The Unapologetic Mathematician

Mathematics for the interested outsider

Regular Outer Measures

As usual, let \mathcal{R} be a ring of sets, \mathcal{S}(\mathcal{R}) be the smallest \sigma-algebra containing \mathcal{R}, and \mathcal{H}(\mathcal{R}) be the smallest hereditary \sigma-algebra containing \mathcal{R}. We’ve asked about the relation between a measure \mu on \mathcal{R}, the outer measure \mu^* it induces on \mathcal{H}(\mathcal{R}), and the measure \bar{\mu} we get by restricting \mu^* to \mathcal{S}(\mathcal{R}). But for now, let’s consider what happens when we start with an outer measure on \mathcal{H}(\mathcal{R}).

Okay, so we’ve got an outer measure \mu^* on a hereditary \sigma-ring \mathcal{H} — like \mathcal{H}(\mathcal{R}). We can define the \sigma-ring \overline{\mathcal{S}} of \mu^*-measurable sets and restrict \mu^* to a measure \bar{\mu} on \overline{\mathcal{S}}. And then we can turn around and induce an outer measure \bar{\mu}^* on the hereditary \sigma-ring \mathcal{H}(\overline{\mathcal{S}}).

Now, in general there’s no reason that these two should be related. But we have seen that if \mu^* came from a measure \mu (as described at the top of this post), then \mathcal{H}(\overline{\mathcal{S}})=\mathcal{H}(\mathcal{R}), and the measure \bar{\mu}^* induced by \bar{\mu} is just \mu^* back again!

When this happens, we say that \mu^* is a “regular” outer measure. And so we’ve seen that any outer measure induced from a measure on a ring is regular. The converse is true as well: if we have a regular outer measure \mu^*=\bar{\mu}^*, then it is induced from the measure \bar{\mu} on \overline{\mathcal{S}}. Induced and regular outer measures are the same.

Doesn’t this start to look a bit like a Galois connection?

April 2, 2010 Posted by | Analysis, Measure Theory | Leave a comment

   

Follow

Get every new post delivered to your Inbox.

Join 401 other followers