[proofplan] We decompose the centered empirical process into the empirical fluctuation about the local alternative plus the deterministic local drift from the alternative to the null. The first term converges by hypothesis, while the second term converges uniformly to the fixed function $h\circ F_0$; Slutsky's theorem in the [Banach space](/page/Banach%20Space) $\ell^\infty(\mathbb R)$ then gives the process-level limit. The Kolmogorov-Smirnov and Cramer-von Mises limits follow by applying the continuous mapping theorem to the supremum functional and the squared-integral functional, respectively, and by using the probability integral transform for the continuous distribution function $F_0$. [/proofplan]

[step:Split the empirical process into fluctuation and deterministic drift]For each $n\in\mathbb N$, define the random element $Z_n:\mathbb R\to\mathbb R$ by \begin{align*} Z_n(x):=\sqrt n\left(F_n(x)-F_0(x)\right). \end{align*} Define also the random element $Y_n:\mathbb R\to\mathbb R$ by \begin{align*} Y_n(x):=\sqrt n\left(F_n(x)-F_{n,\mathrm{alt}}(x)\right). \end{align*} Define the deterministic element $r_n:\mathbb R\to\mathbb R$ by \begin{align*} r_n(x):=\sqrt n\left(F_{n,\mathrm{alt}}(x)-F_0(x)\right). \end{align*} Then, pointwise for every $x\in\mathbb R$, \begin{align*} Z_n(x)=Y_n(x)+r_n(x). \end{align*} Let $g:\mathbb R\to\mathbb R$ be the deterministic function \begin{align*} g(x):=h(F_0(x)). \end{align*} Since $h:[0,1]\to\mathbb R$ is continuous on the compact interval $[0,1]$, $h$ is bounded, and hence $g\in\ell^\infty(\mathbb R)$ because $F_0(\mathbb R)\subset[0,1]$. The second hypothesis says precisely that \begin{align*} \|r_n-g\|_{\ell^\infty(\mathbb R)} = \sup_{x\in\mathbb R}|r_n(x)-g(x)|\to0. \end{align*}[/step]

[guided]The process we want to understand is \begin{align*} Z_n(x):=\sqrt n\left(F_n(x)-F_0(x)\right). \end{align*} The useful decomposition is obtained by adding and subtracting the local-alternative distribution function $F_{n,\mathrm{alt}}$. Define \begin{align*} Y_n(x):=\sqrt n\left(F_n(x)-F_{n,\mathrm{alt}}(x)\right). \end{align*} Define \begin{align*} r_n(x):=\sqrt n\left(F_{n,\mathrm{alt}}(x)-F_0(x)\right). \end{align*} Then, for every $x\in\mathbb R$, \begin{align*} Z_n(x)=\sqrt n\left(F_n(x)-F_0(x)\right). \end{align*} Adding and subtracting $F_{n,\mathrm{alt}}(x)$ gives \begin{align*} Z_n(x)=\sqrt n\left(F_n(x)-F_{n,\mathrm{alt}}(x)\right) +\sqrt n\left(F_{n,\mathrm{alt}}(x)-F_0(x)\right). \end{align*} By the definitions of $Y_n$ and $r_n$, \begin{align*} Z_n(x)=Y_n(x)+r_n(x). \end{align*} This separates the random empirical fluctuation $Y_n$ from the deterministic local drift $r_n$. The deterministic limiting drift is the function $g:\mathbb R\to\mathbb R$ defined by \begin{align*} g(x):=h(F_0(x)). \end{align*} Since $h$ is continuous on the compact interval $[0,1]$, it is bounded, and therefore $g\in\ell^\infty(\mathbb R)$. The second displayed hypothesis is exactly the assertion that $r_n$ converges to $g$ in the uniform norm: \begin{align*} \|r_n-g\|_{\ell^\infty(\mathbb R)} = \sup_{x\in\mathbb R} \left| \sqrt n\left(F_{n,\mathrm{alt}}(x)-F_0(x)\right)-h(F_0(x)) \right| \to0. \end{align*}[/guided]

