[proofplan] The proof identifies the limiting expectation of every bounded continuous [test function](/page/Test%20Function) under $Q_n$. The Radon-Nikodym identity rewrites this expectation as an expectation under $P_n$ with weight $L_n$. Since the limiting likelihood ratio has expectation $1$, the convergence $L_n \xrightarrow{d} L$ loses no mass, which gives uniform integrability of $(L_n)$ under $(P_n)$. We first pass to the limit for truncated likelihood ratios, then remove the truncation uniformly. [/proofplan] [step:Show that the limiting tilted set function is a probability measure] Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space on which the weak limit $(T,L)$ is realised, so $T:(\Omega,\mathcal F)\to(S,\mathcal B(S))$ is measurable and $L:(\Omega,\mathcal F)\to([0,\infty),\mathcal B([0,\infty)))$ is measurable. Define $\mu^Q:\mathcal B(S)\to[0,\infty)$ by \begin{align*} \mu^Q(B) := \mathbb E[\mathbb 1_{\{T\in B\}}L], \qquad B\in\mathcal B(S). \end{align*} For pairwise disjoint sets $B_1,B_2,\dots\in\mathcal B(S)$, the indicators satisfy $\mathbb 1_{\{T\in\bigcup_{k=1}^{\infty}B_k\}}L=\sum_{k=1}^{\infty}\mathbb 1_{\{T\in B_k\}}L$ pointwise, with the partial sums increasing because $L\ge 0$. The [monotone convergence theorem](/theorems/509) gives \begin{align*} \mu^Q\Bigl(\bigcup_{k=1}^{\infty}B_k\Bigr) =\sum_{k=1}^{\infty}\mu^Q(B_k), \end{align*} so $\mu^Q$ is countably additive on $\mathcal B(S)$. Its total mass is \begin{align*} \mu^Q(S)=\mathbb E[\mathbb 1_{\{T\in S\}}L]=\mathbb E[L]=1. \end{align*} Thus $\mu^Q$ is a probability measure on $(S,\mathcal B(S))$. [/step] [step:Deduce uniform integrability of the likelihood ratios from no loss of mass] For each $M>0$, define the truncation map $\varphi_M:[0,\infty)\to[0,M]$ by \begin{align*} \varphi_M(\ell) := \ell\wedge M. \end{align*} The function $\varphi_M$ is bounded and continuous. Since $(T_n,L_n)\xrightarrow{d}(T,L)$ under $P_n$, the marginal convergence $L_n\xrightarrow{d}L$ holds under $P_n$, and therefore \begin{align*} \mathbb E_{P_n}[L_n\wedge M]\to \mathbb E[L\wedge M]. \end{align*} Because $\mathbb E_{P_n}[L_n]=Q_n(\mathcal X_n)=1$ for every $n$, \begin{align*} \mathbb E_{P_n}[(L_n-M)_+] = \mathbb E_{P_n}[L_n-(L_n\wedge M)] = 1-\mathbb E_{P_n}[L_n\wedge M]. \end{align*} Taking $n\to\infty$ for fixed $M$ gives \begin{align*} \lim_{n\to\infty}\mathbb E_{P_n}[(L_n-M)_+] =1-\mathbb E[L\wedge M]. \end{align*} Since $L\ge 0$ and $\mathbb E[L]=1$, the monotone convergence of $L\wedge M \uparrow L$ gives \begin{align*} \mathbb E[L\wedge M]\to 1. \end{align*} Hence the eventual tails of $(L_n)$ vanish. The finitely many initial tails also vanish as $M\to\infty$, so \begin{align*} \lim_{M\to\infty}\sup_{n\in\mathbb N}\mathbb E_{P_n}[(L_n-M)_+]=0. \end{align*} Thus $(L_n)$ is uniformly integrable under $(P_n)$. [guided] The only possible obstruction is that mass in the likelihood ratios might escape to infinity. The condition $\mathbb E[L]=1$ rules this out. For $M>0$, define the truncation map $\varphi_M:[0,\infty)\to[0,M]$ by \begin{align*} \varphi_M(\ell) := \ell\wedge M. \end{align*} This map is bounded and continuous, so [weak convergence](/page/Weak%20Convergence) of $L_n$ to $L$ under $P_n$ implies convergence of the truncated expectations: \begin{align*} \mathbb E_{P_n}[L_n\wedge M]\to \mathbb E[L\wedge M]. \end{align*} The tail above level $M$ is exactly the part