> [!definition] > > Let $X$ be a [[Locally Compact Hausdorff Space|LCH]] space and $\mu$ be a [[Borel Measure|Borel measure]] on $X$. $\mu$ is a **Radon measure** if: > 1. $\mu$ is finite on all [[Compactness|compact]] sets. > 2. [[Regular Measure|Outer regular]] on all Borel sets. > 3. Inner regular on all [[Open Set|open]] sets. > [!definition] > > A signed measure is Radon if its positive and negative variations are Radon, and a complex measure is Radon if its real and imaginary parts are Radon. > [!theorem] > > $\mu$ is inner regular on all of its $\sigma$-finite sets. > > *Proof*. Let $\eps > 0$ and $E \in \cb(X)$ such that $\mu(E) < \infty$. By the outer regularity on open sets, there exists an open set $U \supset E$ such that $\mu(U) < \mu(E) + \eps$, and a compact set $F \subset U$ such that $\mu(F) > \mu(U) - \eps$. Since $\mu(U \setminus E) < \eps$, there exists $V \supset U \setminus E$ open such that $\mu(V) < \eps$. Let $K = F \setminus V$, then $K \subset E$ is compact and > $ > \mu(K) = \mu(F) - \mu(F \setminus V) > \mu(E) - \eps - \mu(V) > \mu(E) - 2\eps > $ > Now suppose that $\mu(E) = \infty$ and $E$ is $\sigma$-finite, then selecting an ascending sequence of sets and using the finite result yields the desired result. > [!theorem] > > $C_c(X)$ is dense in [$L^p(\mu)$](Lp%20Space) if $1 \le p < \infty$. > > *Proof*. Firstly, simple functions are dense in $L^p$. Let $\eps > 0$ and $E \in \cb(X)$, then by the inner and outer regularity of $\mu$, there exists $K \subset E \subset U$ where $K$ is compact and $U$ is open, such that > $ > \mu(K) + \eps > \mu(E) > \mu(U) - \eps > $ > By [[Urysohn's Lemma]], there exists $\phi \in C_c(X)$ such that $\supp{\phi} \subset U$ and $\phi|_K = 1$. This gives that $\norm{\phi - \one_E}_p \le (2\eps)^{1/p}$.