Let $\mu, \mu_1, \mu_2, \dots$ be random measures on a Polish space (separable completely metrizable topological space) $(S, {\mathcal S})$. Suppose I know that

$$\int f d \mu_n \to \int f d\mu$$

in probability for each bounded continuous real-valued function. This would be the definition of weak convergence $\mu_n \Rightarrow \mu$, if I dropped "in probability" and the measures $\mu_n$ were deterministic.

Is there any standard way to extract a weakly convergent subsequence from $(\mu_n)$ consisting of almost all members of $(\mu_n)$? Possibly with additional assumptions? Where can I learn about such things?

Sorry if this is a trivial question.

  • $\begingroup$ I am not really sure what you mean by "consisting of almost all members", can you be more explicit? You can certainly say that there is a subsequence $(\mu_{n_k})$ which converges weakly almost surely; this is a standard fact for real-valued random variables and the proof works for random variables taking values in any metrizable topological space, such as the weak topology on a bounded set of measures on a Polish space. $\endgroup$ Apr 30, 2015 at 15:38
  • $\begingroup$ By "consisting of almost all members" I mean is there, almost surely, a random set A such that $n^{-1} |A \cap \{1, .., n\}| \to 0$ and $\mu_{n, n\not \in A} \Rightarrow \mu$ (or something in this direction). Also I am interested in subsequences that converge to $\mu$ given in the assumption above. $\endgroup$
    – Valentas
    Apr 30, 2015 at 15:54
  • 2
    $\begingroup$ Is it even true for real-valued random variables $X_n$ that if $X_n \to X$ i.p. then you can find an a.s. convergent subsequences which consists of almost all members in your sense? Is it true for the standard "typewriter sequence" counterexample? $\endgroup$ Apr 30, 2015 at 15:58
  • $\begingroup$ @NateEldredge , what is the "standard typewriter sequence"? $\endgroup$
    – Michael
    May 31, 2015 at 0:01
  • 1
    $\begingroup$ @Michael: See Example 4 here. $\endgroup$ May 31, 2015 at 0:02

1 Answer 1


It is interesting if you let the random index set depend on the realizations. For simplicity, restrict attention to random sequences $\{X_1, X_2, X_3, \ldots\}$ that converge to 0 in probability, but not with probability 1. We say that a set $B$ contains almost all positive integers if: $$ \lim_{n\rightarrow\infty} \frac{1}{n}| \{1, \ldots, n\} \cap B| = 1 $$

Claim 1: If $\{X_i\}_{i=1}^{\infty}$ are mutually independent, then with probability 1 there exists a random set $B$ (possibly dependent on $\{X_i\}$) that contains almost all positive integers and such that $X_i$ converges to $0$ over $i \in B$.

Claim 2: There exist examples where $\{X_i\}_{i=1}^{\infty}$ are mutually independent, but for which there exists no deterministic set $B$ that contains almost all positive integers and is such that $X_i$ converges to 0 with probability 1 over $i \in B$.

Claim 3: There are examples where no (possibly random) set $B$ with the desired properties exists. (In view of Claim 1, all such examples must have dependencies between the $X_i$ variables.)

Proof of Claim 1: Suppose $\{X_i\}_{i=1}^{\infty}$ are mutually independent and converge to 0 in probability. Then for all $\epsilon>0$ we have $Pr[|X_i|>\epsilon]\rightarrow 0$. It follows that there is a deterministic sequence of positive numbers $\{\epsilon_1, \epsilon_2, \epsilon_3, \ldots\}$ such that the following two things hold: \begin{align} &\lim_{i\rightarrow\infty} \epsilon_i = 0 \\ &\lim_{i\rightarrow\infty} Pr[|X_i| > \epsilon_i] = 0 \end{align} (As a quick explanation of why: Choose $\epsilon_1=1$. Find an index $n_2>1$ such that $Pr[|X_i|>1/2] \leq 1/2$ for all $i \geq n_2$ and define $\epsilon_i=\epsilon_1$ for all $i \in \{1, \ldots, n_2-1\}$, and define $\epsilon_{n_2}=1/2$. Then find an index $n_3>n_2$ such that $Pr[|X_i|>1/3] \leq 1/3$ for all $i \geq n_3$, and define $\epsilon_i = \epsilon_2$ for all $i \in \{n_2, \ldots, n_3-1\}$, and define $\epsilon_{n_3} = 1/3$, and so on.)

Now define the random set $B$ as follows: Include a positive integer $i$ in the set $B$ if and only if $\{|X_i|\leq \epsilon_i\}$. If the set $B$ has an infinite number of positive integers, then clearly $X_i$ converges to $0$ over $i \in B$. It remains to show that $B$ has almost all positive integers.

