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.


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.

Sorry for bad english.


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...

  • $\begingroup$ 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. $\endgroup$ – Gaston Burrull Apr 1 '13 at 19:07

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