Vtyjkyuk

$mu_alpha$ 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 $sup _alpha int |x|^r, dmu (x) <infty$ 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 $mu_alpha x: leq frac 1 A^r sup _alpha int |x|^r, dmu_alpha (x)$ 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.