# Lebesgue measures are sigma finite

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?

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 $\mu([-k,k]) = 2k < \infty$.

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.

• this is so close to being formatted correctly, why not just add the dollar signs? – Andres Mejia Sep 27 '16 at 20:08
• dollar signs? to where? and thank you, Daniel Buck to editing. I couldn't figure how to show it properly. – Th3mirx Sep 27 '16 at 21:40
• You add dollar signs! Look at the edit – Andres Mejia Sep 27 '16 at 21:42
• 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? – Th3mirx Sep 27 '16 at 21:43
• 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/… – Andres Mejia Sep 27 '16 at 21:44