> [!theorem] > > Let $\seq{X_n}$ be a [[Sequence|sequence]] of [[Random Variable|random variables]] defined on $(\Omega_n, \cf_n, \bp_n)$. If $X_n \xrightarrow{\text d} X$ [[Convergence in Distribution|in distribution]], then there exists a [[Coupling|coupling]] $\seq{Y_n}$ of $\seq{X_n}$ such that $Y_n \to Y$ [[Almost Everywhere|a.e.]]. > > *Proof*. Let $(\Omega, \cf, \bp) = ([0, 1], \cb([0, 1]), m)$ be the target probability space. Let $Y_n: \Omega \to \real$ > $ > Y_n(p) = \inf\bracs{x: F_{X_n}(x) \ge p} > $ > be the "right inverse" of each [[Cumulative Distribution Function|CDF]], then $F_{Y_n} = F_{X_n}$ for all $n$, yielding a coupling. > > Note that the limit $Y_\infty$ is monotone, and can still only have countably many discontinuity points. > > Let $p \in [0, 1]$ be a continuity point of $Y_\infty$, then $F_{X_\infty}$ must be strictly increasing near $p$ (otherwise there would be a jump in $Y_\infty$): > $ > F_{X_\infty}(x) < p < F_{X_\infty}(z) \quad x < Y_\infty(p) < z > $ > Let $y = Y_{\infty}(p)$. Take $\varepsilon > 0$ and $x \in (y - \varepsilon, y)$ and $z \in (y, y + \varepsilon)$ such that $x, z$ are both continuity points. Since $X_n \xrightarrow{d} X_\infty$, $F_{X_n}(x) \to F_{X_\infty}(x) < p$, and $F_{X_n}(x) < p$ eventually. Therefore > $ > Y_n(p) \ge x \ge y - \varepsilon > $ > eventually. Similarly, $F_{X_n}(z) \to F_{X_\infty}(z) > p$, and $F_{X_n}(z) > p$ eventually. Therefore > $ > Y_n(p) \le z < y - \varepsilon > $ > and $Y_n(p) \to Y_\infty(p)$ pointwise. > > Since the discontinuity points here forms a [[Null Set|null set]], $Y_n \to Y_\infty$ [[Almost Everywhere|a.s.]] [[Pointwise Convergence|pointwise]].