# The Unapologetic Mathematician

## 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?

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Posted by John Armstrong | Analysis, Measure Theory

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