Littlewood's First Principle with disjoint intervals

Question

Littlewood's First Principle states that:

If $E$ is a Lebesgue measurable set with $m(E)<\infty$, then for all $\epsilon>0$ there exists a finite collection of open intervals $\{I_1,I_2,\ldots,I_N\}$ such that $m(E \Delta \bigcup_{n=1}^N I_n)<\epsilon$.

My question is:

Can we take the intervals $I_n$ to be pairwise disjoint? And in that case, what do we need to change in the proof?

Answer 1

If $\epsilon>0$ ,there exists a covering with open intervals $\{I_n\}_{n \in \Bbb{N}}$ of $E$ such that $m(E) \leq m( \bigcup_nI_n)\leq \sum_{n=1}^{\infty}m(I_n)< m(E)+\epsilon$

The set $ \bigcup_nI_n$ is open so it can be written as a disjoint union of open intervals.

So $\bigcup_nI_n=\bigcup_nJ_n$ where $J_n$ are disjoint open intervals.

So we have that $$m(E)+ \epsilon > m(\bigcup_nI_n)=m(\bigcup_nJ_n)=\sum_nm(J_n)$$

Note that $\{J_n\}_{n \in \Bbb{N}}$ also is a covering of $E$.

Then you can proceed as in the rest of the proof.