Define $I_i$ as an indicator function that is $1$ if $\{|X_i|>\epsilon_i\}$, and $0$ else. Define $N(k) = \sum_{i=1}^k I_i$ as the number of integers in $\{1, \ldots, k\}$ that are not in the set $B$. Notice that $|I_i-E[I_i]|\leq 1$ for all $i$, so: \begin{align} &E[(I_i-E[I_i])^2] \leq 1\\ &E[(I_i-E[I_i])^4]\leq 1 \end{align}

Define $S(k) = \sum_{i=1}^k E[I_i]$. Then for all $\delta>0$:
\begin{align} Pr\left[\left|\frac{N(k)-S(k)}{k}\right| \geq \delta\right] &=Pr\left[ |N(k)-S(k)| \geq \delta k \right]\\ &\leq Pr[(N(k)-S(k))^4 \geq \delta^4 k^4] \\ &= Pr\left[ \left(\sum_{i=1}^k(I_i-E[I_i])\right)^4 \geq \delta^4 k^4 \right]\\ &\leq \frac{E\left[\left( \sum_{i=1}^k(I_i-E[I_i]) \right)^4 \right]}{\delta^4 k^4} \end{align} where the final inequality holds by the Markov inequality. Because $\{I_i-E[I_i]\}_{i=1}^{\infty}$ are mutually independent and zero mean with second and fourth moments bounded by 1, it holds that there is a number $D>0$ such that: $$ E\left[\left( \sum_{i=1}^k(I_i-E[I_i]) \right)^4\right] \leq Dk + Dk^2 $$ Hence: $$ Pr\left[\left|\frac{N(k)-S(k)}{k}\right| \geq \delta\right] \leq \frac{Dk + Dk^2}{\delta^4 k^4} $$ The right-hand-side is summable, and so with probability 1: $$ \lim_{k\rightarrow\infty} \frac{N(k)-S(k)}{k} = 0 $$ However, $\frac{S(k)}{k} = \frac{\sum_{i=1}^k Pr[|X_i|>\epsilon_i]}{k} \rightarrow 0$, since it is the average of terms that converge to 0. It follows that with probability 1: $$ \lim_{k\rightarrow\infty} \frac{N(k)}{k} = 0 $$ and so (with prob 1) the random set $B$ contains almost all positive integers. $\Box$

Example for Claim 2: Define $\{X_1, X_2, X_3, \ldots\}$ mutually independent with: $$ X_i =\left\{ \begin{array}{ll} 1 &\mbox{ with probability $1/i$} \\ 0 & \mbox{ otherwise} \end{array}\right. $$ This example is well known to converge to 0 in probability, but not with probability 1. Suppose there is a deterministic set $B$ that contains almost all positive integers and for which $X_i$ converges to $0$ over $i \in B$ (we reach a contradiction).

For each positive integer $i$, define $g(i)$ as the number of elements in $\{i, i+1, \ldots, 2i\}$ are are not in $B$. Since $B$ has almost all positive integers, it is not difficult to show that: $$ \lim_{i\rightarrow\infty} \frac{g(i)}{i} = 0 $$ Now for each positive integer $i$, define $\theta_i$ as the probability that there is at least one index $j \in \{i, i+1, \ldots, 2i\} \cap B$ for which $X_j=1$. Then: \begin{align} \theta_i &= 1 - \prod_{j \in \{i, \ldots, 2i\} \cap B}\left(\frac{i-1}{i}\right)\\ &\geq 1 - \frac{\prod_{j=i}^{2i}\left(\frac{i-1}{i}\right)}{(1-1/i)^{g(i)}}\\ &= 1 - \frac{\left(\frac{i-1}{2i}\right)}{(1-1/i)^{g(i)}} \end{align} However, since $g(i)/i\rightarrow 0$ we have: $$ (1-1/i)^{g(i)} = \left((1-1/i)^{i}\right)^{g(i)/i} \approx (1/e)^{g(i)/i}\rightarrow 1 $$ and so: $$ \liminf_{i\rightarrow\infty} \theta_i \geq 1/2 $$ It follows that $\theta_i \geq 1/4$ for all sufficiently large positive integers $i$. Hence, all sufficiently large positive integers $i$ have the property that, with probability at least 1/4, there is an index $j>i$ such that $j\in B$ and $X_j=1$. So $X_i$ cannot converge to 0 with probability 1 over $i \in B$. $\Box$

Example for Claim 3: Consider the same $\{X_1, X_2, X_3, \ldots\}$ sequence from Claim 2, but now form a new (dependent) sequence $\{Y_1, Y_2, Y_3, \ldots\}$ by: $$ \{X_1, X_1, \: \: X_2, X_2, X_2, X_2, \: \: X_3, X_3, X_3, X_3, X_3, X_3, X_3, X_3, \ldots\} $$ Specifically, the $Y_i$ elements are filled in over frames, where each frame $k \in \{1, 2, 3, \ldots\}$ has size $2^k$ and consists of the same value $X_k$. It is clear that $Y_i$ converges to $0$ in probability (since $X_i$ converges to $0$ in probability).

Now take any (potentially random) set $B$ that contains almost all positive integers (the set $B$ is allowed to depend on the $\{X_i\}$ realizations). For all sufficiently large positive integers $k$, this set $B$ must contain at least half of the indices in frame $k$. But, with probability 1, $X_k=1$ for an infinite number of integers $k$. It follows that, with probability 1, $Y_i=1$ for an infinite number of elements $i \in B$. So, with probability 1, $Y_i$ does not converge to 0 over $i \in B$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .