Convergence to a constant in probability but not almost surely Please give a example that a sequence of random variables that converge to a
constant $c$ in probability but fail to converge to $c$ with probability $1$.
Thanks very much.
 A: My apologies, I had posted an answer earlier which was wrong. So here is the correct one:
Define $\Omega = [0,1]$. Let $A_1=[0,\frac{1}{2}],A_2= [\frac{1}{2},1],A_3=[0,\frac{1}{4}], A_4=[\frac{1}{4},\frac{1}{2}]... $
In general $$ A_{2^n + k - 1} = [\frac{k-1}{2^n},\frac{k}{2^n}] \quad k \in \{1,..2^n\}$$
It is of note that $P(A_k) \geq P(A_{k+1})$. In fact $\lim_{n \to \infty} P(A_k) = 0$.
Define $X_k = 1_{A_k}$. Now $X_k \to 0 \quad i.p$ because $\forall \epsilon > 0$
$$P(X_m > \epsilon) = P(X_m = 1) = P(A_m)$$
Take limits to conclude that $X_n \to 0$ in probability.
Now for the a.s debunking. $X_n \to 0$ a.s is equivalent to showing $\forall \epsilon > 0$
$$ P(\cap_{k\geq 1}\cup_{n \geq k}\{X_n \geq \epsilon\}) = 0$$
$$ \iff P(\cap_{k\geq 1}\cup_{n \geq k}A_n) = 0$$
Now $\cup_{n \geq m}A_n = \Omega$ because if you pick a m, find a j s.t $m < 2^j$ and then
$$\cup_{l=1}^{2^j-1}A_{2^j + l} = \Omega \subseteq \cup_{n \geq m}A_n$$ Thus
$$P(\cap_{k\geq 1}\cup_{n \geq k}A_n) = 1 $$
Thus $X_n$ does not converge to 0 a.s.
