Convergence in probability of different uniform distribution R.V.s A question in my book is as follows:

Suppose $X_n$ is uniformly distributed on the set of points $\{1/n, 2/n, \ldots, 1 \}$. Does $X_n \xrightarrow{P} X$, where $X$ is
  $\mathcal{U}(0,1)$?

The solution the book gives says

From the information given, one cannot tell whether $X_n$ converges in
  probability to $X$ because the joint distribution of $X_n$ and $X$ has
  not been defined.

I don't see why this joint distribution is not defined. I know that the joint distribution is required in order to subtract random variables as required in the definition of convergence in probability, yet I don't see why these random variables do not have a joint distribution. Can someone explain this to me?
 A: Suppose that $X_n\to X$ in probability to some $X$ for each sequence $(X_n)$ satisfying the assumption of the problem.
We could replace for odd numbers the random variable $X_n$ by $1-X_n+1/n$, which has the same distribution. This of course changes the distribution of the sequence $(X_n)_{n\geqslant 1}$. The modified sequence still satisfies the assumptions of the problem. We thus have $X_{2n}\to X$ in probability but $X_{2n+1}\to 1-X$ in probability. This is a contradiction, since we cannot have $X=1/2$ (the limiting distribution is uniform).
A: One possibility is that $X_1,\ldots,X_n$ are independent.
Another possibility is this: Suppose $Y$ has a continuous uniform distribution on $[0,1]$, and then let
$$
X_n = \frac{\lfloor 1+ nx\rfloor} n.
$$
(For example, if $n=6$, then multiply $Y$ by $6$ and add $1$, getting a number in $[1,7]$, then take its integer part, getting a number in $\{1,2,3,4,5,6\}$, then divide that by $6$, getting a number called $X_6$ in $\{1/6,\,2/6,\, 3/6,\, 4/6,\, 5/6,\,6/6\}$.)
In this second possibility, the random variables $X_n$ each have the same distribution as in the first possibility, but they are very far from independent; they are highly correlated.
Thus different joint distributions are consistent with the same given marginal distributions.
