In [3], Marczewski proved that convergence almost surely (a.s.) is equivalent to convergence in probability precisely when is the (at most) countable union of disjoint atoms (i.e. purely atomic). In particular, convergence a.s. is defined by a topology when is purely atomic. It is well known that convergence a.s. need not be defined by a topology when is not purely atomic. Try to construct a counter-example on the unit interval with Borel -algebra and Lebesgue measure.

Today, I want to give an extension of Marczewski’s result which characterizes when convergence a.s. is defined by a topology. Using arguments similar to those in my last post, we will prove the following theorem.

LetTheorem 1be a finite measure space. There exists a topology on the space of measurable functions coinciding with convergence a.s. if and only if is purely atomic.

*Proof:* Marczewski’s result proves sufficiency, so we prove necessity. If is not purely atomic, then there exists a measurable subset which is not an atom and on which satisfies the intermediate value property. In particular, for each , there exists a partition of into measurable sets with . Consider the sequence formed by indicator functions of the :

I claim that converges to zero in measure. Indeed, is supported on , for some and . Note tht as and implies . For any , for infinitely many , whence does not converge to zero a.s. Recall that if every subsequence of has a further subsequence which converges to some in a topology , then in . But convergence in measure implies convergence of a subsequence a.s. to the same limit, which is a contradiction.

- S.K. Berberian,
*Lectures in Functional Analysis and Operator Theory*, Spring Verlag 1974. -
- V.I. Bogachev,
*Measure Theory Volume I,*Springer 2007.

- V.I. Bogachev,
- E. Marczewski, “Remarks on the Convergence of Measurable Sets and Measurable Functions”,
*Colloquium Math.***2**(1955), 118-124.