3
$\begingroup$

How can I show a Lebesgue measure on $\mathbb{R}$ is $\sigma$-finite? I know that a measure $\mu$ on $(\mathbb{R},\mathfrak{B}(\mathbb{R}))$is a Lebesgue measure on $R$ if $\mu (A)$ is the length of A for every interval of A. But how do I show such a measure is $\sigma$ finite?

$\endgroup$
4
$\begingroup$

Remember that a space is $\sigma$-finite if it is the countable union of $\mu$-finite subsets. We can write $\mathbb R$ as a countable union of increasing intervals $$ \mathbb R = \bigcup_{k=1}^\infty [-k, k]. $$ And we know $\mu([-k,k]) = 2k < \infty$.

$\endgroup$
1
$\begingroup$

Let's say $(\Omega,\operatorname{Borel})$ and $\mu ((a,b])=b-a$

$\Omega=\Bbb{R}=\bigcup_{i=0}^\infty A_i,\quad\mu(A_i)<\infty$

$A_i=(-i,-i+1] \bigcup (i-1,1],\quad\mu(A_i)<2 \infty$

$A_1=(-1,0] \bigcup (0,1]= (-1,1]$

$A_1 \bigcup A_2 =(-2,2]$ goes to $\bigcup_{i=0}^k A_i= (-k,k]\to \Bbb{R}$

$\lim_{k\to \infty} \bigcup_{i=0}^k A_i =\Bbb{R}$ then we say that Lebesgue measure is also $\sigma$-finite.

$\endgroup$
  • 3
    $\begingroup$ this is so close to being formatted correctly, why not just add the dollar signs? $\endgroup$ – Andres Mejia Sep 27 '16 at 20:08
  • $\begingroup$ dollar signs? to where? and thank you, Daniel Buck to editing. I couldn't figure how to show it properly. $\endgroup$ – Th3mirx Sep 27 '16 at 21:40
  • $\begingroup$ You add dollar signs! Look at the edit $\endgroup$ – Andres Mejia Sep 27 '16 at 21:42
  • $\begingroup$ I just saw that, I didn't know it before! so I should start with dollar sign to formula and close with dollar sign again? $\endgroup$ – Th3mirx Sep 27 '16 at 21:43
  • $\begingroup$ Yes, if you do that, your answer will format. Welcome to SE by the way! Here is a basic guide: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Andres Mejia Sep 27 '16 at 21:44

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.