# Convergence of measures — revisited

In this thread, I asked a question about the convergence of measures. The conjecture I posed there, which turned out to be false, was supposed to be a lemma that I wanted to use to prove a proposition, outlined as follows.

Consider the measurable space $(\mathbb R,\mathscr B_{\mathbb R})$. Suppose that $\{\mu_n\}_{n=1}^{\infty}$ is a sequence of finite Radon measures and $\mu$ is a finite Radon measure on $\mathbb R$. Suppose also that $$\lim_{n\to\infty}\mu_n(\mathbb R)=\mu(\mathbb R)\tag{1}$$ and that $\{\mu_n\}_{n=1}^{\infty}$ converges to $\mu$ vaguely, in the sense that $$\lim_{n\to\infty}\int f\,\mathrm d\mu_n=\int f\,\mathrm d\mu$$ for any $f\in C_c$, where $C_c$ is the space of compactly supported, continuous, real-to-real functions. For any $x\in\mathbb R$ and $n\in\mathbb N$, define \begin{align*}F_n(x)\equiv&\,\mu_n((-\infty,x]),\\F(x)\equiv&\,\mu((-\infty,x]).\end{align*}

Proposition:$\quad$ For any $a\in\mathbb R$ and $\varepsilon>0$, $$\limsup_{n\to\infty}F_n(a)\leq F(a+\varepsilon).$$

I figured out a proof that I think is correct, which I now share in order to seek feedback. Your thoughts are much appreciated.

Proof:$\quad$ Let $a\in\mathbb R$ and $\varepsilon>0$ be given. Also, fix $\delta>0$. Since $\mu$ is a finite Radon measure, there exists a compact set $C\subset\mathbb R$ such that $$\mu(\mathbb R)\geq\mu(C)>\mu(\mathbb R)-\frac{\delta}{5}.$$ Since compact sets are bounded, one can choose $K>0$ so large that $$-K<a<a+\varepsilon<K,$$ $C\subseteq[-K,K]$, and $$\mu(\mathbb R)\geq\mu([-K,K])>\mu(\mathbb R)-\frac{\delta}{5}.\tag{2}$$

Pick any $L>K$. Clearly, $[-K,K]\subseteq[-L,L]$. Construct a continuous function $h$ such that $h$ is 1 on $[-K,K]$, it vanishes outside $[-L,L]$, and it is linear on $[-L,-K]$ and on $[K,L]$. Then, $h\in C_c$. By vague convergence, there exists some $N_1\in\mathbb N$ such that $$\int h\,\mathrm d\mu<\int h\,\mathrm d\mu_n+\frac{2\delta}{5}\quad\forall n\geq N_1.\tag{3}$$

Also, by (1), there exists some $N_2\in\mathbb N$ such that $$\mu(\mathbb R)+\frac{\delta}{5}>\mu_n(\mathbb R)>\mu(\mathbb R)-\frac{\delta}{5}\quad\forall n\geq N_2\tag{4}.$$

Now, if $n\geq\max\{N_1,N_2\}$, then, on the one hand, \begin{align*}\mu([-K,K])=&\,\int_{[-K,K]}h\,\mathrm d\mu\leq\int h\,\mathrm d\mu<\int h\,\mathrm d\mu_n+\frac{2\delta}5\\=&\,\int_{[-L,L]}h\,\mathrm d\mu_n+\frac{2\delta}5\leq\mu_n([-L,L])+\frac{2\delta}5;\end{align*} and, on the other hand, $$\mu_n([-L,L])\leq\mu_n(\mathbb R)<\mu(\mathbb R)+\frac{\delta}{5}<\mu([-K,K])+\frac{2\delta}{5},$$ by (2) and (4). Therefore, $$\mu_n([-L,L])-\frac{2\delta}{5}<\mu([-K,K])<\mu_n([-L,L])+\frac{2\delta}{5}\quad\forall n\geq\max\{N_1,N_2\}\tag{5}.$$

Let $M>L$ be arbitrary. Construct a continuous function $g$ such that $g$ is 1 on $[-L,a]$, it vanishes outside $[-M,a+\varepsilon]$, and it is linear on $[-M,-L]$ and on $[a,a+\varepsilon]$. Then, $g\in C_c$. By vague convergence, there exists some $N_3\in\mathbb N$ such that $$\int g\,\mathrm d\mu_n<\int g\,\mathrm d\mu+\frac{\delta}{5}\quad\forall n\geq N_3.\tag{6}$$

Let $n\geq N^*\equiv\max\{N_1,N_2,N_3\}$. Then, \begin{align*} F_n(a)=&\,\mu_n((-\infty,a])=\mu_n((-\infty,-L))+\mu_n([-L,a])\\ =&\,\mu_n(\mathbb R)-\mu_n([-L,L])-\mu_n((L,\infty))+\mu_n([-L,a])\\ \leq&\,\mu_n(\mathbb R)-\mu_n([-L,L])+\mu_n([-L,a])\\ \underbrace{<}_{(5)}&\,\mu_n(\mathbb R)-\mu([-K,K])+\frac{2\delta}5+\mu_n([-L,a])\\ \underbrace{<}_{(4)}&\,\mu(\mathbb R)-\mu([-K,K])+\frac{3\delta}5+\mu_n([-L,a])\\ \underbrace{<}_{(2)}&\,\frac{4\delta}5+\mu_n([-L,a])=\int_{[-L,a]}g\,\mathrm d\mu_n+\frac{4\delta}{5}\\ \leq&\,\int g\,\mathrm d\mu_n+\frac{4\delta}{5}\underbrace{<}_{(6)}\int g\,\mathrm d\mu+\delta=\int_{[-M,a+\varepsilon]}g\,\mathrm d\mu+\delta\\ \leq&\,\mu([-M,a+\varepsilon])+\delta\leq\mu((-\infty,a+\varepsilon])+\delta=F(a+\varepsilon)+\delta. \end{align*}

