Let $(\Omega, \mathcal{B}, \mu)$ be a measure space where $\Omega \subset \mathbb{R}^K$ is the unit sphere in the $K$-dimensional space, and $\mu$ is a finite, positive measure (not necessarily a probability measure). Let $\mu(\Omega)=m \in (0,\infty)$. The CDF of $\mu$ -- let us call it $F$ -- admits a continuous density $f$ which is strictly positive on $[0,1]$. I want a construct a sequence of finitely supported measures $\mu_n$, with supports contained in $\Omega$, converging (weakly) to $\mu$, ensuring the following property: If $x,x'$ are in the support of both $\mu_n$ and $\mu$ then we must have $\mu_n({x})/\mu_n({x'})=f({x})/f({x'})$.
I have constructed the distribution sequence as follows. Define $\Omega_n = \{x \in [0,1]: x = k/n\;\text{for some}\;k \in \mathbb{Z}\}$. Define:
$$f_n(x)=\begin{cases} &\frac{mf(x)}{\sum\limits_{x'\in \Omega_n}f(x')}\;\text{if}\;x \in \Omega_n,\\ &0,\;\text{otherwise}. \end{cases}$$
Fix $x\in \Omega$.
\begin{align*} \therefore\;F_n(x)&=\frac{m\sum\limits_{x'\in \Omega_n, x' \leq x}f(x')}{\sum\limits_{x'\in \Omega_n}f(x')}\\ &=\frac{m\sum\limits_{x'\in \Omega_n, x' \leq x}f(x')\times\frac{1}{n^K}}{\sum\limits_{x'\in \Omega_n}f(x')\times\frac{1}{n^K}} \end{align*}
where $x' \leq x$ indicates componentwise inequality.
By the definition of the Reimann integral, the limit of the numerator and the denominator as $n \rightarrow \infty$ are $m\int\limits_{x' \in \Omega, x'\leq x}f(x')dx'$ and $\int\limits_{x' \in \Omega}f(x')dx'$ respectively, which gives us pointwise convergence of the CDF's.
I have two questions:
- Is the argument for showing pointwise convergence of the sequence of CDF's correct?
- Is pointwise convergence of the sequence of CDF's sufficient for weak convergence of the corresponding sequence of measures, like it is in case of probability measures?
Any help is most appreciated.