## Continuity of Measures

Again we start with definitions. An extended real-valued set function on a collection of sets is “continuous from below” at a set if for every increasing sequence of sets — that is, with each — for which — remember that this limit can be construed as the infinite union of the sets in the sequence — we have . Similarly, is “continuous from above” at if for every decreasing sequence for which and which has for at least one set in the sequence we have . Of course, as usual we say that is continuous from above (below) if it is continuous from above (below) at each set in its domain.

Now I say that a measure is continuous from above and below.

First, if is an increasing sequence whose limit is also in , then . Let’s define and calculate

where we’ve used countable (and finite) additivity to turn the disjoint union into a sum and back.

Next, if is a decreasing sequence whose limit is also in , and if at least one of the has finite measure, then . Indeed, if has finite measure then by monotonicity, and thus the limit must have finite measure as well. Now is an *increasing* sequence, and we calculate

And thus a measure is continuous from above and from below.

On the other hand we have this partial converse: Let be a finite, non-negative, additive set function on an algebra . Then if either is continuous from below at every or is continuous from above at , then is a measure. That is, either one of these continuity properties is enough to guarantee countable additivity.

Since is defined on an algebra, which is closed under finite unions, we can bootstrap from additivity to finite additivity. So let be a countably infinite sequence of pairwise disjoint sets in whose (disjoint) union is also in , and define the two sequences in :

If is continuous from below, is an increasing sequence converging to . We find

On the other hand, if is continuous from above at , then is a decreasing sequence converging to . We find