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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.

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

1 Answer 1

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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.

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  • $\begingroup$ Very nice answer, thanks a lot! $\endgroup$
    – Stanisla
    May 27, 2021 at 8:54

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