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Let $S$ be a metric space with be Borel $\sigma$-algebra $\Sigma$. Let $\boldsymbol{P}(S)$ be the set of probability measures on $(S,\Sigma)$. According to wikipedia, "the weak topology is generated by the following basis of open sets:

$$\left\{ U_{\phi, x, \delta} \,\left|\, \begin{array}{c} \phi \colon S \to \mathbb{R} \text{ is bounded and continuous,} \\ x \in \mathbb{R} \text{ and } \delta > 0 \end{array} \right. \right\}$$

where

$$U_{\phi, x, \delta} := \left\{ \mu \in \boldsymbol{P}(S) \,:\, \left| \int_{S} \phi \, \mathrm{d} \mu - x \right| < \delta \right\}."$$

I want to verify that this is in fact a basis. It is easy to see that any $\nu\in\boldsymbol{P}(S)$ is contained in some basis element (for any $\delta$ and $\phi$, take $x=\int \phi d\nu$). The next property is that

  • if $\nu$ belongs to the intersection of two basis elements $B_{1}$ and $B_{2}$, then there is a basis element $B_{3}\ni\nu$ such that $B_{3}\subseteq B_{1}\cap B_{2}$.

In other words, given that

(1) $\displaystyle\left| \int_{S} \phi_1 \, \mathrm{d} \nu - x_1 \right| < \delta_1$; and

(2) $\displaystyle\left| \int_{S} \phi_2 \, \mathrm{d} \nu - x_2 \right| < \delta_2$,

I need to find $\delta^\ast, x^\ast\in\mathbb{R}$ and a continuous, bounded, real-valued function $\phi^\ast$ such that

$\displaystyle\left| \int_{S} \phi^\ast \, \mathrm{d} \nu - x^\ast \right| < \delta^\ast$ implies (1) and (2).

I would appreciate any suggestions on how to proceed. Thanks!

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1 Answer 1

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Put $\varphi_1:=\phi_1-x_1$ and $\varphi_2:=\phi_2-x_2$ and put $\varphi^*:=|\varphi_1|+|\varphi_2|=|\phi_1-x_1|+|\phi_2-x_2|$, which is continuous and bounded since so are $\phi_1$ and $\phi_2$. Take $\delta^*:=\min(\delta_1,\delta_2)$ and $x^*=0$. We get that if $\displaystyle\left|\int_S\varphi^*\mathrm d\nu-x^*\right|\leq\delta^*$ then $$\left|\int_S\phi_1\mathrm d\nu-x_1\right| =\left|\int_S(\phi_1-x_1)\mathrm d\nu\right|\leq\int_S|\phi_1-x_1|\mathrm d\nu\leq\int_S \varphi^*\mathrm d\nu\leq\delta^*\leq\delta_1$$ and $$\left|\int_S\phi_2\mathrm d\nu-x_2\right| =\left|\int_S(\phi_2-x_2)\mathrm d\nu\right|\leq\int_S|\phi_2-x_2|\mathrm d\nu\leq\int_S \varphi^*\mathrm d\nu\leq\delta^*\leq\delta_2,$$ which is the expected result.

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