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Let X be a Hausdorff topological space $X$ and $\mathcal B(X)$ its Borel $\sigma$-algebra. Let $\mu$ be a non-negative Borel measure and $B \in\mathcal B(X)$.

  • $\mu$ is tight on $B$ iff $$\mu(B) = \sup \{\mu(K) \mid K \subset B, K \text{ is compact}\}.$$
  • $\mu$ is inner regular on $B$ iff $$\mu(B) = \sup \{\mu(C) \mid C \subset U, C \text{ is closed}\}.$$
  • $\mu$ is outer regular on $B$ iff $$\mu(B) = \inf \{\mu(U) \mid B \subset U, U \text{ is open}\}.$$
  • $\mu$ is locally finite iff every point of $X$ has a neighborhood $U$ for which $\mu(U)$ is finite.

First, we have theorems:

Theorem 1: If $\mu$ is tight on $B$, then $\mu$ is inner regular on $B$.

Theorem 2: If $\mu$ is finite, then $\mu$ is tight on any Borel set iff $$\forall \varepsilon>0, \exists K \text{ compact}: \mu(K^c) < \varepsilon.$$ Theorem 3: If $X$ is a metric space and $\mu$ finite, then $\mu$ is inner regular and outer regular on any Borel set.

  1. Are there some conditions of $X$ that [$\mu$ is tight on any open set] implies [$\mu$ is tight on any Borel set].

  2. Are there some conditions of $X$ that [$\mu$ is inner regular on any open set] implies [$\mu$ is inner regular on any Borel set].

In Wikipedia page,

  • If $\mu$ is locally finite, then $\mu$ is finite on compact sets, and for locally compact Hausdorff spaces, the converse holds, too. Thus, in this case, local finiteness may be equivalently replaced by finiteness on compact subsets.
  • $\mu$ is called a Radon measure if it is locally finite and tight on any open set. In many situations, such as finite measures on locally compact spaces, this also implies outer regularity.

In a lecture note,

$\mu$ is called a Radon measure (on a metric space) iff $\mu$ is tight on any open set, outer regular on any Borel set, and finite on compact sets.

Metric spaces are not necessarily locally compact, so being finite on compact sets does not necessarily imply being locally finite.

  1. Could you confirm if the def of Radon measure in this lecture note is more general that of Wikipedia?

  2. By Theorem 2, if $\mu$ is finite, the outer regular condition in the lecture note can be safely removed. Is my observation correct?

Analyst
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  • Starting with measures on general topological spaces is rather artificial. Most often one have linear maps $\lambda$ on $C(X)$ or $C_{00}(X)$. Under some regularity conditions ($X$ Hausdorff, locally compact) and certain properties of $\lambda$ (positivity for example) then there is a measure $\mu$ with all the regularity properties of your posting such that $\lambda f=\int f,d\mu$ for all $f\in C_{00}(X)$. Those linear functionals are the (nonnegative) Radon measure. Additional properties on $\lambda$ give rise to complex Radon measures. The Lebesgue measure is one example (to be continue) – Mittens Apr 24 '22 at 16:44
  • (continue) the Lebesgue measure for example followed from considering the linear functional $\lambda(f)=\int f(s)ds$ for $f\in\mathcal{C}_{00}(\mathbb{R})$ where $\int$ stands for the Riemann integral. – Mittens Apr 24 '22 at 16:47

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