## An Extension of a Theorem of Marczewski

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

Theorem 1 Let ${(X,\mathcal{A},\mu)}$ be a finite measure space. There exists a topology ${\tau}$ on the space of measurable functions ${L^{0}(\mu)}$ coinciding with convergence a.s. if and only if ${X}$ is purely atomic.

Proof: Marczewski’s result proves sufficiency, so we prove necessity. If ${X}$ is not purely atomic, then there exists a measurable subset ${A}$ which is not an atom and on which ${\mu}$ satisfies the intermediate value property. In particular, for each ${n\in\mathbb{N}}$, there exists a partition of ${A}$ into ${m_{n}}$ measurable sets ${A_{1}^{n},\ldots,A_{m_{n}}^{n}}$ with ${0<\mu(A_{i}^{n})\leq n^{-1}}$. Consider the sequence ${f_{k}}$ formed by indicator functions of the ${A_{i}^{n}}$:

$\displaystyle 1_{A_{1}^{1}},\ldots,1_{A_{m_{1}}^{1}},1_{A_{1}^{2}},\ldots,1_{A_{m_{2}}^{2}},1_{A_{1}^{3}},\ldots,$

I claim that ${f_{k}}$ converges to zero in measure. Indeed, ${f_{k}}$ is supported on ${A_{i}^{n}}$, for some ${n}$ and ${1\leq i\leq m_{n}}$. Note tht ${\mu(A_{i}^{n})\rightarrow 0}$ as ${n\rightarrow\infty}$ and ${k\rightarrow\infty}$ implies ${n\rightarrow\infty}$. For any ${x\in X}$, ${f_{k}(x)=1}$ for infinitely many ${k}$, whence ${f_{k}}$ does not converge to zero a.s. Recall that if every subsequence of ${f_{k}}$ has a further subsequence which converges to some ${f}$ in a topology ${\tau}$, then ${f_{k}\rightarrow f}$ in ${\tau}$. But convergence in measure implies convergence of a subsequence a.s. to the same limit, which is a contradiction. $\Box$

1. S.K. Berberian, Lectures in Functional Analysis and Operator Theory, Spring Verlag 1974.
1. V.I. Bogachev, Measure Theory Volume I, Springer 2007.
2. E. Marczewski, “Remarks on the Convergence of Measurable Sets and Measurable Functions”, Colloquium Math. (1955), 118-124.