# Lebesgue measure of an open set is the sum of lengths of components

If $$O\subset\mathbb{R}$$ is an open set with unique disjoint decomposition in intervals $$O=\bigcup I_i$$, where $$l(a,b)=b-a$$. Define $$\tau(O)=\sum l(I_i)$$. And define the Lebesgue outer measure $$m$$ of a set $$E\subset \mathbb{R}$$ as $$m(E)=\inf_{\displaystyle E\subset \bigcup J_k} \sum l(J_k)$$ where $$J_k$$ must be open intervals (or open sets) (I proved that open sets not change this definition replacing $$l$$ by $$\tau$$).

Show that $$m(O)=\tau(O)$$ if $$O$$ is open.

Attempt:

Since $$O\subset O=\bigcup I_k$$ we have that $$m(O)\leq \sum l(I_i)=\tau(O)$$.

I tried $$3$$ hours to prove that $$m(O)\geq\tau(O)$$. I just proved that if $$O\subset \bigcup J_k$$ then exists a collection $$\{R_k\}$$ of open intervals such that $$\bigcup J_k=\bigcup R_j$$ where $$R_j$$ are disjoint intervals. And I proved that $$\tau(O)\leq\sum l(R_j)$$ which is inmediate by the fact that each $$I_i$$ must be a subset of one $$R_j$$.

But I can't prove that $$\sum l(R_j)\leq \sum l(J_k)$$. I spend a lot of time trying to prove it, it's not as trivial as it seems.

Assuming $$\sum l(R_j)\leq \sum l(J_k)$$ I'm done since this implies that $$\tau(O)\leq \sum l(J_k)$$ and then $$\tau(O)\leq m(O)$$.

I also proved that for open and closed intervals $$m(I)=l(I)$$. But this fact was not useful for me.

How did you get $R_j$'s? Just follow what happens to the $l$ measures along the procedure..
$R_1:=J_1$. Then check whether $J_2$ overlaps any of the so far defined $R_j$'s or not, now it's with $R_1$ only. If overlaps, redefine $R_1:=J_1\cup J_2$, then $l(R_1) < l(J_1)+l(J_2)$, if not, start an $R_2:=J_2$, then $l(R_1)+l(R_2)=l(J_1)+l(J_2)$. And so on...
• I didn't build the intervals $R_j$ from the old $J_k$ in recursive form, I used that $\cup J_k$ is open then there exist a representation of disjoint intervals, and I was trying to prove that $\sum l(R_k)\leq \sum l(J_k)$ which is hard without your constructive approach. – Gaston Burrull Apr 1 '13 at 19:07