# An exemple that convergence in probability does not imply convergence almost surely

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

• Such a subset cannot exist because it would contradict the convergence in probability. I would suggest looking for a subset on which $\limsup X_n(\omega) = 1$ instead. Aug 15 at 19:09
• To add to user624's point, you do not need to show the limit goes to 1 for certain $\omega$ of positive measure. It is enough to show the limit does not exist for certain $\omega$ of positive measure. Actually you can show that for every $\omega \in [0,1]$, the limit does not exist since $X_n(\omega)$ infinitely bounces between 0 and 1 as $n\rightarrow\infty$. Aug 15 at 19:55
• @user6247850 Thank you for the comment ! However I do not see where the existence of such a subset would contradict the convergence in probability since it is not used on what I wrote for proving the convergence in probability in the sense that the limit of the lebesgue measure of the set I describe inside the probability (corresponding to the event $\lvert X_n\rvert > \varepsilon$) is still $0$ no ? Am I missing something ? Aug 16 at 17:56
• @coboy Well, if $\lim X_n(\omega) = 1$ for $\omega \in E$ where $\mathbb{P}(E) > 0$, then $\lim \mathbb{P}(|X_n| > \varepsilon) \ge \mathbb{P}(E) > 0$, so $X_n$ could not converge to $0$ in probability. This uses essentially the same proof that almost sure convergence implies convergence in probability. Aug 16 at 18:42
• @coboy I'm not sure what kind of link to convergence of functions you mean. Random variables can, of course, be seen as functions on $\Omega$, so every mode of convergence for random variables is equivalent to a mode of convergence for functions, but I don't think that's quite what you had in mind Aug 16 at 21:11

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 $$\liminf_n X(\omega) = 0, \limsup_n X(\omega) = 1$$.