When a Radon measure finite in bounded sets

measure-theorymetric-spaces

Let $(X,d)$ be a locally compact separable complete metric space (locally compact Polish metric space) and $\sigma$ a Radon measure i.e. a Borel measure finite in compact sets $\sigma(K)<\infty$ and regular: For every $A \in \mathcal{B}(X)$
$$\sigma(A)=\inf\{ \sigma(O): A \subset O \text{ open} \},$$
$$\sigma(A)=\sup\{ \sigma(K): A \supset K \text{ compact} \}$$
Can I guarantee in this case that $\sigma$ is finite in bounded sets? They are bounded w.r.t. the distance $d$.

I know that this is true in $\mathbb{R}^d$ because closed balls are compact. In a metric space I can't use this fact. I proved that $\sigma$ is $\sigma$-finite because I can get a increasing sequence of compact $K_n$ such that $X = \cup_{n \geq 1} K_n$. Could you give me any hint? Thanks a lot!

Best Answer

For every metric $d(x, y)$ you can define $\bar{d}(x, y)=min\{d(x, y), 1\}$ that is a metric inducing the same topology of $d(x, y)$ for wich every set is bounded. Hence I say no, elsewhere every set would have a finite measure.

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