Since this is true for all $n\geq N^*$, it follows that $$\limsup_{n\to\infty} F_n(a)=\inf_{k\in\mathbb N}\sup_{n\geq k}F_n(a)\leq \sup_{n\geq N^*}F_n(a)\leq F(a+\varepsilon)+\delta.$$ Given that $\delta>0$ can be made arbitrarily small, the claim follows. $\blacksquare$

Note: this proposition is needed to prove Proposition 7.19(b) in Folland (1999). The earlier printings report the result without the assumption (1), in which case the proposition I presented can be shown to be false. The errata sheet points out that the claim is true if (1) is assumed, but gives no proof. That's why I attempted to construct one. Any comments are welcome.

• If additionally $\mu_n(\mathbb{R}) = \mu(\mathbb{R})$ for all sufficiently large $n$, then $F_n(x) \rightarrow F(x)$ at any continuity point of $F$, and $\limsup_{n\rightarrow\infty} \mu_n(A) \leq \mu(A)$ for all closed sets $A$. A good reference is Billingsley's Convergence of Probability Measures, which addresses the type of question you're asking about. – snar Jul 11 '14 at 0:57
• @Snarski Even better: the assumption $\mu_n(\mathbb R)=\mu(\mathbb R)$ for a large $n$ and beyond is not even needed. This is because a somewhat similar argument shows that $\liminf_{n\to\infty} F_n(a)\geq F(a-\varepsilon)$. (In fact, this argument is simpler and does not even rely on the extra assumption (1)). This shows that if $F$ is continuous at $a$, then $\lim_{n\to\infty} F_n(a)=F(a)$. – triple_sec Jul 11 '14 at 1:01
• Neat, thanks. I prefer to be careful as pathological things can happen. For instance, if $\mu_n \Rightarrow \mu$ on $\mathbb{R}$, $X$ is a measurable subset of $\mathbb{R}$ such that $\mu(X) = \mu(\mathbb{R})$, and $f : X \rightarrow \mathbb{R}$ is continuous on $X$, then one expects $\mu_n f^{-1}\Rightarrow \mu f^{-1}$. This is false unless $\mu_n(X) \equiv \mu(X)$ for all $n$ (or except for finitely many $n$). – snar Jul 11 '14 at 4:20
• You will definitely need (1) if you only test against compactly supported functions. Otherwise, take $\mu_n=\delta_{-n}$, which converges to $\mu=0$, but $F_n(a)=1$, $F(a)=0$. – user138530 Jul 11 '14 at 4:23
• @ChristianRemling That's exactly the counterexample brought up on the textbook errata sheet. Assuming (1), though, can prevent the mass from “escaping to infinity.” Intuitively, if (1) is true, then, by regularity, there will exist a compact set on which vague convergence makes the $\{\mu_n\}_{n=1}^{\infty}$ “uniformly tight.” This ensures that only a tiny fraction of the mass can escape from this compact set, and hence the desired approximation argument using compactly supported functions can then be worked out. – triple_sec Jul 11 '14 at 6:04

This turned out to be too long to comment so I'm writing it as an answer even though it is a different question. In this case, it is not the escape of mass but the "placement" of mass. Usually we try to simplify weak convergence results by restricting to a nice subset. To do this we really require $\mu_n(X) = 1$ for all sufficiently large $n$.
To see this, take $X = \mathbb{Q} \cap [0,1]$ and define $\mu_n$ on $[0,1]$ such that $\mu_n(X) = 0$, $\mu_n([0,1]) = 1$ and take $\mu(X) = \mu([0,1]) = 1$.
(For instance, one can take $\{r_i\}_{i \geq 1}$ to be an enumeration of the rationals, set $\mu = \sum_{i=1}^\infty \delta_{r_i}2^{-i}$, $\mu_n = \sum_{i=1}^\infty \delta_{r_i + \pi/n}2^{-i}$.)
Notice that for any $f \in C_b([0,1])$, $$\delta_{r_i + \pi/n} f = f(r_i + \pi/n) \rightarrow f(r_i) = \delta_{r_i}f,$$ and since we can bound the tail of $\mu$ and $\mu_n$, $\mu_n \Rightarrow \mu$. From the mapping theorem, one expects $\mu_n h^{-1} \Rightarrow \mu h^{-1}$ for $h$ such that $\mu(D_h) = 0$, where $D_h$ is the set of discontinuities of $h$.
$$h(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & x \in [0,1] \setminus\mathbb{Q}\end{cases}.$$ Notice that $h|_X$ is continuous in the sense that for any $x_n \rightarrow x$ for $x_n, x \in X$, we have $h(x_n) \rightarrow h(x)$. However, $$1 = \mu_n([0,1]\setminus \mathbb{Q}) = \mu_n h^{-1}(\{1\}) \geq \mu h^{-1}(\{1\}) = 0.$$ Since $\{1\}$ is a closed set, we do not have $$\limsup_{n\rightarrow\infty} \mu_n h^{-1}(F) \leq \mu h^{-1}(F)$$ for all closed $F$, so $\mu_n h^{-1} \not\Rightarrow \mu h^{-1}.$