# Showing the lebesgue measure is sigma finite on sets

I want to show that the lebesgue measure is $\sigma$-finite on the following;

• $C_\mathrm{open} = \{ A \subset \mathbb{R} : A\,\,\mathrm{open}\}$

• $C_\mathrm{closed} = \{ B \subset\mathbb{R}: B\,\,\mathrm{closed}\}$.

Usually, for example for $C = \{(a,b): -\infty\leq a\leq b\leq \infty \}$ I took the interval $(-i,i)$ and used infinite unions etc to write it as half open so that I could take the lebesgue measure of it, and then saw that the lebesgue measure was $2i$ which is finite as $i<\infty$ , but in this case I dont know how to take elements of these sets?

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What is your definition of $\sigma$-finiteness? Usually, it is defined as a property of a measure space, not a collection of sets. – Michael Greinecker May 20 '12 at 19:31
Do you want to write $\Bbb R$ as countable union of open/closed sets of finite measure? – leo May 20 '12 at 19:35
By the way, welcome to math.SE. Please consider read the FAQ to pick some LaTeX tips. Learn LaTeX is always a good thing. – leo May 20 '12 at 19:37

Well $(-i,i)$ is open, so what you did is right: $\mathbb R = \bigcup_{i \in \mathbb N} (-i,i)$.
Similarly, $\mathbb R = \bigcup_{i \in \mathbb Z} [i, i+ 1]$.
@Rosie Not every set in $C_{open}$ is of the form $(a,b)$. For example, $(a,b) \cup (c,d)$ is an open set not of this form. Similarly for $C_{closed}$. – Rudy the Reindeer May 20 '12 at 20:15
You have $C \subset C_{open}$ but not $C \supset C_{open}$. – Rudy the Reindeer May 20 '12 at 20:16