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I'm trying to prove below equivalence of weak convergence of finite Borel measures.

Let $(E, d)$ be a metric space and $\mu, \mu_1, \mu_2,\ldots$ finite Borel measures on $E$. Let $g:E \to \mathbb R$ be bounded continuous. If $\mu_i(A) \to \mu(A)$ for all Borel set $A \subseteq E$ with $\mu(\partial A) = 0$, then $\int_E g \mathrm d \mu_i \to \int_E g \mathrm d \mu$.

Could you verify if my attempt is fine?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

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1 Answer 1

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Assume $|g(x)| \le \alpha/2$ for all $x\in E$. Let $\nu := g_{\sharp} \mu$ be the induced finite Borel measure on $\mathbb R$. Then $\nu$ has at most countably many atoms. For each $m$, there are $-\alpha = t_1 < \cdots < t_m = \alpha$ such that $\nu (\{t_j\}) = 0, t_{j+1} - t_{j} < 1/m$, and $\nu([t_1, t_m]) = \mu(X)$.

Let $A_j := g^{-1}([t_j, t_{j+1}))$. Then $\overline {A_j} \subseteq g^{-1}([t_j, t_{j+1}])$ and $\mathring {A_j} \subseteq g^{-1}((t_j, t_{j+1}))$. Then $\partial A_j \subseteq g^{-1} (\{t_j, t_{j+1}\})$ and thus $\mu(\partial A_j) \le \nu(\{t_j, t_{j+1}\})=0$. This implies $\mu_i(A_j) \to \mu(A_j)$ as $i \to \infty$. Let $$ g_m := \sum_{j=1}^{m-1} t_j 1_{A_j}. $$

Notice that $|g_m (x) - g(x)| \le 1/m$ for all $x\in E$. We have \begin{align} & \left |\int_E g \mathrm d \mu_i - \int_E g \mathrm d \mu \right | \\ = &\left |\int_E (g-g_m) \mathrm d \mu_i + \int_E (g_m-g) \mathrm d \mu+ \left [ \int_E g_m \mathrm d \mu_i - \int_E g_m \mathrm d \mu \right ] \right | \\ \le & \int_E |g-g_m| \mathrm d \mu_i + \int_E |g_m-g| \mathrm d \mu+ \sum_{j=1}^{m-1} \left | \int_{A_j} g_m \mathrm d \mu_i - \int_{A_j} g_m \mathrm d \mu \right | \\ \le &\frac{\mu_i(E)}{m} + \frac{\mu(E)}{m} + \alpha\sum_{j=1}^{m-1} |\mu_i(A_j)-\mu(A_j)|. \end{align}

We first take the limit $i \to \infty$ and then $m \to \infty$. This completes the proof.

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