$\{\mu_{\alpha}\}$ is a series of probability measure corresponding to random variables $\{X_{\alpha}\}$. If $\exists r>0,M>0,s.t.\forall \alpha,\mathbb E|X_{\alpha}|^r<M,$ then $\mu_{\alpha}$ is relatively compact.
I know that if $X_{\alpha}$ is defined in a complete separable space, then I just need to prove $\mu_{\alpha}$ is tight based on 'prohorov theorem' and it will be easier .
But generally, how can I get this statement?
Thanks for your help.
If I understand the question correctly $\mu_{\alpha}$ is a family of Borel probability measures on $\mathbb R$ with $\sup _{\alpha} \int |x|^{r}\, d\mu (x) <\infty$ and you want to show that $\{\mu_{\alpha}\}$ is tight. The domain of $X_{\alpha}$'s does not come i not the picture at all. (You are probably saying that the basic probability space is a complete separable metric space with some given Borel probability measure but this has no realtion to the tightness question). Since $\mu_{\alpha} \{x:|x|>A\} \leq \frac 1 {A^{r}} \sup _{\alpha} \int |x|^{r}\, d\mu_{\alpha} (x)$ we see that $\{\mu_{\alpha}\}$ is tight. PS: if the random variables take values in a metric space then $E|X_{\alpha}|^{r}$ does not even make sense, so I am assuming that $X_{\alpha}$'s are real valued.