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Let $(\Omega,\mathcal{F},P)$ be a probability space. Suppose $(X_n)\subseteq (\mathbb{R}^k)^{\Omega}$ is a sequence of random vectors converging in probability to the vector $c\in \mathbb{R}^k$. Let $(\Psi_n)$ be a sequence of functions $\Psi_n:\mathbb{R}^k\times \Omega\to \mathbb{R}$, such that for all $n$ and all $x\in \mathbb{R}^k$, $\Psi_n(x,\cdot)$ is measurable. Consider also a continuous function $\Psi:K\to \mathbb{R}$ where $K$ is a compact subset of $\mathbb{R}^k$ containing $c$. Do we have that: $$ \forall \varepsilon>0,\ \lim\limits_nP(\sup\limits_{x\in K}|\Psi_n(x,\cdot)-\Psi(x)|>\varepsilon)=0\Rightarrow \forall \varepsilon>0,\ \lim\limits_nP(|\Psi_n(X_n,\cdot)-\Psi(c)|>\varepsilon)=0\ ? $$

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    $\begingroup$ I think you might need a condition that $c$ is interior to $K$. It seems the statement is incorrect e.g. if $K=\{c\}$. $\endgroup$ Jun 19 at 17:07
  • $\begingroup$ Indeed, that's the problem I had and it is the reason for the question mark up there xD $\endgroup$ Jun 19 at 20:28
  • $\begingroup$ @Arizabalaga Does your question in anyway relate to my post (Click here)? $\endgroup$
    – Arbuja
    Jun 20 at 1:32
  • $\begingroup$ I could not say, I will think more about it $\endgroup$ Jun 20 at 18:02
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In retrospect, as pointed in the comment section, the hypothesis should allow $c$ to be an interior point $K$.
In that case, the answer is affirmative.


Let $\tilde{\Psi}$ be any extension of $\Psi$ to an continuous function on $\mathbb{R}^k$.
Because $c$ is an interior point of $K$ hence there is a positive number $r>0$ such that $$B_r(c) \subset K$$.

So if $|X_n-c|<r$, we have $$\begin{align}|\Psi_n(X_n,\cdot)-\Psi(c)| &\le |\Psi_n(X_n,\cdot)-\Psi(X_n)|+|\Psi(X_n) - \Psi(c)| \\ & \le \sup_{x \in K} |\Psi_n(x,\cdot)-\Psi(x)|+|\Psi(X_n) - \Psi(c)| \\ &= \sup_{x \in K} |\Psi_n(x,\cdot)-\Psi(x)|+|\tilde{\Psi}(X_n) - \tilde{\Psi}(c)| \end{align}$$ Thus if $\epsilon< |\Psi_n(X_n,\cdot)-\Psi(c)|$ and $|X_n-c|<r$, we have either $$\epsilon/2 < \sup_{x \in K} |\Psi_n(x,\cdot)-\Psi(x)| $$ or $$\epsilon/2 <|\tilde{\Psi}(X_n) - \tilde{\Psi}|(c)| $$ So : $$\begin{align} \mathbb{P}(\epsilon< |\Psi_n(X_n,\cdot)-\Psi(c)|) &\le \mathbb{P}( |X_n-c| \ge r)+ \mathbb{P} \left( \epsilon/2 < \sup_{x \in K} |\Psi_n(x,\cdot)-\Psi(x)|\right)\\ &+ \mathbb{P} \left( \epsilon/2 <|\tilde{\Psi}(X_n) - \tilde{\Psi}(c)|\right) \end{align} $$ Hence the conclusion.

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