> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]], and $\mu: \cb(X) \to [0, \infty]$ be a Borel [[Measure Space|measure]], then $\mu$ is **regular** if > - $\mu(K) < \infty$ for every [[Compactness|compact]] $K$. > - **Outer Regular:** $\mu(E) = \inf\bracs{\mu(U): U \in \topo, U \supset E}$ for all $E \in \cb(X)$. > > If $\nu$ is a [[Signed Measure|signed measure]] or [[Complex Measure|complex measure]], then $\nu$ is regular if $\abs{\nu}$ is. > [!theorem] > > Let $f \in L^+(\real^n)$, then the measure $fdm$ is regular if and only if $f \in \loci$ is [[Locally Integrable|locally integrable]]. > > *Proof*. Firstly $fdm$ satisfies the compact requirement if and only if $f \in \loci$. Suppose that $f \in \loci$ and let $E \in \cb(\real)$ be bounded. Let $\delta > 0$, then by the outer regularity of the Lebesgue measure, there exists $U \supset E$ bounded such that $m(U \setminus E) < \delta$. Since $fdm \ll dm$, for any $\varepsilon > 0$, there exists $\delta > 0$ such that $\int_{U \setminus E}fdm < \varepsilon$ whenever $m(U \setminus E) < \delta$. Hence we have found an open set $U \supset E$ such that $\int_{U}fdm > \int_{E}fdm + \varepsilon$. As $\varepsilon$ is arbitrary, $fdm$ satisfies outer regularity. > [!theorem] > > Let $\nu$ be a regular measure on $\real^n$, and > $ > d\nu = d\lambda + fdm > $ > be its [[Lebesgue-Radon-Nikodym Theorem|Lebesgue decomposition and Radon-Nikodym derivative]]. Then $d\lambda$ and $fdm$ are both regular. > > *Proof*. If $K$ is compact, then $\nu(K) = \lambda(K) + \int_K fdm < \infty$, so $\lambda(K)$ and $\int_{K}fdm$ are both bounded. By restricting to the domains of $d\lambda$ and $fdm$, both measures are outer regular.