[step:Combine the Brownian bridge fluctuation with the uniform drift limit]Define the random element $Y:\mathbb R\to\mathbb R$ by \begin{align*} Y(x):=B(F_0(x)). \end{align*} By the first hypothesis, $Y_n\xrightarrow{d}Y$ in $\ell^\infty(\mathbb R)$. Since $h$ is bounded on $[0,1]$ and $\|r_n-g\|_{\ell^\infty(\mathbb R)}\to0$, there exists $N\in\mathbb N$ such that $r_n\in\ell^\infty(\mathbb R)$ for all $n\geq N$. The finitely many earlier indices do not affect the limiting distribution, so we apply the argument for $n\geq N$. By the previous step, $r_n\to g$ in $\ell^\infty(\mathbb R)$, hence $r_n\xrightarrow{\mathbb P}g$ in $\ell^\infty(\mathbb R)$. Define the addition map $A:\ell^\infty(\mathbb R)\times\ell^\infty(\mathbb R)\to\ell^\infty(\mathbb R)$ by $A(f_1,f_2)=f_1+f_2$. This map is continuous for the uniform norm because \begin{align*} \|A(f_1,f_2)-A(k_1,k_2)\|_{\ell^\infty(\mathbb R)} \leq \|f_1-k_1\|_{\ell^\infty(\mathbb R)}+ \|f_2-k_2\|_{\ell^\infty(\mathbb R)}. \end{align*} The [Slutsky's Lemma](/theorems/1850) metric-space form applies to $Y_n\xrightarrow{d}Y$ and $r_n\xrightarrow{\mathbb P}g$, and gives \begin{align*} Z_n=Y_n+r_n\xrightarrow{d}Y+g \end{align*} in $\ell^\infty(\mathbb R)$. Therefore \begin{align*} x\mapsto \sqrt n\left(F_n(x)-F_0(x)\right) \xrightarrow{d} x\mapsto B(F_0(x))+h(F_0(x)). \end{align*}[/step]

[guided]The first hypothesis gives the random fluctuation limit: \begin{align*} Y_n\xrightarrow{d}Y \end{align*} in $\ell^\infty(\mathbb R)$, where the random element $Y:\mathbb R\to\mathbb R$ is defined by \begin{align*} Y(x):=B(F_0(x)). \end{align*} The previous step converted the local-alternative assumption into the deterministic norm convergence \begin{align*} \|r_n-g\|_{\ell^\infty(\mathbb R)}\to0, \end{align*} where $g:\mathbb R\to\mathbb R$ is defined by $g(x):=h(F_0(x))$. Deterministic convergence in the metric of $\ell^\infty(\mathbb R)$ implies convergence in probability, so \begin{align*} r_n\xrightarrow{\mathbb P}g \end{align*} in $\ell^\infty(\mathbb R)$. We now use the [Slutsky's Lemma](/theorems/1850) metric-space form in $\ell^\infty(\mathbb R)$. The operation being used is addition, so define the map $A:\ell^\infty(\mathbb R)\times\ell^\infty(\mathbb R)\to\ell^\infty(\mathbb R)$ by $A(f_1,f_2)=f_1+f_2$. This map is continuous for the uniform norm because, for $f_1,f_2,k_1,k_2\in\ell^\infty(\mathbb R)$, \begin{align*} \|A(f_1,f_2)-A(k_1,k_2)\|_{\ell^\infty(\mathbb R)} \leq \|f_1-k_1\|_{\ell^\infty(\mathbb R)}+ \|f_2-k_2\|_{\ell^\infty(\mathbb R)}. \end{align*} Slutsky's theorem therefore gives \begin{align*} Y_n+r_n\xrightarrow{d}Y+g \end{align*} in $\ell^\infty(\mathbb R)$. Since the first step proved $Z_n=Y_n+r_n$, this is exactly \begin{align*} x\mapsto \sqrt n\left(F_n(x)-F_0(x)\right) \xrightarrow{d} x\mapsto B(F_0(x))+h(F_0(x)). \end{align*}[/guided]

