I'm curious if this is a valid argument. It makes sense to me, but I doubt my abilities with induction etc. Also, I'm not sure if I am making a jump from the definition of a measurable set, or if it is a sound progression.
The set-up Given a countable set of disjoint open intervals, $I_k$, prove the outer measure of their union equals the sum of their lengths, ie: $\sum_{k=1}^{\infty} l(I_k) = |\cup_{k=1}^{\infty} I_k|$
Thanks
We know that the outer measure of an open interval is just its length, that is $$|(a,b)|=(b-a)$$ We know that a set $E$ is measurable if for each set A, $$|A|=|A\cap E|+|A\cap E^c|\hspace{.5cm}(1)$$
We can use this to show equality with an induction argument.
For $k=1,|(I_1)|=l(I_1)$ as our open interval $I_1$ is measurable. Assume that $$|\cup_{k=1}^{N-1} I_k|=\sum_{k=1}^{N-1} l(I_k)$$
Now take (1) with $A=I_1\cup \dots \cup I_N$ and $E=I_N$ Since $I_N$ is an interval, it is measurable and $$ |\cup_{k=1}^{N} I_k|= |(\cup_{k=1}^{N} I_k)\cap I_N| + (|\cup_{k=1}^{N} I_k\cap I_N^c|=|I_N|+|\cup_{k=1}^{N-1} I_k|$$
Now $|I_N|+|\cup_{k=1}^{N-1} I_k|$ is the sum we are looking for and as $$|\cup_{k=1}^{N} I_k|\subset |\cup_{k=1}^{\infty} I_k|$$ we have that $$|\cup_{k=1}^{N} I_k|\le |\cup_{k=1}^{\infty} I_k|$$ for any positive $N$ by Axler 2.5 (outer measure preserves order), then by our hypothesis$$|\cup_{k=1}^{\infty} I_k|\ge |\cup_{k=1}^{N} I_k|\ge |\sum_{k=1}^{N} l(I_k)|$$ for any positive $N$. Take $N \rightarrow \infty$ to get $$|\cup_{k=1}^{\infty} I_k|\ge |\sum_{k=1}^{\infty} l(I_k)|$$ Now the other direction comes from the countable subadditivty of outer measure and we have $$|\cup_{k=1}^{\infty} I_k|= |\sum_{k=1}^{\infty} l(I_k)|$$ as desired