finite expectation on compact metric space

Question

I would like to show that if $(E,d)$ is a compact metric space with a Borel $\sigma$-Algebra $\Sigma$ then any probability measure $\mu$ on $E$ has finite expectation, which means $\int d(0,x)\mu(dx)<\infty$.

I know from the compact metric space that $E$ is closed and bounded, if I take $E\subset\mathbb{R}$. Furthermore I know that $E$ has finite expectation if $\int d(0,x) \mu^+(dx)<\infty$ and $\int d(0,x)\mu^-(dx)<\infty$. How can I show this?

Answer 1

If the metric space $E$ is compact, then it is bounded. (In fact it is totally bounded, but this isn't important.)

Therefore, there exists an $M > 0$ such that $d(x,0) < M$ for all $x \in E$.

But then, if I understood your question correctly, $$ \int d(x,0) d\mu \leq M \int d\mu = M,$$ since $\mu(E) = 1$ as $\mu$ is a probability measure.