Questions tagged [borel-measures]

Use this tag for questions related to Borel measures, which, on a topological space, are measures defined on all open sets.

A Borel measure on a topological space is a measure defined on all open sets (and thus on all Borel sets).

Formally, let X be a locally-compact Hausdorff space, and let $\mathfrak B$(X) be the smallest $\sigma$-algebra that contains the open sets of X, which is called the $\sigma$-algebra of Borel sets. A Borel measure is any measure $\mu$ defined on the $\sigma$-algebra of Borel sets. Some authors require also that $\mu$ be locally compact, meaning that $\mu$(C) < $\infty$ for every compact set C. If a Borel measure $\mu$ is inner regular and outer regular, it is called a regular Borel measure (some authors also require it to be tight). If $\mu$ is inner regular and locally finite, it is called a Radon measure.

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Applications of Riesz-Markov-Kakutani Theorem.

I wanna ask, if you know some interesting (or not) applications of the theorem mentioned in the title. I mean this one when we want to represent a functional as an integral over some measure. If you have some informations, proofs, exercises, books,…
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Borel measures in S.-T. Yau College Student Mathematics Contests 2018

This problem is from S.-T. Yau College Student Mathematics Contests 2018. I have som basic knowledge of Lebesgue measures, but I have not systematically studied Borel measures. I don't have any idea about this problem. Let $\mu$ be a Borel measure…
Piquancy
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Absolutely continuous measures

We have two measures with $\mu_F(]-\infty,x])=F(x)$ $\mu_G(]-\infty,x])=G(x)$ $G(x)=F(x)^2$ Prove that $\mu_G$ is absolutely continuous with respect to $\mu_F$. So we have to prove that $\mu_F(A)=0\Rightarrow\mu_G(A)=0$. I was able to prove this…
milfor
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Prove that this function is Borel Measurable.

I have this function and I want to prove it is Borel measurable. $$\begin{equation} f:\mathbb{R}^2\rightarrow\mathbb{R}:f(x,y)=\begin{cases} \sin(\frac{1}{x-y}) &\text{ for } x>y \\\ x^2+y^2 &\text{ for } x\leq y \end{cases}\end{equation}$$ I…
wisk
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