Squeeze theorem for convergence in distribution

Let $(X_n)_{n\in\mathbb{N}}$, $(X'_n)_{n\in\mathbb{N}}$, $Y$ real random variables such that $$X_n\leq Y\leq X'_n \quad\forall n\in\mathbb{N}\ .$$ Suppose that the sequences $X_n$ e $X'_n$ converge to the same limit in distribution (i.e. in law, weakly), that is $$X_n\xrightarrow[n\to\infty]{d} X \quad\text{e}\quad X'_n\xrightarrow[n\to\infty]{d} X\ .$$ Then is it true that $$Y\ \overset{\,d}=\ X\ ?$$

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Limit in distribution $Z_n\to Z$ for real valued random variables is equivalent to the fact that $\mathbb P(Z_n\leqslant x)\to\mathbb P(Z\leqslant x)$ for every $x$ at which the function $F_Z:z\mapsto\mathbb P(Z\leqslant z)$ is continuous.
In the present case, for every $x$, $\mathbb P(X'_n\leqslant x)\leqslant F_Y(x)\leqslant\mathbb P(X_n\leqslant x)$, and both $\mathbb P(X'_n\leqslant x)$ and $\mathbb P(X_n\leqslant x)$ converge to $F_X(x)$ when $n\to\infty$, for every $x$ at which $F_X$ is continuous. Hence $F_Y=F_X$ at every point of continuity of $F_X$. Both $F_X$ and $F_Y$ are continuous to the right hence $F_X=F_Y$ everywhere, that is, $Y\overset{\,d}=X$.
faster than I was :) I was thinking about the last step, there you also use, that a CDF has at most countable many discontinuities, right? Or how are you sure to find a decreasing sequence of points of continuity of $F_X$? – math Dec 4 '12 at 14:42
@math I think you can always find such a sequence. I would justify it like this: the continuity points are dense in $\mathbb{R}$ because their complementary set is countable. Is it correct? – qwertyuio Dec 4 '12 at 14:52