4
$\begingroup$

In Varadhan's probability theory text, it is noted that countable additivity is key for showing that convergence a.e. implies convergence in measure. I'm wondering if there is a salient example of a finitely additive "probability" measure (i.e. a measure that has all the properties of a probability measure except is merely finitely and not countably additive), such that convergence a.e. no longer implies convergence in probability.

Of course, if we don't require that the entire space has finite measure, then we can come up with examples like $f_n = \chi_{n,n+1}$, but this sort of example doesn't work in a probability space, since of course the tails have to vanish, so in fact this sort of r.v. does in fact converge in probability to zero.

I'm more interested in a measure that otherwise acts like a probability measure outside of the countable additivity, in the hope that it will give me further intuition for why countable additivity is so crucial here (and please don't just refer to the proof for "intuition" -- I am familiar with the proof, and I understand in theory why countable additivity is important, but I think an illustrative example would make the phenomenon more salient).

$\endgroup$
3
$\begingroup$

Yes. Let us consider the following three restricted axioms:

  • Axiom 1: $P[A]\geq 0$ for all events $A$.

  • Axiom 2: $P[S]=1$.

  • Axiom 3: $P[A \cup B] = P[A]+P[B]$ whenever events $A,B$ are disjoint.

Consider $S = \mathbb{N} = \{1, 2, 3, ...\}$ and $P:2^{\mathbb{N}}\rightarrow\mathbb{R}$ defined as follows for subsets $A \subseteq\mathbb{N}$: $$ P[A] = \lim_{n\rightarrow\infty} \frac{|\{1, ..., n\} \cap A|}{n}$$ whenever the limit exists (where $|B|$ denotes the number of elements of set $B$). Consistently fill in the probabilities $P[A]$ for all sets $A$ for which the limit does not exist using the theory of Banach limits.

It can be shown that this $P$ function satisfies Axioms 1-3. It is easy to see that $P[A]=0$ for all finite sets $A \subseteq\mathbb{N}$. Then:

1) Countable additivity fails: $$1=P[S] = P[\cup_{i=1}^{\infty} \{i\}] \neq \sum_{i=1}^{\infty}\underbrace{P[\{i\}]}_{0}=0$$ So this is not a valid probability measure.

2) For $\omega \in \mathbb{N}$ define $$X_n(\omega)= \left\{ \begin{array}{ll} 1 &\mbox{ if $\omega \in \{1, ..., n\}$} \\ 0 & \mbox{ otherwise} \end{array} \right.$$ It holds that $\lim_{n\rightarrow\infty} X_n(\omega) = 1$ for all $\omega \in \mathbb{N}$. So $X_n$ converges to 1 surely (even stronger than "almost surely"). However for $\epsilon=1/2$ and for each $n \in \{1, 2, 3, ...\}$ we have $$ P[|X_n-1|>\epsilon] = P[X_n=0] = P[\{1, ..., n\}^c] =1-P[\{1, ..., n\}] = 1-0=1$$ So $X_n$ does not converge to $1$ in probability.

$\endgroup$
  • 1
    $\begingroup$ If you don't want to use Banach limits, you can restrict $P[A]$ to sets $A$ in the algebra of subsets of $\mathbb{N}$ that are either finite or have finite complements. $\endgroup$ – Michael Mar 4 at 5:52
  • 1
    $\begingroup$ This is great, thanks! $\endgroup$ – rem Mar 4 at 14:39
  • 1
    $\begingroup$ So really what this is exemplifying is the fact that without countable additivity, we don't get continuity results, and so even if the sets where X_n are far from X are getting smaller, their measure might not be decreasing. $\endgroup$ – rem Mar 4 at 14:44
  • $\begingroup$ Yes, countable additivity gives you "continuity of probability" in the sense that $$A_n\nearrow A \implies P[A_n] \nearrow P[A]$$ As an aside, the above three axioms (with $S$ as sample space) imply that if $\{A_i\}_{i=1}^{\infty}$ are disjoint events then $P[\cup_{i=1}^n A_i] = \sum_{i=1}^n P[A_i]$ for any positive integer $n$, and $$\mbox{$P[\cup_{i=1}^{\infty}A_i] \geq \sum_{i=1}^{\infty} P[A_i]$}$$ We cannot conclude equality holds without the countable additivity axiom. $\endgroup$ – Michael Mar 4 at 15:34
1
$\begingroup$

Let $(\Omega,\mathcal{F})=(\mathbb{N},2^{\mathbb{N}})$ and for $A\in \mathcal{F}$ let $(\delta_A)_n=\mathbf{1}_A(n)$. Then we define $$ \mathsf{P}(A):=T(\delta_A), $$ where $T$ is the Banach limit on $\ell_{\infty}$. By the properties of $T$, $\mathsf{P}$ is a finitely additive probability measure on $(\Omega,\mathcal{F})^{*}$ such that $\mathsf{P}(\{n,n+1,\ldots\})=1$ for any $n\in \mathbb{N}$.

Consider $f_n(\omega):=1_{[n,\infty)}(\omega)$. The sequence $\{f_n\}$ converges pointwise to $0$. However, $\mathsf{P}(\{f_n>\epsilon\})\to 1$ for any $\epsilon\in (0,1)$.


${}^*$ We need to check 3 conditions:

  1. For any $A\subset \mathbb{N}$, $0\le\liminf_{n\to\infty}\mathbf{1}_A(n) \le T(\delta_A)\le\limsup_{n\to\infty}\mathbf{1}_A(n)\le1$;

  2. $T(\delta_{\mathbb{\Omega}})=1$ ($\because \delta_{\mathbb{N}}$ is a constant sequence);

  3. If $A,B\subset\mathbb{N}$ are disjoint, then $T(\delta_{A\cup B})=T(\delta_A+\delta_B)=T(\delta_A)+T(\delta_B)$.

$\endgroup$
  • $\begingroup$ I think you are trying to do somehting similar to my answer. However, your $\delta(A)$ is not well defined. What is $\delta(\{2, 4, 6, 8, ...\})$? $\endgroup$ – Michael Mar 4 at 5:43
  • $\begingroup$ @Michael Thank's! I saw your answer and tried to come up with a "trivial" example. However, it seems that one needs some sophisticated construction anyway. $\endgroup$ – d.k.o. Mar 4 at 8:19

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.