[proofplan] The proof is a direct random-index consequence of stochastic equicontinuity. We fix an error level and split the event where the empirical-process increment at the random index is large according to whether the random index lies in a small $d_P$-ball around $m_{\theta_0}$. On that local event, the random increment is bounded by the local supremum appearing in the definition of stochastic equicontinuity; on the complementary event, convergence of $d_P(m_{\hat\theta_n},m_{\theta_0})$ to zero controls the probability. Letting the local radius be chosen by stochastic equicontinuity gives the desired convergence in outer probability $\mathbb P_{\mathrm{out}}$. [/proofplan] [step:Control the random increment by a local supremum and a localization error] Fix $\varepsilon>0$. For each $\delta>0$ and $n\in\mathbb N$, define the event \begin{align*} A_{n,\delta}:=\{\omega\in\Omega:d_P(m_{\hat\theta_n(\omega)},m_{\theta_0})<\delta\}. \end{align*} Define the local empirical-process supremum \begin{align*} S_{n,\delta}:=\sup_{\theta\in\Theta:\ d_P(m_\theta,m_{\theta_0})<\delta}|G_n(m_\theta)-G_n(m_{\theta_0})|. \end{align*} On $A_{n,\delta}$, the random function $m_{\hat\theta_n}$ is one of the functions included in the supremum defining $S_{n,\delta}$. Therefore \begin{align*} |G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|\le S_{n,\delta} \end{align*} on $A_{n,\delta}$. Hence, by [monotonicity and subadditivity](/theorems/1081) of outer probability, \begin{align*} \mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon) \le \mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon)+\mathbb P_{\mathrm{out}}(A_{n,\delta}^c). \end{align*} [guided] Fix $\varepsilon>0$. The purpose of the argument is to replace the random index $\hat\theta_n$ by a deterministic local supremum. For a radius $\delta>0$, define the localization event \begin{align*} A_{n,\delta}:=\{\omega\in\Omega:d_P(m_{\hat\theta_n(\omega)},m_{\theta_0})<\delta\}. \end{align*} This event records exactly when the random criterion function $m_{\hat\theta_n}$ falls inside the $d_P$-ball of radius $\delta$ around the base function $m_{\theta_0}$. Define also \begin{align*} S_{n,\delta}:=\sup_{\theta\in\Theta:\ d_P(m_\theta,m_{\theta_0})<\delta}|G_n(m_\theta)-G_n(m_{\theta_0})|. \end{align*} The definition of $S_{n,\delta}$ is deterministic in the index set but random through the empirical process $G_n$. If $\omega\in A_{n,\delta}$, then $m_{\hat\theta_n(\omega)}$ satisfies the condition appearing under the supremum. Therefore the particular increment at the random index is bounded by the supremum over all such local increments: \begin{align*} |G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|\le S_{n,\delta} \end{align*} on $A_{n,\delta}$. Consequently, if the random increment is larger than $\varepsilon$, then either the local supremum is larger than $\varepsilon$ or the localization event has failed. In set notation, \begin{align*} \{|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon\}\subseteq \{S_{n,\delta}>\varepsilon\}\cup A_{n,\delta}^c. \end{align*} Taking outer probabilities and using monotonicity and subadditivity gives \begin{align*} \mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon) \le \mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon)+\mathbb P_{\mathrm{out}}(A_{n,\delta}^c). \end{align*} This is the key reduction: the first term is controlled by stochastic equicontinuity, and the second by the assumed convergence of the random index in the semimetric $d_P$. [/guided] [/step] [step:Choose the localization radius using stochastic equicontinuity] Let $\eta>0$. By stochastic equicontinuity at $m_{\theta_0}$, there exists $\delta>0$ such that \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon)<\eta. \end{align*} For this fixed $\delta$, the assumption \begin{align*} d_P(m_{\hat\theta_n},m_{\theta_0})\xrightarrow{\mathbb P_{\mathrm{out}}}0 \end{align*} implies \begin{align*} \lim_{n\to\infty}\mathbb P_{\mathrm{out}}(A_{n,\delta}^c)=0. \end{align*} Indeed, \begin{align*} A_{n,\delta}^c=\{\omega\in\Omega:d_P(m_{\hat\theta_n(\omega)},m_{\theta_0})\ge\delta\}. \end{align*} [guided] Let $\eta>0$. The preceding step reduced the problem to two probabilities, so we now choose the radius $\delta$ to make the local-supremum probability small. Stochastic equicontinuity at $m_{\theta_0}$ states that the limiting upper probability of the local supremum becomes arbitrarily small as the radius decreases. Therefore there exists $\delta>0$ such that \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon)<\eta. \end{align*} With this radius fixed, the localization error is exactly the event that the random index has not entered the $d_P$-ball of radius $\delta$ around $m_{\theta_0}$. By definition of $A_{n,\delta}$, \begin{align*} A_{n,\delta}^c=\{\omega\in\Omega:d_P(m_{\hat\theta_n(\omega)},m_{\theta_0})\ge\delta\}. \end{align*} The assumed convergence \begin{align*} d_P(m_{\hat\theta_n},m_{\theta_0})\xrightarrow{\mathbb P_{\mathrm{out}}}0 \end{align*} means precisely that, for every fixed positive radius, the outer probability of this complement tends to zero. Hence \begin{align*} \lim_{n\to\infty}\mathbb P_{\mathrm{out}}(A_{n,\delta}^c)=0. \end{align*} [/guided] [/step] [step:Pass to the limit and obtain convergence in outer probability] Using the estimate from the first step with the $\delta$ chosen in the second step gives \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon) \le \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon). \end{align*} The right-hand side is strictly smaller than $\eta$. Since $\eta>0$ was arbitrary, we obtain \begin{align*} \lim_{n\to\infty}\mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon)=0. \end{align*} Because this holds for every $\varepsilon>0$, it is precisely \begin{align*} G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})\xrightarrow{\mathbb P_{\mathrm{out}}}0. \end{align*} If all displayed quantities are measurable, the same proof with ordinary probability $\mathbb P$ in place of outer probability proves the measurable-probability variant. This proves the theorem. [guided] We now combine the two bounds. From the first step, for the radius $\delta$ chosen using stochastic equicontinuity, \begin{align*} \mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon) \le \mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon)+\mathbb P_{\mathrm{out}}(A_{n,\delta}^c). \end{align*} Taking the limit superior in $n$ and using \begin{align*} \lim_{n\to\infty}\mathbb P_{\mathrm{out}}(A_{n,\delta}^c)=0 \end{align*} from the preceding step gives \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon) \le \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon). \end{align*} The choice of $\delta$ gives \begin{align*} \limsup_{n\to\infty}\mathbb P_{\mathrm{out}}(S_{n,\delta}>\varepsilon)<\eta. \end{align*} Since $\eta>0$ was arbitrary, the only possible nonnegative value of the left-hand side is $0$. Therefore \begin{align*} \lim_{n\to\infty}\mathbb P_{\mathrm{out}}(|G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})|>\varepsilon)=0. \end{align*} This is exactly convergence to zero in outer probability: \begin{align*} G_n(m_{\hat\theta_n})-G_n(m_{\theta_0})\xrightarrow{\mathbb P_{\mathrm{out}}}0. \end{align*} When all random variables and local suprema are measurable, each invocation of outer probability above may be replaced by ordinary probability $\mathbb P$, so the ordinary-probability statement follows by the identical argument. [/guided] [/step]