# The Unapologetic Mathematician

## Inner Measures

A quick one for today.

In analogy with the outer measure $\mu^*$ induced on the hereditary $\sigma$-ring $\mathcal{H}(\mathcal{S})$ by the measure $\mu$ on the $\sigma$-ring $\mathcal{S}$, we now define the “inner measure” $\mu_*$ that $\mu$ induces on the same hereditary $\sigma$-ring $\mathcal{H}(\mathcal{S})$. We’ve seen that the outer measure is

$\displaystyle\mu^*(E)=\inf\{\mu(F)\vert E\subseteq F\mathcal{S}\}$

Accordingly, the inner measure is a set function defined by

$\displaystyle\mu_*(E)=\sup\{\mu(F)\vert E\supseteq F\mathcal{S}\}$

In a way, the properties of $\mu_*$ are “dual” to those of $\mu^*$. The easy ones are the same: it’s non-negative, monotone, and $\mu_*(0)=0$.

We could also define $\mu_*$ in terms of the completed measure. Since $\mathcal{S}\subseteq\overline{\mathcal{S}}$, it’s clear

$\displaystyle\mu_*(E)=\sup\{\mu(F)\vert E\supseteq F\mathcal{S}\}\leq\sup\{\bar{\mu}(F)\vert E\supseteq F\overline{\mathcal{S}}\}$

On the other hand, the definition of the completion says that for every $F\in\overline{\mathcal{S}}$ there is a $G\in\mathcal{S}$ with $G\subseteq F$ and $\mu(G)=\bar{\mu}(F)$, and so this is actually an equality.

April 8, 2010 - Posted by | Analysis, Measure Theory

## 3 Comments »

1. [...] and measurable kernels. These will allow us to move statements we want to show about outer and inner measures to the realm of proper measures, where we can use nice things like [...]

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2. [...] a subset , we write for the image of under the transformation — . I say that the outer and inner Lebesgue measures are both nicely behaved under the transformation [...]

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3. [...] a measure space , we will routinely use without comment the associated outer measure and inner measure on the hereditary -ring [...]

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