From Terrence Tao's blog
Exercise 23 (Implications and equivalences) Let ${X_n, X}$ be random variables taking values in a ${\sigma}$-compact metric space ${R}$.
(ii) Show that if ${X_n}$ converges in distribution to ${X}$, then ${X_n}$ has a tight sequence of distributions.
(iv) Show that ${X_n}$ converges in distribution to ${X}$ if and only if ${\mu_{X_n}}$ converges to ${\mu_X}$ in the vague topology (i.e. ${\int f\ d\mu_{X_n} \rightarrow \int f\ d\mu_X}$ for all continuous functions ${f: R \rightarrow {\bf R}}$ of compact support).
(v) Conversely, if ${X_n}$ has a tight sequence of distributions, and ${\mu_{X_n}}$ is convergent in the vague topology, show that ${X_n}$ is convergent in distribution to another random variable (possibly after extending the sample space). What happens if the tightness hypothesis is dropped?
Isn't (v) already part of (iv)? What do I miss to see? Thanks!
Best Answer
I believe the discussion of this old mathexchange post clarifies what is meant (I don't find the way the exercise was written particularly clear): Definition of convergence in distribution
See in particular the comment by Chris Janjigian. In (iv), the limiting distribution is assumed to be that of a R.V.; in (v), all that is given is that we have vague convergence (which need not be to a R.V., per the discussion of the link). However, if we add tightness, it will be - and this is exercise (v).