Probability I, Fall 2018

The course Probability I (M385C) was taught by Dr. Gordan Žitković. His lecture notes for the class can be found here.

In the fourth problem set, there is a problem about a sequence of measurable functions $\{f_n\}$ with the following sub-subsequence property:

Given any subsequence of $\{f_n\}$ there exists a further subsequence which converges almost everywhere to a measurable function $f$ .

Given a sequence with this property, we can immediately get some convergence results for the whole sequence under certain assumptions. If the measure space we are working over is finite, convergence a.e. implies convergence in measure. Hence the same sub-subsequence property is true for the $\{f_n\}$ with convergence in measure. Now, convergence in measure is metrizable on the space of measurable functions modulo a.e. equality, and hence by the general sub-subsequence lemma for metric spaces we have that $f_n \xrightarrow{m} f$ .

Note that this general sub-subsequence lemma for metric spaces shows that convergence a.e. is not metrizable, i.e. there does not exists a metric $d$ on the space of measurable functions in which convergence in $d$ is equivalent to convergence a.e.

To see this, let $\{A_n\}$ be a sequence of sliding, shrinking dyadic intervals in $[0,1]$ , and let $f_n = \chi_{A_n}$ . Then, given any subsequence $\{f_{n_k}\}$ , we know that $\Vert f_{n_k} \Vert_1 \rightarrow 0$ as $k \rightarrow \infty$ . Then, there exists a further subsequence which converges to $0$ a.e. If there did exists a metric (or even pseudometric) $d$ which metrizes convergence a.e., then by the sub-subsequence lemma, we would have that the whole sequence $f_n \rightarrow 0$ a.e. But this is a contradiction since $f_n$ does not converge anywhere on $[0,1]$ .