In my lecture notes, there is the following example to study in order to show that convergence in probability does not imply convergence almost surely.
We consider the measurable space $([0,1], \mathcal{B}([0,1]), \lambda)$ and the random variable defined by $X_n(\omega) = \mathbb{1}_{[\frac{n}{2^{k(n)}}, \frac{n+1}{2^{k(n)}}]}(\omega)$. $k(n)$ is defined as the unique integer such that $2^{k(n)-1}\leq n < 2^{k(n)}$ and we notice that $k(n)$ form an increasing sequence and start at $1$.
The idea is to consider $\varepsilon\in]0,1[$ and notice that
$$
\mathbb{P}(\lvert X_n \rvert > \varepsilon) = \mathbb{P}\left(\{\omega\in[0,1] : \frac{n}{2^{k(n)}}\leq \omega\leq \frac{n+1}{2^{k(n)}}\}\right) = \frac{1}{2^{k(n)}}
$$
Which tends to $0$ as $n$ goes to infinity by the preceding remark on $k(n)$. So we conclude that $X_n$ converges in probability to $0$. Until here, that's fine. However, when I want to show that this is not a convergence almost surely, I am not so sure on what to do.
Using the definition of convergence almost surely, I would like to find a subset of $[0,1]$ with measure strictly positive on which $lim_{n\to\infty}X_n(\omega) = 1$.
Has someone any idea on how to proceed, please ?
Thank you a lot
Best Answer
The example hides the point of the exercise with unnecessary computation.
Show that $k(n) = \lfloor \log_2 n \rfloor +1$.
Let $I_n = [ {n \over 2^{k(n)} } ,{n+1 \over 2^{k(n)} } )$. Show that $I_{2^m},...,I_{2^{m+1}-1}$ form a partition of $[{1 \over 2},1)$.
Hence for any $\omega \in [{1 \over 2},1)$, we see that $\liminf_n X(\omega) = 0, \limsup_n X(\omega) = 1$.