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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!

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What is the formal definition that you are referring to ? –  nonlinearism Feb 15 '13 at 17:03
    
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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
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@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
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@DavideGiraudo apologies, I posted my comment as an answer. –  guy Feb 18 '13 at 22:47

1 Answer 1

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.

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