Convergence in probability over compact set

Question

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: $$ \forall \varepsilon>0,\ \lim\limits_nP(\sup\limits_{x\in K}|\Psi_n(x,\cdot)-\Psi(x)|>\varepsilon)=0\Rightarrow \forall \varepsilon>0,\ \lim\limits_nP(|\Psi_n(X_n,\cdot)-\Psi(c)|>\varepsilon)=0\ ? $$

Answer 1

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.

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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.