Example of convergence in distribution but not in probability While I was looking for an example of a sequence of random variables which converges in distribution, but doesn't converge in probability, I have read that it should be enough to consider a sequence of independent and identically distributed non-degenerate random variables. I don't understand why... Can someone explain (or correct if it isn't right)?
Thank you
 A: Choose the probability space $([0,1],\mathscr{B},m)$. ($\mathscr{B}$ consists of all Borel sets of $[0,1]$, $m$ is the Lebesgue measure.)  
Let $X_{2n}(\omega)=\omega$, $X_{2n-1}(\omega)=1-\omega$.  
Show that $X_{2n}$ and $X_{2n-1}$ have the same distribution. (Uniform distribution.)  
Show that $\{X_n\}$ does not converge in probability.
A: Here is an example of a sequence of random variables $\{X_n\}_{n\geq1}$ defined on $\Omega$ which converges in distribution to $X:\Omega\to\mathbb{R}$ but does not converge in probability to $X$.
Consider the abstract space $\Omega=\{0,1\}$ and
\begin{cases}
X_n(0)=0\\
 X_n(1)=1\\
X(1)=0\\
 X(0)=1  
\end{cases}
all with probability $\frac{1}{2}$.
Then we have
$$X_n\xrightarrow[]{\text{dist}}X$$
because $F_n=F$, where $F_n$ and $F$ are the distribution functions of $X_n$ and $X$ respectively.
However, $$X_n\xrightarrow[]{\text{prob}}X$$ does not hold since we have $|X_n(\omega)-X(\omega)|=1$, for $ \omega\in\Omega$. Thus
$$P(\omega \in \Omega: |X_n(\omega)-X(\omega)|>\epsilon)\not\to 0.$$
