$\{X_{n}\}$ be a sequence of independent random variables defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Prove that the probability sequence $\{X_{n}\}$ converges is either one or zero.

I am studying for an exam that I have and this is one problem that I am stuck on. I tried to argue by contradiction (i.e. suppose $X_{n} \rightarrow c $ for some $0 < c < 1$), and I tried to use different inequalities but I haven't had any luck. I'm also familiar with Kolmogorov's Zero-One Law, and Borel-Cantelli Lemmas, but I'm not sure how to apply any of them here.

I will appreciate your help in this problem


$(X_n)$ converges iff $(X_k,X_{k+1},...)$ converges, Hence the event $\{\lim_n X_n \text {exists}\}$ is measurable w.r.t. sigma field generated by $X_k,X_{k+1},...$. This is true for each $k$ and hence the event is measurable w.r.t. the tail sigma field. By $0-1$ Law the event has probabilty $0$ or $1$.

  • $\begingroup$ Does the mode of convergence matter here? Or does Kolmogorov's $0-1$ law apply regardless? $\endgroup$ – Math1000 Dec 11 '19 at 6:01
  • $\begingroup$ @Math1000 It is understood here that the event under consideration is the event that the sequences converges at every point. It doesn't even make sense to talk about the probability that $(X_n)$ converges in some other sense. $\endgroup$ – Kavi Rama Murthy Dec 11 '19 at 6:03
  • $\begingroup$ Almost sure convergence, you mean? $\endgroup$ – Math1000 Dec 11 '19 at 6:04
  • $\begingroup$ To assign a probability yo have to have an event. What other event can you think of here? Ev en almost sure convergence does not make sense. Only thing that make sense is $P(\omega: \lim_{n \to \infty} X_n(\omega) \, exists)$ $\endgroup$ – Kavi Rama Murthy Dec 11 '19 at 6:07
  • 1
    $\begingroup$ @hom It is a basic fact in measure theory that if sequence of functions is measurable w.r.t. some sigma field then the set of points where the sequence converges belongs to the sigma field. $\endgroup$ – Kavi Rama Murthy Dec 11 '19 at 6:09

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