lost by truncation: \begin{align*} (L_n-M)_+ = L_n-(L_n\wedge M). \end{align*} Since $L_n=dQ_n/dP_n$ and $Q_n$ is a probability measure, \begin{align*} \mathbb E_{P_n}[L_n]=\int_{\mathcal X_n} L_n\,dP_n = Q_n(\mathcal X_n)=1. \end{align*} Therefore \begin{align*} \mathbb E_{P_n}[(L_n-M)_+] =1-\mathbb E_{P_n}[L_n\wedge M]. \end{align*} Letting $n\to\infty$ with $M$ fixed gives \begin{align*} \lim_{n\to\infty}\mathbb E_{P_n}[(L_n-M)_+] =1-\mathbb E[L\wedge M]. \end{align*} Now $L\wedge M$ increases pointwise to $L$ as $M\to\infty$, and $L$ is nonnegative with $\mathbb E[L]=1$. Hence \begin{align*} \mathbb E[L\wedge M]\to \mathbb E[L]=1. \end{align*} So the limiting upper tail is arbitrarily small for large $M$. For each fixed finite set of indices $n$, the ordinary integrability of $L_n$ gives \begin{align*} \mathbb E_{P_n}[(L_n-M)_+]\to 0 \end{align*} as $M\to\infty$. Combining the finite initial indices with the eventual bound gives \begin{align*} \lim_{M\to\infty}\sup_{n\in\mathbb N}\mathbb E_{P_n}[(L_n-M)_+]=0. \end{align*} This is precisely uniform integrability of the likelihood ratios. [/guided] [/step] [step:Pass to the limit for bounded continuous test functions] Let \begin{align*} f:S\to\mathbb R \end{align*} be bounded and continuous, and define its supremum norm by \begin{align*} \|f\|_\infty := \sup_{s\in S}|f(s)|. \end{align*} For $M>0$, define $h_M:S\times[0,\infty)\to\mathbb R$ by \begin{align*} h_M(s,\ell) := f(s)(\ell\wedge M). \end{align*} The function $h_M$ is bounded and continuous. Hence joint convergence under $P_n$ gives \begin{align*} \mathbb E_{P_n}[f(T_n)(L_n\wedge M)]\to \mathbb E[f(T)(L\wedge M)]. \end{align*} Moreover, \begin{align*} \left|\mathbb E_{P_n}[f(T_n)L_n]-\mathbb E_{P_n}[f(T_n)(L_n\wedge M)]\right| &\le \|f\|_\infty \mathbb E_{P_n}[(L_n-M)_+]. \end{align*} The uniform integrability just proved implies that this error is uniformly small in $n$ for large $M$. Also, \begin{align*} \left|\mathbb E[f(T)L]-\mathbb E[f(T)(L\wedge M)]\right| &\le \|f\|_\infty \mathbb E[L-(L\wedge M)]\to 0. \end{align*} Therefore \begin{align*} \mathbb E_{P_n}[f(T_n)L_n]\to \mathbb E[f(T)L]. \end{align*} [/step] [step:Identify the weak limit under the alternative measures] Since $L_n=dQ_n/dP_n$, the Radon-Nikodym identity gives, for every bounded continuous $f:S\to\mathbb R$, \begin{align*} \mathbb E_{Q_n}[f(T_n)] = \int_{\mathcal X_n} f(T_n(x))\,dQ_n(x) = \int_{\mathcal X_n} f(T_n(x))L_n(x)\,dP_n(x) = \mathbb E_{P_n}[f(T_n)L_n]. \end{align*} By the previous step, \begin{align*} \mathbb E_{Q_n}[f(T_n)]\to \mathbb E[f(T)L]. \end{align*} By the definition of $\mu^Q$, \begin{align*} \mathbb E[f(T)L]=\int_S f(s)\,d\mu^Q(s). \end{align*} Thus the laws of $T_n$ under $Q_n$ converge weakly to $\mu^Q$, meaning \begin{align*} T_n\xrightarrow{d}T^Q \end{align*} under $Q_n$ for any $S$-valued random element $T^Q$ with law $\mu^Q$. [/step] [step:Recover convergence on continuity sets] Let $B\in\mathcal B(S)$ satisfy $\mu^Q(\partial B)=0$. Since the laws of $T_n$ under $Q_n$ converge weakly to $\mu^Q$, the continuity-set characterization of weak convergence on metric spaces gives \begin{align*} Q_n(T_n\in B)\to \mu^Q(B). \end{align*} Using the definition of $\mu^Q$, this is \begin{align*} Q_n(T_n\in B)\to \mathbb E[\mathbb 1_{\{T\in B\}}L]. \end{align*} This is the asserted characterization on continuity sets and completes the proof. [/step]