[step:Apply the supremum functional to obtain the Kolmogorov-Smirnov limit]Define the supremum functional $S:\ell^\infty(\mathbb R)\to\mathbb R$ by \begin{align*} S(f):=\sup_{x\in\mathbb R}|f(x)|. \end{align*} Let $K_n$ denote the corresponding Kolmogorov-Smirnov-type statistic, defined by \begin{align*} K_n:=S(Z_n). \end{align*} For $f_1,f_2\in\ell^\infty(\mathbb R)$, the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |S(f_1)-S(f_2)| \leq \sup_{x\in\mathbb R}|f_1(x)-f_2(x)| = \|f_1-f_2\|_{\ell^\infty(\mathbb R)}. \end{align*} Thus $S$ is continuous. By the [Continuous Mapping Theorem](/theorems/1847), \begin{align*} K_n=S(Z_n)\xrightarrow{d}S(Y+g). \end{align*} Since $F_0$ is a continuous distribution function, $F_0(\mathbb R)$ is dense in $(0,1)$ after adjoining its endpoint limits, and since $B+h$ has continuous sample paths on $[0,1]$, we have almost surely \begin{align*} S(Y+g)=\sup_{x\in\mathbb R}|B(F_0(x))+h(F_0(x))|. \end{align*} The density of the image of $F_0$ and continuity of $t\mapsto B(t)+h(t)$ imply \begin{align*} S(Y+g)=\sup_{0\leq t\leq1}|B(t)+h(t)|. \end{align*} Therefore \begin{align*} K_n\xrightarrow{d}\sup_{0\leq t\leq1}|B(t)+h(t)|. \end{align*}[/step]

[guided]We turn the process convergence into the Kolmogorov-Smirnov statistic by applying the supremum functional. Define $S:\ell^\infty(\mathbb R)\to\mathbb R$ by \begin{align*} S(f):=\sup_{x\in\mathbb R}|f(x)|. \end{align*} Let $K_n$ denote the Kolmogorov-Smirnov-type statistic associated to $Z_n$, defined by \begin{align*} K_n:=S(Z_n). \end{align*} This is a valid real-valued map because each $f\in\ell^\infty(\mathbb R)$ is bounded. For $f_1,f_2\in\ell^\infty(\mathbb R)$, the reverse triangle inequality gives \begin{align*} |S(f_1)-S(f_2)| \leq \sup_{x\in\mathbb R}|f_1(x)-f_2(x)| = \|f_1-f_2\|_{\ell^\infty(\mathbb R)}. \end{align*} Thus $S$ is Lipschitz continuous with Lipschitz constant $1$. The process limit proved in the previous step is \begin{align*} Z_n\xrightarrow{d}Y+g \end{align*} in $\ell^\infty(\mathbb R)$. Since $S$ is continuous, the [Continuous Mapping Theorem](/theorems/1847) gives \begin{align*} K_n=S(Z_n)\xrightarrow{d}S(Y+g). \end{align*} It remains to identify $S(Y+g)$ with the standard Brownian-bridge drift supremum on $[0,1]$. Since $F_0$ is a continuous distribution function, its image over $\mathbb R$, together with the endpoint limits $0$ and $1$, is dense in $[0,1]$. Since the Brownian bridge has continuous sample paths and $h$ is continuous, the map $t\mapsto B(t)+h(t)$ is almost surely continuous on $[0,1]$. Therefore taking the supremum over the dense set reached by $F_0$ gives the same value as taking the supremum over all of $[0,1]$: \begin{align*} S(Y+g)=\sup_{0\leq t\leq1}|B(t)+h(t)|. \end{align*} Combining these conclusions yields \begin{align*} K_n\xrightarrow{d}\sup_{0\leq t\leq1}|B(t)+h(t)|. \end{align*}[/guided]

