Vague convergence of absolutely continuous measures to discrete, or vice versa

Question

1. Can a sequence of absolutely continuous probability measures converge vaguely to a discrete probability measure?

2. Can a sequence of discrete probability measures converge vaguely to an absolutely continuous probability measure?

This is an exercise from 'A Course in Probability Theory' by Chung, but I fail to solve. I know a sequence ${\mu_n, n \geq 1}$ of s.p.m’s is said to converge vaguely to a s.p.m. $\mu$ iff there exists a dense subset $D$ of $R$ such that, $\mu_n(a,b]\longrightarrow{}\mu(a,b]$ for all $a\in D, b\in D, a<b$.

Answer 1

Hints:

1. Consider $$F_n(x) := \begin{cases} 0, & x \leq 0, \\ x^n, & x \in [0,1], \\ 1, & x >1. \end{cases}$$ Show that $F_n$ is a distribution function of an absolutely continuous probability measure. Prove that $F_n$ converges vaguely to the (discrete) probability measure $\delta_1$. (Here $\delta_1$ denotes the Dirac measure centered at $1$.)
2. Define $$F_n(x) := \begin{cases} 0 & x< 0, \\ \frac{1}{n}, & x \in \big[0, \frac{1}{n} \big), \\ \frac{2}{n}, & x \in \big[ \frac{1}{n}, \frac{2}{n} \big), \\ \vdots & \\ 1, & x \geq 1. \end{cases}$$ Show that $F_n$ is the distribution function of a discrete probability measure. Prove that $F_n$ converges vaguely to an absolutely continuous probability measure.

Remarks

• In fact, one can show that any discrete probability measure is a vague limit of a sequence of absolutely continuous probability measures.
• A very typical example for the first question is also the Normal distribution with mean $0$ and variance $\frac{1}{n}$ which converges weakly to $\delta_0$. However, using the definition of vague convergence you provided in your question, this is more difficult to show than the example above.