What is the difference between weak convergence and convergence in measure? example let $\mu_{n}\Rightarrow \mu$ on $[0,1]$ (the space where all measures are defined is $[0,1]$.) How does this contrast with the statement that we have a sample space $\Omega$ on which are defined random variables $X_{1},\ldots,$ which map $\Omega $ to $[0,1]$ and $X_{n}$ converges in measure to some random variable $X$.


2 Answers 2


Weak convergence is weaker than convergence in measure. A very easy and illuminating example is to note that for weak convergence, the only requirement is that the distributions of the random variable converges. So if we pick for instance $X$ to be Gaussian (or something symmetric), and consider the sequence $X,-X,X,-X,\ldots$ (Ok, for your case you may need to pick a $[0,1]$ valued r.v. symmetric about $0.5$ and use $X,1-X,X,1-X,\ldots$). This sequence trivially converges weakly (as the distributions in the sequence are identical) but it's clear that this sequence does not converge in measure.

PS if the weak convergence is to a degenerate distribution (constant), then the convergence is also in measure, if I remember correctly.

If I interpret your remarks correctly, you are asking about the induced measures on $[0,1]$, throwing out the original sample space. Then you talk about the same notion of convergence, which is that $\mu_n(A) \to \mu(A)$ for sets $A$ without atoms of $\mu$, where these are measures on $[0,1]$. This notion relates to the weak* convergence of measures as linear functionals on the space of continuous functions. This is all in the realm of real analysis.

But then you ask about a sample space for probability and notions of convergence in measure (probability). For these, the sample space matters a lot! (As my example shows). To even talk about a sequence of random variables generically you have to use a sample space of the form $\Omega^\mathbb{N}$.

If we go back to just $[0,1]$, are you then asking how it relates to other notions of convergence? (ptwise, and in measure convergence make sense only for functions, say densities, but not for generic measures).

  • $\begingroup$ fine, the main difference is that one speaks of convergence of measures while the other is a convergence of actual functions......but under weak convergence, via application of Prokhorov's theorem I can make a sequence of random variables which converge almost surely....Let us say that in case of analyzing weak convergence since anyways I am only analysing the outcomes on space $[0,1]$ (which is the range of random variables), so it is essentially a difference of pointwise vs convergence in measure. PS: I am assuming that I can pick a sample space $\Omega$ as I want to $\endgroup$
    – user24367
    Sep 6, 2013 at 5:48
  • $\begingroup$ @user17523 I don't follow. Prokhorov is about extracting a subsequence that converges weakly, isn't it? Moreover, almost sure convergence is stronger than convergence in measure. There is a result that allows extraction of an almost surely convergent subsequence of a sequence converging in measure though. $\endgroup$
    – Evan
    Sep 6, 2013 at 5:57
  • $\begingroup$ sorry I wanted to mean Skorokhod........ $\endgroup$
    – user24367
    Sep 6, 2013 at 6:02
  • $\begingroup$ ooh ok. i see it now... interesting... $\endgroup$
    – Evan
    Sep 6, 2013 at 6:05
  • $\begingroup$ @ Evan...i wanted to mean that since I am anyways just observing the measurements from an experiment (which is the value of random outcome)...so there is no way to tell what my outcome $\omega\in\Omega$ was given the measurement $X(\omega)$ since in complicated processes no way you can get $\omega$ $\endgroup$
    – user24367
    Sep 6, 2013 at 6:16

Weak convergence is the convergence of measures given the weak* topology of $ C([0,1]) $ by Riesz-Representation theorem, which implies $ \mu_n \overset{*}{\rightharpoonup} \mu $ iff for all measurable functions $f$ we have $$ \int_0^1 fd\mu_n \rightarrow \int_0^1 fd\mu $$ Whereas convergence in measure is convergence of measurable functions, $ f_n \overset{\mu}{\rightarrow} f $ iff as $ n\rightarrow \infty $ we have $$ \mu(\{x \in [0,1]\ |\ |f_n(x)-f(x)|> \epsilon \}) \rightarrow 0 $$ Thus $ X_n \overset{\mu}{\rightarrow} X $ iff $ \mathbb{P}(|X_n-X|>\epsilon) \rightarrow 0 $ as $ n \rightarrow \infty $

  • $\begingroup$ I know the definitions.......can you, in the example above which I showed, comment as to whether the are the same or not? $\endgroup$
    – user24367
    Sep 6, 2013 at 5:31
  • $\begingroup$ @user17523 When you say $ X_n \rightarrow X $ in measure, it implies automatically that they are random variables of same measure space with a single measure $ \mu $, I don't see where do you come up with convergence of measures in this aspect. $\endgroup$
    – smiley06
    Sep 6, 2013 at 7:51
  • $\begingroup$ @smiley06 $X_n$ have associated push-forward measures $\mu_{X_n}$ defined by $\mu_{X_n}(A) = P(X_n \in A)$. $\endgroup$ Sep 26, 2014 at 23:37

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