[step:Apply the squared-integral functional on measurable bounded paths to obtain the Cramer-von Mises limit]Let $\mu_0$ denote the probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ whose distribution function is $F_0$. Let $\mathcal M$ be the set of bounded Borel-[measurable functions](/page/Measurable%20Functions) $f:\mathbb R\to\mathbb R$, equipped with the metric inherited from $\ell^\infty(\mathbb R)$. The space $\mathcal M$ is closed in $\ell^\infty(\mathbb R)$ because a uniform limit of bounded Borel-measurable functions is bounded and Borel-measurable. Define the squared-integral functional $Q:\mathcal M\to\mathbb R$ by \begin{align*} Q(f):=\int_{\mathbb R}f(x)^2\,d\mu_0(x). \end{align*} This definition is well-defined because $f^2$ is bounded and Borel-measurable whenever $f\in\mathcal M$. The empirical distribution functions and the deterministic distribution functions are Borel-measurable, so $Z_n\in\mathcal M$ for every $n\in\mathbb N$. Also $Y+g\in\mathcal M$ almost surely, since the Brownian bridge has continuous sample paths on $[0,1]$, $F_0$ is Borel-measurable, and $h$ is continuous. If $f_m\to f$ in the uniform metric on $\mathcal M$, then \begin{align*} |Q(f_m)-Q(f)| \leq \int_{\mathbb R}|f_m(x)-f(x)|\,|f_m(x)+f(x)|\,d\mu_0(x). \end{align*} Since $\mu_0(\mathbb R)=1$, this gives \begin{align*} |Q(f_m)-Q(f)| \leq \|f_m-f\|_{\ell^\infty(\mathbb R)} \left(\|f_m\|_{\ell^\infty(\mathbb R)}+\|f\|_{\ell^\infty(\mathbb R)}\right). \end{align*} [Uniform convergence](/page/Uniform%20Convergence) implies that $\|f_m\|_{\ell^\infty(\mathbb R)}$ is bounded for all sufficiently large $m$, so the right-hand side tends to $0$. Hence $Q$ is continuous on $\mathcal M$. Although the process convergence is stated in $\ell^\infty(\mathbb R)$, the statistic only uses the closed Borel-measurable subspace $\mathcal M$: the paths $Z_n$ belong to $\mathcal M$ for every $n\in\mathbb N$, and $Y+g$ belongs to $\mathcal M$ almost surely. Since $\mathcal M$ is closed, the Borel $\sigma$-algebra on $\mathcal M$ is the subspace Borel $\sigma$-algebra inherited from $\ell^\infty(\mathbb R)$; equivalently, convergence in distribution in the ambient metric restricts to convergence in distribution in $\mathcal M$ for random elements supported on $\mathcal M$. Therefore the [Continuous Mapping Theorem](/theorems/1847) applies to the continuous map $Q:\mathcal M\to\mathbb R$. Let $C_n$ denote the Cramer-von Mises-type statistic, defined by \begin{align*} C_n:=Q(Z_n). \end{align*} Then \begin{align*} C_n=Q(Z_n)\xrightarrow{d}Q(Y+g). \end{align*} Finally, because $F_0$ is continuous, if $X$ is a real-valued [random variable](/page/Random%20Variable) with distribution function $F_0$, then $F_0(X)$ has the uniform distribution on $[0,1]$. Almost surely, the map $t\mapsto (B(t)+h(t))^2$ is bounded and Borel-measurable on $[0,1]$, because $B$ has continuous sample paths and $h$ is continuous. Applying the probability-integral-transform change of variables pathwise gives \begin{align*} Q(Y+g)=\int_{\mathbb R}\left(B(F_0(x))+h(F_0(x))\right)^2\,d\mu_0(x). \end{align*} Therefore \begin{align*} Q(Y+g)=\int_0^1\left(B(t)+h(t)\right)^2\,d\mathcal L^1(t). \end{align*} Consequently \begin{align*} C_n\xrightarrow{d}\int_0^1\left(B(t)+h(t)\right)^2\,d\mathcal L^1(t). \end{align*} This completes the proof of the process limit and of both EDF-test statistic limits.[/step]

