Functions and convergence in law Let $X$ be a random variable taking values in some metric space $M$. Let $\{\phi_n\}$ be a sequence of measurable functions from $M$ to another metric space $\tilde M$. Suppose that $\phi_n(X)$ converges in law to a random variable $Y$. Must it be the case that the pairs $(X, \phi_n(X))$ converge in law to a pair $(X, \phi(X))$, where $\phi$ is a measurable function such that $\phi(X)$ has the same law as $Y$?
 A: Generally speaking, the answer is no. Follows a counter-example with Bernoulli random variables (if you prefer continuous variables, you can construct something similar with the central limit theorem, see the end of my answer).
Let $X$ be a sequence of independent random variables with uniform distribution on the discrete space $\Bbb Z_2$ and let $\phi_n(x) = x_1 + \dots + x_n$. There is trivially convergence in law for $\phi_n(X)$ because for every $n$, the sum $X_1+\dots+X_n$ has uniform distribution on $\Bbb Z_2$.
Suppose that there is some random variable $Y$ such that $(X,\phi_n(X))$ converges in law to $(X,Y)$.
Consider an arbitrary function $f \colon \Bbb Z_2 \to \Bbb R$ and an arbitrary event $A \in \sigma(X_1,\dots,X_k)$. For every $n \geq k+1$, the random variables $1_A$ and $\phi_n(X)$ are independent. Taking the identity
$$
\Pr(A)\cdot E[f(\phi_n(X))] = E[1_A\cdot f(\phi_n(X))]
$$
to the limit, we see that $\Pr(A)\cdot E[f(Y)] = E[1_A\cdot f(Y)]$.
As a consequence, the random variables $X$ and $Y$ are independent. If $Y$ was to be some $\phi(X)$, then it would be constant. This is absurd.

Let $X$ be a sequence of i.i.d. real random variables with $E(X_1)=0$ and $E(X_1^2)=1$. The central limit theorem asserts that
$$
\phi_n(X) = \frac{1}{\sqrt{n}}\sum_{k=1}^n X_k \xrightarrow[n\to\infty]{law} Y \sim \mathcal{N}(0,1)
$$
This would require a proper proof but I think that the following is true (because $1/\sqrt{n}$ "kills" $X_1,\dots,X_k$ for finite $k$):
If $(X,\phi_n(X))$ converges in law to $(X,Y)$, then $X$ and $Y$ are independent.
Another consequence of this is that in the CLT, there is never almost sure convergence (or convergence in probability) for the sequence $\phi_n(X)$. Notice that this is not in contradiction with the Almost Sure CLT.
A: A more elementary counterexample: Let $X$ have uniform distribution over $[0,1]$ and define $\phi_n(x):=x$ when $n$ is odd, and $\phi_n(x):=1-x$ when $n$ is even. It's clear that every $\phi_n(X)$ has the same distribution (namely, that of $X$), hence we have convergence in law. However, the pair $(X,\phi_n(X))$ doesn't converge in law to anything, since its distribution keeps flipping back and forth.
