This post is a bit long, so please don't be annoyed with me :). I have a question regarding convergence in probability of conditional characteristic functions, which I describe below:
Let $X_n$ be a sequence of random vectors, and let $Y$ and $Z$ be two random vectors, such that $$\mathbb{P}(X_n\in A\big| Z) \xrightarrow{\mathbb{P}} \mathbb{P}(Y \in A)$$ for all rectangular sets $A$. Here, $\xrightarrow{\mathbb{P}}$ denotes convergence in probability.
My question is, does this imply that: $$\mathbb{E}\left(e^{i\mathbf{t}'X_n}\big|Z\right) \xrightarrow{\mathbb{P}} \mathbb{E}\left(e^{i\mathbf{t}'Y}\big|Z\right)$$ for all vectors $\mathbf{t}$?
$\textbf{My attempt:}$
I think I have a solution, where I seem to have proved that the implication is true, but the proof looks a bit unnecessarily complicated (and I am not 100% sure whether it is correct, too). I will describe the sketch of my proof below.
Let $\Omega$ be the sample space on which $Z$ is defined. Let the dimension of the $X_n$ s and $Y$ be $d$. Let $\mathcal{Q}$ denote the set of all points in $\mathbb{R}^d$ with all coordinates rational. For each $q = (q^{(1)},\cdots,q^{(d)})\in \mathcal{Q}$, let $$I_q := (-\infty,q^{(1)}] \times \cdots \times (-\infty,q^{(d)}]~,$$ in other words, the rectangle joining $(-\infty,\cdots,-\infty)$ and $q$.
Order the the countable set $\mathcal{Q}$ as $\{q_1,q_2,\cdots\}$ and fix $\mathbf{t} \in \mathbb{R}^d$ to begin with.
Now, let $n_0$ be a subsequence of the natural numbers. We will show that $n_0$ has a further subsequence $\tilde{n}$, such that $$\mathbb{E}\left(e^{i\mathbf{t}'X_n}\big|Z\right) \xrightarrow{\mathbb{a.s.}} \mathbb{E}\left(e^{i\mathbf{t}'Y}\big|Z\right)$$ where $\xrightarrow{a.s.}$ denotes almost sure convergence, which is equivalent to solving the problem.
What I do now, is to follow Cantor's diagonalization method. To elaborate, get a subsequence $n_1$ of $n_0$ and a probability $1$ set $\Omega_1$, such that for all $\omega \in \Omega_1$, $$\mathbb{P}(X_{n_1} \in I_{q_1}|Z)(\omega) \rightarrow \mathbb{P}(Y \in I_{q_1})~.$$ Next, get a subsequence $n_2$ of $n_1$ and a probability $1$ set $\Omega_2$, such that for all $\omega \in \Omega_2$, $$\mathbb{P}(X_{n_2} \in I_{q_2}|Z)(\omega) \rightarrow \mathbb{P}(Y \in I_{q_2})~.$$ Continue this way. Now define $\tilde{n} := (n_1(1),n_2(2),n_3(3),\cdots)$ and $\tilde{\Omega} := \bigcap_{i\geq 1} \Omega_i$. Then, for all $\omega \in \tilde{\Omega}$, $$\mathbb{P}(X_{\tilde{n}} \in I_{q}|Z)(\omega) \rightarrow \mathbb{P}(Y \in I_{q_2})$$ for all $q \in \mathcal{Q}$. Since $\mathcal{Q}$ is dense in $\mathbb{R}^d$, it follows that for all $\omega \in \tilde{\Omega}$, $$\mathbb{P}(X_{\tilde{n}} \in A|Z)(\omega) \rightarrow \mathbb{P}(Y \in A)$$ for all rectangles $A$, which is enough to conclude that for all $\omega \in \tilde{\Omega}$, $$\mathbb{E}\left(e^{i\mathbf{t}'X_\tilde{n}}\big|Z\right)(\omega) \rightarrow \mathbb{E}\left(e^{i\mathbf{t}'Y}\big|Z\right)~,$$ completing the proof.
$\textbf{Question:}$
Can anyone be kind enough to verify to me, whether my proof is correct?
I do not feel that the proof should be so complicated, and I think there should be a simpler proof. If so, can anyone kindly give me such a proof?
Is there a simpler proof, if we are given the stronger hypothesis that:
$$\sup_{A\in \mathcal{C}} \Big|\mathbb{P}(X_n\in A\big| Z) - \mathbb{P}(Y \in A)\Big| \xrightarrow{\mathbb{P}} 0~,$$ where $\mathcal{C}$ denotes the class of all rectangles?
$\textbf{Actually, I will be very happy if you can just provide a simpler solution to my question 3 only.}$
Many thanks.