# If the sequence of distribution functions weakly converge, the sequence of corresponding subprobability measures converges weakly, too

Let

• $\mu,\mu_n$ be subprobability measures on $\left(\mathbb{R},\mathcal{B}\left(\mathbb{R}\right)\right)$
• $F,F_n$ be the distribution functions of $\mu,\mu_n$ with $$\lim_{n\to\infty}F_n(x)=F(x)\;\;\;\text{for all continuity points }x\in\mathbb{R}\text{ of }F$$ and $$F(\infty)\ge\limsup_{n\to\infty}F_n(\infty)$$ $\color{blue}{\text{where }g(\infty):=\displaystyle\lim_{x\to\infty}g(x)}$
• $f:\mathbb{R}\to [0,1]$ be Lipschitz continuous with Lipschitz constant $1$
• $\varepsilon>0$

I've got the following two questions:

1. Why can we choose $N\in\mathbb{N}$ and $N+1$ continuity points $y_0<\ldots<y_N$ of $F$ such that $$F(y_0)<\varepsilon\;\;\;\text{and}\;\;\;F(y_N)>F(\infty)-\varepsilon\;\;\;\text{and}\;\;\;y_i-y_{i-1}<\varepsilon$$
2. Why do we've got $$\int f\;d\mu_n\le\left(F_n(y_9)+F_n(\infty)-F_n(y_N)\right)+\sum_{i=1}^N\left(f(y_i)+\varepsilon\right)\left(F_n(y_i)-F_n(y_{i-1})\right)$$

1. Note that $F$ has at most countable many discontinuity points (this follows from the fact that $F$ is càdlàg, i.e. has left-limits and is right-continuous). By the continuity of the measure, there exist $r_0, R_0 \in \mathbb{R}$ such that $$F(r) = \mu(-\infty,r] < \epsilon \qquad \qquad F(R) > F(\infty)-\epsilon$$ for all $r \leq r_0$, $R \geq R_0$. Since there are at most countably many discontinuity points, we can choose $y_0 \leq r_0$ and, inductively, $y_i$ such that $|y_i-y_{i-1}| < \epsilon$ and each $y_i$ is a continuity point. For $N$ sufficiently large, we have $y_N \geq R$ and we are done.
2. Note that \begin{align*} \int f \, d\mu_n &= \int_{(-\infty,y_0]} f \, d\mu_n + \sum_{i=1}^N \int_{(y_{i-1},y_i]} f \, d\mu_n + \int_{(y_N,\infty)} f \, d\mu_n \end{align*} For the first and third term on the right-hand side use that $|f| \leq 1$ and for the second one the Lipschitz continuity of $f$.
• How exactly does the existence of $r_0$ follow from the continuity of the measure? – 0xbadf00d Mar 28 '15 at 13:59
• @oxbadfood Since $A_n := (-\infty,-n]$ satisfies $\bigcap_n A_n = \emptyset$ and $A_n \downarrow$, it follows that $$0 = \lim_{n \to \infty} \mu(A_n).$$ – saz Mar 28 '15 at 14:15
• @oxbadfood Use $$\int_{(y_{i-1},y_i]} (f(x)-f(y_i))+f(y_i) \, d\mu_n(x) \leq \left(\sup_{y \in (y_{i-1},y_i]} |f(y)-f(y_{i})| + |f(y_i)| \right) \int_{(y_{i-1},y_i]} 1 \, \mu_n(dx).$$ – saz Mar 29 '15 at 14:23
• Now, $$S:=\limsup_{n\to\infty}\int f\;d\mu_n\le\underbrace{F(y_0)}_{< \varepsilon}+\underbrace{F(\infty)-F(y_N)}_{< \varepsilon}+\sum_{i=1}^N(f(y_i)+ \varepsilon)\left(F(y_i)-F(y_{i-1})\right)$$I've read that it holds $$S\le 4\varepsilon+\int f\;d\mu\tag{1}$$However, the best estimation that I get is $$\begin{matrix}S&\le&2 \varepsilon+\sum_{i=1}^N\left(f(y_i)+\varepsilon\right)\underbrace{\left(F(y_i)-F(y_{i-1})\right)}_{=:T_i\le 1}\\&<&(N+2)\varepsilon+\sum_{i=1}^Nf(y_i)\left(F(y_i)-F(y_{i-1})\right)\\&<&2(N+1)\varepsilon+\int f\;d\mu\end{matrix}\tag{2}$$ – 0xbadf00d Mar 29 '15 at 16:45