[guided]For the Cramer-von Mises statistic we need an integral, so we restrict attention to paths for which the integral is defined. Let $\mu_0$ denote the probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ whose distribution function is $F_0$. Let $\mathcal M$ be the set of bounded Borel-measurable functions $f:\mathbb R\to\mathbb R$, equipped with the metric inherited from $\ell^\infty(\mathbb R)$. The space $\mathcal M$ is closed in $\ell^\infty(\mathbb R)$: if $f_m\in\mathcal M$ and $f_m\to f$ uniformly, then $f$ is bounded and Borel-measurable as a uniform limit of Borel-measurable functions. Define $Q:\mathcal M\to\mathbb R$ by \begin{align*} Q(f):=\int_{\mathbb R}f(x)^2\,d\mu_0(x). \end{align*} This integral is well-defined because $f^2$ is bounded and Borel-measurable and because $\mu_0$ is a probability measure. We verify continuity of $Q$. If $f_m\to f$ uniformly in $\mathcal M$, then \begin{align*} |Q(f_m)-Q(f)| \leq \int_{\mathbb R}|f_m(x)-f(x)|\,|f_m(x)+f(x)|\,d\mu_0(x). \end{align*} Since $\mu_0(\mathbb R)=1$, this implies \begin{align*} |Q(f_m)-Q(f)| \leq \|f_m-f\|_{\ell^\infty(\mathbb R)} \left(\|f_m\|_{\ell^\infty(\mathbb R)}+\|f\|_{\ell^\infty(\mathbb R)}\right). \end{align*} Uniform convergence implies that the norms $\|f_m\|_{\ell^\infty(\mathbb R)}$ are bounded for all sufficiently large $m$, and therefore the right-hand side tends to $0$. Hence $Q$ is continuous on $\mathcal M$. The subspace issue is important: $Q$ is not defined on all of $\ell^\infty(\mathbb R)$, because arbitrary bounded functions need not be Borel-measurable. Here the empirical distribution functions and deterministic distribution functions are Borel-measurable, so $Z_n\in\mathcal M$ for every $n\in\mathbb N$. Also $Y+g\in\mathcal M$ almost surely because the Brownian bridge has continuous sample paths on $[0,1]$, $F_0$ is Borel-measurable, and $h$ is continuous. Since $\mathcal M$ is closed in $\ell^\infty(\mathbb R)$, its Borel $\sigma$-algebra is exactly the subspace Borel $\sigma$-algebra, and the ambient convergence in distribution restricts to convergence in distribution for these $\mathcal M$-valued random elements. The [Continuous Mapping Theorem](/theorems/1847) therefore applies to $Q:\mathcal M\to\mathbb R$. Let $C_n$ denote the Cramer-von Mises-type statistic associated to $Z_n$, defined by \begin{align*} C_n:=Q(Z_n). \end{align*} Then \begin{align*} C_n=Q(Z_n)\xrightarrow{d}Q(Y+g). \end{align*} It remains to rewrite $Q(Y+g)$ as an integral over $[0,1]$. Since $F_0$ is continuous and $X$ is a real-valued random variable with distribution function $F_0$, the probability integral transform gives that $F_0(X)$ has the uniform distribution on $[0,1]$. Almost surely, $t\mapsto (B(t)+h(t))^2$ is bounded and Borel-measurable on $[0,1]$. Applying this change of variables pathwise yields \begin{align*} Q(Y+g)=\int_0^1\left(B(t)+h(t)\right)^2\,d\mathcal L^1(t). \end{align*} Therefore \begin{align*} C_n\xrightarrow{d}\int_0^1\left(B(t)+h(t)\right)^2\,d\mathcal L^1(t). \end{align*}[/guided]