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).
Best Answer
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.