0
$\begingroup$

Let $\mathcal {B}$ be the $\sigma$-algebra generated by the set of open subsets of $\mathbb{R}$. A Borel measure $\nu$ on $\mathbb{R}$ is a measure on $\mathcal {B}$ such that $\nu(K) \lt \infty$ for every compact subset $K$. The only Borel measures I know are essentially as follows.

1) Let $f$ be a non-negative $\mathcal {B}$-measurable function such that $\int_K f d\mu \lt \infty$ for every compact subset $K$ where $\mu$ is the Lebesgue measure. We write $\nu(M) = \int_M f d\mu$ for $M\in \mathcal B$. Then $\nu$ is a Borel measure.

2) Let $E$ be a countable subset of $\mathbb R$. Let $f: E \rightarrow \mathbb R$ be a non-negative function such that $\sum_{x\in K\cap E} f(x) \lt \infty$ for every compact subset $K$. We write $\nu(M) = \sum_{x \in M\cap E} f(x)$ for $M\in \mathcal B$. Then $\nu$ is a Borel measure.

I would like to know other non-trivial examples of Borel measures on $\mathbb R$.

$\endgroup$
  • $\begingroup$ The Dirac delta measure? $\endgroup$ – Prahlad Vaidyanathan Apr 2 '15 at 4:47
  • $\begingroup$ There are also singular measures, together with 2 types you named, that's essentially all there could be. Lebesgue's decomposition theorem. $\endgroup$ – Jorkug Apr 2 '15 at 6:19
  • $\begingroup$ @PrahladVaidyanathan The Dirac delta measure is of type 2. $\endgroup$ – Makoto Kato Apr 2 '15 at 23:31
  • $\begingroup$ @Jorkug A singular measure is not necessarily of type 2. $\endgroup$ – Makoto Kato Apr 2 '15 at 23:31
  • $\begingroup$ never said it was. $\endgroup$ – Jorkug Apr 3 '15 at 5:31
0
$\begingroup$

Let $C$ be a cantor set. Let $f:C \to [0,1]$ be a Borel isomorphism. We put $\lambda(X)=\mu(f(X \cap C))$ for each $ X \in B(R)$. Then $\lambda$ is other non-trivial example(singular, following William Curtis remark) of Borel measure on $R$.

$\endgroup$
  • $\begingroup$ What is the Borel isomorphism? $\endgroup$ – Makoto Kato Apr 4 '15 at 1:20
  • $\begingroup$ Let $X$ and $Y$ be some Borel subsets of Polish topological space. A bijective mapping $f: X \to Y$ is called Borel isomorphism if both $f$ and $f^{-1}$ are Borel mappings. $\endgroup$ – Gogi Pantsulaia Apr 4 '15 at 4:07
  • $\begingroup$ Would you explain why there exists a Borel isomorphism $f:C \to [0, 1]$? $\endgroup$ – Makoto Kato Apr 5 '15 at 1:15
  • $\begingroup$ see prac.im.pwr.edu.pl/~cichon/Materialy/BOOK.pdf, Theorem 3.14, p. 61 $\endgroup$ – Gogi Pantsulaia Apr 5 '15 at 5:13
  • 1
    $\begingroup$ You have stated the question and obtained the answer, is not it? $\endgroup$ – Gogi Pantsulaia Apr 7 '15 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.