# Weak convergence of Borel probability measure

Let $$f:[a, b] \to [0, \infty)$$ be a measurable function such that $$\int_a^b f(x) dx = 1$$.

Let a sequence of Borel probability measures on $$\mathbb R$$ be defined as $$\mu_n := \sum_{i=1}^{n} a_i \delta_{x_i}$$, where $$x_i = a + \frac{i}{n} (b-a), \ a_i = \int_{x_i-1}^{x_i} f(x) dx$$ and $$\delta_{x_i}$$ is the probability measure with all the mass in $$x_i$$.

Why does $$\mu_n \to \mu$$ converge weakly, where $$\mu$$ is the probability measure definded by $$\mu(A) = \int_{A \cap [a, b]} f(x) d(x)$$.

My attempt is as follows:

First, we test against the constant function $$1$$, explicitly

$$1 \equiv \int_{A \cap [a, b]} 1\,\mathrm{d}\mu_n \to \int_{A \cap [a, b]} 1\,\mathrm{d}\mu =\mu(A \cap [a, b])$$

It remains to check that $$\mu$$ is a positive measure. By Hahn's decomposition theorem, write $$X=A \cap [a, b]$$ which is a disjoint union. This means $$\mu=\mu_+-\mu_-$$ where $$\mu_+,\mu_-$$ are the positive measures

$$\mu_\pm(A) := \mu(A\cap X_\pm)$$

Then, for any compact $$K\subset X_-$$ and $$N\in\mathbb{N}_+$$, consider the non-negative, bounded continuous function

$$f_{K,N} := \max\{1-N\,\mathrm{dist}(x,K),0\}$$

By weak convergence,

$$\int_Xf_{K,N}\,\mathrm{d}\mu = \lim_{n\to\infty}\int_Xf_{K,N}\,\mathrm{d}\mu_n\ge0$$

So

$$\int_Xf_{K,N}\,\mathrm{d}\mu_+ - \int_Xf_{K,N}\,\mathrm{d}\mu_-\ge0$$

Now take $$N\to\infty$$ and apply dominated convergence theorem,

$$-\mu_-(K) = \mu_+(K) - \mu_-(K) \ge0$$

Since $$K$$ is an arbitrary subset of $$X_-$$, by regularity of finite Borel measures on Polish spaces we see that $$\mu_-$$ is the zero measure. Hence $$\mu=\mu_+$$ is a positive measure.

Is this possible and correct? Any advice, hint or correction is appreciated.

• $\mu$ is positive because $f \geq 0$. Did you prove that $\mu_n \to \mu$ weakly? May 27, 2021 at 5:03

Notice that the propposed limit distribution, $$\mu(dx)=\mathbb{1}_{[a,b]}(x)f(x)\,dx$$ is a positive measure since $$f\geq0$$, so $$\mu=\mu_+$$ ($$\mu_-\equiv0$$).
It is enough to check the behavior of $$\mu_n$$ acting on continuous functions over $$[a,b]$$ (this is because the $$\mu_n$$'s and the propose weak limit measure $$\mu(dx)=f(x)\,dx$$ are supported on $$[a,b]$$).
Let $$g\in\mathcal{C}[a,b]$$. Given $$\varepsilon$$, there is $$\delta>0$$ such that $$|x-y|<\delta$$ implies $$|g(x)-g(y)|<\varepsilon$$.
$$|\mu_n(g)-\mu(g)|=\Big|\sum^n_{j=1}\Big(a_jg(x_j)-\int^{x_j}_{x_{j-1}}f(t)g(t)\,dt\Big)\Big|$$ For all $$n$$ large enough (so that $$(b-a)/n <\delta$$), we have $$\big|a_jg(x_j)-\int^{x_j}_{x_{j-1}}f(t)g(t)\,dt\big|\leq \int^{x_j}_{x_{j-1}} f(t)|g(x_j)-g(t)|\,dt\leq \varepsilon\int^{x_j}_{x_{j-1}}f(t)\,dt$$ Thus $$|\mu_n(g)-\mu(g)|\leq \varepsilon\int^b_af(t)\,dt=\varepsilon$$
This shows that $$\mu_n(g)\xrightarrow{n\rightarrow\infty}\mu(g)$$ for all $$g\in\mathcal{C}[a,b]$$, which is precisely what weak convergence means.