Let $\mu_n: \mathcal{R}^k \to [0,1 ]$ be a sequence of probability measures on the Borel sets of $\mathbb{R}^k$ such that $\mu_n \to \mu$ weakly where $\mu:\mathcal{R}^k \to [0,1]$ is a probability measure.s it true that $\{\mu_n\}_{n=1}^\infty$ is tight?
I guess it is: By Prohorov's theorem, it suffices to check that every subsequence of $\{\mu_n\}$ has a weakly convergent subsequence. But this is obvious since all subsequences converge to $\mu$ as well.
Thus $\{\mu_n\}$ is tight.
Is this correct?
You are correct. (One of) Prohorov's theorem(s) shows that relative compactness (the subsequential statement) implies tightness when the base space is a complete, separable metric space. The base space here is Euclidean space, so the theorem applies and any weakly convergent sequence is relatively compact, hence tight. See pp.35-37 of Billingsley's Convergence of Probability Measures.