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 $\mathbb R$.

  • $\begingroup$ 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

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 $R$.

  • $\begingroup$ 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
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    $\begingroup$ You have stated the question and obtained the answer, is not it? $\endgroup$ – Gogi Pantsulaia Apr 7 '15 at 8:11

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