1
$\begingroup$

The definition of lebesgue measure (in my textbook):

The set-function $\lambda^{n}$ on ($\mathbb{R}^{n}, \mathcal{B}(\mathbb{R}^{n})$) that assigns every half-open $[[a,b)) = [a_{1},b_{1}) \times \dots \times [a_{n},b_{n})\in\mathcal{J}$ the value: $ \lambda^{n}([[a,b))):=\prod_{j=1}^{n}(b_{j}-a_{j}) $ is called n-dimensional Lebesgue measure.

On the next page the book mentions that the Lebesque Measure is a measure on the Borel sets $\mathcal{B}(\mathbb{R})$

My question/problem:

According to the definition of a measure in the book. The measure must be a map between $\mathcal{B}(\mathbb{R})$ and $[0,\infty]$. But how can the defined $\lambda^{n}$ maps all the elements in $\mathcal{B}(\mathbb{R})$? When it only maps half-open rectangles? Can $\lambda^{n}$ map other kinds of set in $\mathcal{B}(\mathbb{R})$?

$\endgroup$
  • 4
    $\begingroup$ This is not only a definition, but also a theorem : there exists a unique measure on Borel sets which gives this value to each half-open rectangles. This is not really obvious, and is a consequence of Caratheodory's extension theorem. You should find details in textobooks on measure theory. $\endgroup$ – Ahriman Sep 9 '12 at 14:43
  • $\begingroup$ Usually, the Lebesgue measure is defined not only on Borel sets but also on all Lebesgue-measurabe sets. Note that the measure must be $\sigma$-additive. If you define it only on sets $[[a,b))$, its value on other sets will be defined uniquely. $\endgroup$ – Yury Sep 9 '12 at 14:52
  • $\begingroup$ en.wikipedia.org/wiki/Hahn-Kolmogorov_theorem $\endgroup$ – t.b. Sep 10 '12 at 10:10

Your Answer

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

Browse other questions tagged or ask your own question.