probability - How to show two random variables have the same distribution?

Let $\{X_n\}$ and $\{Y_n\}$ be sequences of random variables such that the pairs $(X_i,X_j)$ and $(Y_i,Y_j)$ have the same distributions for all $i,j$. If $X_n \rightarrow X$ in probability, show that $Y_n$ converges in probability to some limit $Y$ having the same distribution

I am trying to prove it in the following steps:

Step1: Show that $P(|Y_n-Y_m|> \epsilon) = P(|X_n-X_m|> \epsilon)$ for arbitrary $\epsilon > 0$. This is true since $(X_i,X_j)$ and $(Y_i,Y_j)$ have the same distributions for all $i,j$.

Step2: $X_n \rightarrow X$ in probability implies $X_n$ is Cauchy convergent in probability(meaning that for all $\epsilon >0 $, $P(|X_n-X_m|> \epsilon)$ goes to zero as m,n go to $\infty$). So $Y_n$ is Cauchy convergent in probability as well.

Step3: Show that there exists a random variable $Y$ such that $Y_n \rightarrow Y$ in probability. Actually, $Y(\omega) = \lim_{k\rightarrow \infty} Y_{n_k}(\omega)$ for a subsequence such that $P(|Y_{n_k} - Y_{n_{k+1}} | > \epsilon) \leq 2^{-k}$. And we can show $Y_n \rightarrow Y$ in probability using Borel Cantelli Lemma.

Step4: Show $Y$ have the same distribution as $X$.

I am having trouble about Step4. So do $(X_i,X_j)$ and $(Y_i,Y_j)$ have the same distributions imply $X_n$ and $Y_n$ are identically distributed?

Answer

$(X_n,X_{n+1}) \to (X,X)$ in probability and $(Y_n,Y_{n+1}) \to (Y,Y)$ in probability. Hence $(X,X)$ has same distribution as $(Y,Y)$ which implies that $X$ has the same distribution a $Y$.