[proofplan] The empirical distribution function converges uniformly almost surely to the true distribution function by the [Glivenko-Cantelli Theorem](/theorems/2004). The Kolmogorov-Smirnov distance from $F_n$ to $F_0$ is therefore forced to converge to the positive separation $\Delta$ between $F$ and $F_0$. Multiplying by $\sqrt{n}$ sends this statistic to infinity almost surely, while the critical values remain bounded. This implies that rejection occurs eventually on an event of probability one, so the rejection probabilities converge to $1$. [/proofplan]

[step:Use Glivenko-Cantelli to control the empirical distribution uniformly] Let $(\Omega, \mathcal{F}, \mathbb{P})$ be the probability space on which the sample is defined, and let $(X_n)_{n \in \mathbb{N}}$ be the i.i.d. real-valued random variables with distribution function $F$ whose empirical distribution functions are $(F_n)_{n \in \mathbb{N}}$. For each $n \in \mathbb{N}$, define the [random variable](/page/Random%20Variable) $E_n:\Omega \to [0,\infty]$ by \begin{align*} E_n(\omega)=\sup_{x \in \mathbb{R}} |F_n(x,\omega)-F(x)|. \end{align*} The supremum is a measurable random variable; this follows from the standard empirical distribution function measurability argument, reducing the supremum over $\mathbb{R}$ to a supremum over a countable dense set together with the one-sided continuity properties of distribution functions. By the [Glivenko-Cantelli Theorem](/theorems/2004), applied to the i.i.d. sequence $(X_n)_{n \in \mathbb{N}}$ with distribution function $F$, there exists an event $\Omega_0 \in \mathcal{F}$ with $\mathbb{P}(\Omega_0)=1$ such that \begin{align*} E_n(\omega) \to 0 \end{align*} for every $\omega \in \Omega_0$. [/step]

[step:Show the Kolmogorov-Smirnov distance converges to the fixed separation]For each $n \in \mathbb{N}$, define the random variable $D_n:\Omega \to [0,\infty]$ by \begin{align*} D_n(\omega)=\sup_{x \in \mathbb{R}} |F_n(x,\omega)-F_0(x)|. \end{align*} Since $F_0$ and $F_n(\cdot,\omega)$ are distribution functions, their one-sided continuity and monotonicity give the same countable-supremum measurability argument for $D_n$; no continuity of $F_0$ is needed. Fix $\omega \in \Omega_0$. For every $x \in \mathbb{R}$, the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |F_n(x,\omega)-F_0(x)| \geq |F(x)-F_0(x)| - |F_n(x,\omega)-F(x)|. \end{align*} Taking the supremum over $x \in \mathbb{R}$ on the left-hand side and using $|F_n(x,\omega)-F(x)| \leq E_n(\omega)$ for every $x \in \mathbb{R}$ gives \begin{align*} D_n(\omega) \geq \Delta - E_n(\omega). \end{align*} The ordinary triangle inequality also gives, for every $x \in \mathbb{R}$, \begin{align*} |F_n(x,\omega)-F_0(x)| \leq |F_n(x,\omega)-F(x)| + |F(x)-F_0(x)|. \end{align*} Taking suprema yields \begin{align*} D_n(\omega) \leq E_n(\omega) + \Delta. \end{align*} Since $E_n(\omega) \to 0$, the squeeze estimate \begin{align*} \Delta - E_n(\omega) \leq D_n(\omega) \leq \Delta + E_n(\omega) \end{align*} implies \begin{align*} D_n(\omega) \to \Delta. \end{align*}[/step]

