*} \inf S_p=a. \end{align*} By the definition of the generalized [quantile function](/page/Quantile%20Function), \begin{align*} F^{-1}(p)=\inf\{x\in\mathbb R:F(x)\ge p\}=\inf S_p=a. \end{align*} This proves the asserted quantile plateau identity for every $p\in(F(a-),F(a)]$. [/step]