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

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    $\begingroup$ 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. $\endgroup$ Commented Aug 15, 2023 at 19:09
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    $\begingroup$ 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 $n\rightarrow\infty$. $\endgroup$
    – Michael
    Commented Aug 15, 2023 at 19:55
  • $\begingroup$ @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 ? $\endgroup$
    – G2MWF
    Commented Aug 16, 2023 at 17:56
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    $\begingroup$ @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. $\endgroup$ Commented Aug 16, 2023 at 18:42
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    $\begingroup$ @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 $\endgroup$ Commented Aug 16, 2023 at 21:11

1 Answer 1


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


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