Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recently I am studying the measure theory and question of definition of convergence in measure came into my mind. The usual definition of it is given as: $\forall\, \epsilon > 0,\; \lim_{n\to\infty}\mathbb{P}[|X-X_n|>\epsilon] = 0$. What I can't understand is the difference of between the formal definition and expression like $\lim_{n\to\infty}\mathbb{P}[|X-X_n|>0]$. To me it seems quite similar. Are these two actually the same of not? Great thanks!

share|cite|improve this question
What is the formal definition that you are referring to ? – nonlinearism Feb 15 '13 at 17:03
No, they are not the same. Suppose $X_n \downarrow X$, strictly. Then $[|X - X_n| > 0] = [X_n > X]$ which has probability $1$, so $P(|X - X_n| > 0)$ does not go to $0$. – guy Feb 15 '13 at 17:05
@guy To maintain the integrity of this site, I think you should make your comment an answer. Nice answer, btw. – Quinn Culver Feb 15 '13 at 19:15
@DavideGiraudo apologies, I posted my comment as an answer. – guy Feb 18 '13 at 22:47
up vote 1 down vote accepted

No, they are not the same. The event $[|X_n - X| > 0]$ is the same as the event $[X_n \ne X]$ and it is clearly possible for $X_n \to X$ however you like while having $P(X_n \ne X) = 1$. An example is if $X_n (\omega) \downarrow X(\omega)$ strictly where $X_n$ and $X$ are rvs defined on a probability space $(\Omega, \mathcal F, P)$. An explicit example is $X_n = X + \frac 1 n$, in which case $|X_n - X| = \frac 1 n$, so $P(|X_n - X| > 0) = 1$ but $P(|X_n - X| > \epsilon) = 0$ for sufficiently large $n$ so that $X_n \to X$ in probability (in fact, $X_n \to X$ almost surely).

This isn't very different from having $x_n \to x$ when $x_n$ is a sequence of real numbers; we don't need $|x_n - x| = 0$ for the limit to be $x$, we just need $|x_n - x| < \epsilon$ for suitably large $n$. Convergence in probability requires that $|x_n - x| < \epsilon$ with high probability for sufficiently large $n$, but as far as illustrating the distinction we are addressing the real number case makes the same point.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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