When a.e. convergence does not imply convergence in probability

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.