Consider the following two statements about a random sequence $X_n$:

(1) $X_n \stackrel{a.e.}{\rightarrow} X$.

(2) $\mathrm{P}\{|X_n-X|>\epsilon, \ i.o.\} = 0, \ \forall \epsilon>0$.

(a.e. stands for "almost everywhere" and i.o. stands for "infinitely often".)

What is the difference between statements (1) and (2), or are they equivalent? Does one imply the other?

I appreciate any help!

  • 1
    $\begingroup$ In (2) you shouldn't write a limit since the event $\{ \vert X_n-X\vert>\varepsilon,\, i.o\}$ does not depend on $n$. $\endgroup$
    – Etienne
    Jul 18, 2013 at 9:17
  • $\begingroup$ You're right Etienne. I edited (2) accordingly. $\endgroup$
    – user86780
    Jul 18, 2013 at 11:02

1 Answer 1


Ok, we first show that an event in $(2)$ is very similar to the convergence, so that $(2)$ follows from $(1)$. Moreover, you can express a convergence as a monotonic intersection of such events as in $(2)$. That would imply that $(1)$ follows from $(2)$. Now some technical stuff:

Note that $\{|X_n - X|>\varepsilon \text{ i.o.}\} = \{|X_n - X|\leq\varepsilon \text{ e.a.}\}$ where $\mathrm{e.a.}$ stands for eventually always: $$ \{\omega:\exists N(\omega,\varepsilon) \text{ such that }|X_n(\omega) - X(\omega)|\leq \varepsilon \text{ for all }n\geq N(\omega,\varepsilon)\}. $$ Now let $A_m:=\{\omega:|X_n - X|\leq\frac1m \text{ e.a.}\}$ so that $P(A_m) = 1$ and $A_{m+1}\subseteq A_m$. As a result, since $$ \bigcap_mA_m = \{\omega:|X_n - X|\leq \frac1m \text{ e.a. for any }m \} = \{\omega:X_n \to X\} $$ we obtain that $(2)$ implies $$ P\{\omega:X_n \to X\} = P\left(\bigcap_mA_m\right) = \lim_m P(A_m) = 1 $$ so that $(2)\implies (1)$. The reverse direction is much easier to show.

  • $\begingroup$ Thanks Ilya! My intuition was also saying that (1) and (2) should be equivalent. $\endgroup$
    – user86780
    Jul 18, 2013 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.