[guided]The statistic in the test compares $F_n$ with the null distribution function $F_0$, but the theorem we can use directly compares $F_n$ with the true distribution function $F$. The bridge is the triangle inequality: if $F$ is separated from $F_0$ by $\Delta$, and $F_n$ is uniformly close to $F$, then $F_n$ must also be close to distance $\Delta$ from $F_0$. Recall from the previous step that $E_n:\Omega \to [0,\infty]$ is defined by \begin{align*} E_n(\omega)=\sup_{y \in \mathbb{R}} |F_n(y,\omega)-F(y)|, \end{align*} and that there is an event $\Omega_0 \in \mathcal{F}$ with $\mathbb{P}(\Omega_0)=1$ such that $E_n(\omega) \to 0$ for every $\omega \in \Omega_0$. For each $n \in \mathbb{N}$, define $D_n:\Omega \to [0,\infty]$ by \begin{align*} D_n(\omega)=\sup_{x \in \mathbb{R}} |F_n(x,\omega)-F_0(x)|. \end{align*} Since $F_0$ and the empirical distribution functions are distribution functions, their one-sided continuity and monotonicity make $D_n$ measurable by reducing the supremum to a countable supremum over a dense set adapted to their continuity points. Fix $\omega \in \Omega_0$. For a fixed point $x \in \mathbb{R}$, apply the reverse triangle inequality to the [real numbers](/page/Real%20Numbers) $F_n(x,\omega)-F_0(x)$ and $F(x)-F_0(x)$: \begin{align*} |F_n(x,\omega)-F_0(x)| \geq |F(x)-F_0(x)| - |F_n(x,\omega)-F(x)|. \end{align*} Because $E_n(\omega)$ is defined as the supremum of $|F_n(y,\omega)-F(y)|$ over all $y \in \mathbb{R}$, we have \begin{align*} |F_n(x,\omega)-F(x)| \leq E_n(\omega) \end{align*} for every $x \in \mathbb{R}$. Hence \begin{align*} |F_n(x,\omega)-F_0(x)| \geq |F(x)-F_0(x)| - E_n(\omega). \end{align*} Taking the supremum over $x \in \mathbb{R}$ gives \begin{align*} D_n(\omega) \geq \Delta - E_n(\omega). \end{align*} The matching upper bound uses the ordinary triangle inequality. For every $x \in \mathbb{R}$, \begin{align*} |F_n(x,\omega)-F_0(x)| \leq |F_n(x,\omega)-F(x)| + |F(x)-F_0(x)|. \end{align*} Taking the supremum over $x \in \mathbb{R}$ and bounding the two terms separately gives \begin{align*} D_n(\omega) \leq E_n(\omega) + \Delta. \end{align*} Thus \begin{align*} \Delta - E_n(\omega) \leq D_n(\omega) \leq \Delta + E_n(\omega). \end{align*} Since $\omega \in \Omega_0$, the previous step gives $E_n(\omega) \to 0$. Therefore the [squeeze theorem](/theorems/627) for real sequences implies \begin{align*} D_n(\omega) \to \Delta. \end{align*} This is the key consistency mechanism: the unscaled statistic does not shrink to zero under the fixed alternative; it converges to the positive separation $\Delta$.[/guided]

[step:Scale the positive limiting separation past the bounded critical values] Since $(c_n)_{n \in \mathbb{N}}$ is bounded, choose $M \in [0,\infty)$ such that \begin{align*} |c_n| \leq M \end{align*} for every $n \in \mathbb{N}$. Fix $\omega \in \Omega_0$. Since $D_n(\omega) \to \Delta$ and $\Delta>0$, there exists $N_1(\omega) \in \mathbb{N}$ such that \begin{align*} D_n(\omega) \geq \frac{\Delta}{2} \end{align*} for every $n \geq N_1(\omega)$. Choose $N_2 \in \mathbb{N}$ such that \begin{align*} \sqrt{n}\frac{\Delta}{2} > M \end{align*} for every $n \geq N_2$. For each $n \in \mathbb{N}$, define the rejection event \begin{align*} A_n:=\{\omega \in \Omega : \sqrt{n}D_n(\omega)>c_n\}. \end{align*} Since $D_n$ is measurable, each $A_n$ belongs to $\mathcal{F}$. Then for every $n \geq \max\{N_1(\omega),N_2\}$, \begin{align*} \sqrt{n}D_n(\omega) \geq \sqrt{n}\frac{\Delta}{2} > M \geq c_n. \end{align*} Thus $\omega \in A_n$ for all sufficiently large $n$. Hence \begin{align*} \mathbb{P}\left(\liminf_{n \to \infty} A_n\right)=1. \end{align*} [/step]

[step:Convert almost sure eventual rejection into consistency] For each $N \in \mathbb{N}$, define the event \begin{align*} B_N := \bigcap_{n=N}^{\infty} A_n. \end{align*} The events $(B_N)_{N \in \mathbb{N}}$ are increasing in $N$, and \begin{align*} \bigcup_{N=1}^{\infty} B_N = \liminf_{n \to \infty} A_n. \end{align*} By continuity from below of probability measures, \begin{align*} \lim_{N \to \infty} \mathbb{P}(B_N) = \mathbb{P}\left(\liminf_{n \to \infty} A_n\right) = 1. \end{align*} For every $n \geq N$, the inclusion $B_N \subset A_n$ gives \begin{align*} \mathbb{P}(A_n) \geq \mathbb{P}(B_N). \end{align*} Therefore \begin{align*} \liminf_{n \to \infty} \mathbb{P}(A_n) \geq \mathbb{P}(B_N) \end{align*} for every $N \in \mathbb{N}$. Letting $N \to \infty$ gives \begin{align*} \liminf_{n \to \infty} \mathbb{P}(A_n) \geq 1. \end{align*} Since $\mathbb{P}(A_n) \leq 1$ for every $n \in \mathbb{N}$, we conclude that \begin{align*} \mathbb{P}(A_n) \to 1. \end{align*} This is exactly consistency of the Kolmogorov-Smirnov test against the fixed alternative $F$. [/step]