Conditonal convergence implies convergence?

Note : All measures below are probability measures.

Let $\mu_n(X,Y)$ be a random probability measure on $\mathbb C$ depending on two random variables X and Y with values in $\mathbb{R}^N$.

Actually, it comes from random matrices, $\mu_n$ is t he empirical measure of a matrix model : $\mu_n = \mu_{\frac{1}{\sqrt n}X_nY_n}$ where $X_n = Diag(X_1,..,X_n)$. and $Y = (Y_{ij})_{1 \leq i,j \leq n}$ and I note X the infinite sequence $X =(X_i)_{i \in \mathbb{N}}$ and Y the infinite array $Y = (Y_{ij})_{i,j \in \mathbb{N}^2}$

Fact is if I freeze $X = x$ and consider it is deterministic, I have a sequence of measures

$$\mu_n^x = \mu_n(x,Y) = \mu_{\frac{1}{\sqrt n}Diag(x_1,..,x_n)Y_n}$$

I proved that :

$$\text{For almost all x}, \quad \mu_n^x \overset{\mathcal{P}}\rightsquigarrow \mu_{\infty}$$

where $\mathcal{P}$ denotes the convergence in probability (with respect to $Y$). $$\mathbf{\mu_\infty} \hspace{0.1cm}\textbf{does not depend on x}$$

Now, $X$ and $Y$ are independent, is it true that $$\mu_n\overset{\mathcal{P}}\rightsquigarrow \mu_{\infty} ?$$

It seems to me that yes, but I could not write it properly. I started this way :

Say $X$ is defined on some probabilized space $(\Omega_1, \mathbb P_1)$ and $Y$ on some $(\Omega_2, \mathbb P_2)$. Write $\Omega = \Omega_1 \times \Omega_2$ endowed with the probability $\mathbb P = \mathbb P_1 \otimes \mathbb P_2$ (independence).

Let $f \in C_k(\mathbb{C})$ be a test function (weak, vague or tight convergence are equivalent on $\mathcal P(\mathbb C)$).

Let $A_n^\epsilon(f) = \{\omega / \quad \vert \int fd\mu_n - \int fd\mu_\infty\vert > \epsilon \}$. We want to show $\mathbb P(A_n^\epsilon(f)) \underset{n \to \infty}\rightarrow 0$.

$P(A_n^\epsilon(f)) = \int_{\Omega_1}\int_{\Omega_2}\mathbb{1}_{\{\omega \in A_n^\epsilon(f)\}}d\mathbb P(w) = \int_{\Omega_1 \setminus N} \left( \int_{\Omega_2}\mathbb{1}_{\{(\omega_1,\omega_2) \in A_n^\epsilon(f)\}}d\mathbb P_2(w_2) \right) d\mathbb P_1(w_1)$

where N denotes the zero-measure set out of which we have $\mu_n^x \overset{\mathcal{P}}\rightsquigarrow \mu_{\infty}$.

So on $\Omega_1 \setminus N$, $\omega_1 \mapsto \int_{\Omega_2}\mathbb{1}_{\{(\omega_1,\omega_2) \in A_n^\epsilon(f)\}}d\mathbb P_2(w_2) \underset{n \to \infty}\rightarrow 0$ pointwise. And we conclude by dominated convergence.

Is it correct (is the set $A_n^\epsilon(f)$ really measurable?) ? Is there a more direct or more general way to prove this, without defining such a probability space for instance ?