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 $$\mu(B) = \inf \{\mu(U) \mid B \subset U, U \text{ is open}\}.$$
It is shown in this lecture note that
Theorem: If $X$ is a metric space and $\mu$ is finite, then the collection $$ \mathcal R := \{B \in\mathcal B(X) \mid \mu \text{ is inner regular and outer regular on } B\} $$ is a $\sigma$-algebra.
In the proof, the author does not use any topological property of a metric space, so I think below result still holds.
Statement 1: If $\mu$ is finite, then the collection $$ \mathcal R := \{B \in\mathcal B(X) \mid \mu \text{ is inner regular and outer regular on } B\} $$ is a $\sigma$-algebra.
I would like to ask if any of below generalizations is true.
- Statement 2: The collection $$ \mathcal R := \{B \in\mathcal B(X) \mid \mu \text{ is inner regular and outer regular on } B\} $$ is a $\sigma$-algebra.
- Statement 3: The collection $$ \mathcal R := \{B \in\mathcal B(X) \mid \mu \text{ is inner regular on } B\} $$ is a $\sigma$-algebra.
- Statement 4: The collection $$ \mathcal R := \{B \in\mathcal B(X) \mid \mu \text{ is outer regular on } B\} $$ is a $\sigma$-algebra.
Thank you so much!