To prove that the Lebesgue outer measure $m^*(I)$ of a closed interval $I$ is less or equal to its length $l(I)$ is easy enough. The converse however starts with a step that seems obvious but that I can not seem to prove:
For an arbitrary $\epsilon>0$, and closed cover $\{I_n\}$ of $I$, it is said (Measure integral and probability, P.22) that $$\sum_{n=1}^{\infty}l(I_n)\leq m^*(I) + \frac{\epsilon}{2}$$
What I know from the outer measure being a infimum is that $$m^*(I)\leq \sum_{n=1}^{\infty}l(I_n)$$ so I could write $$m^*(I)\leq \sum_{n=1}^{\infty}l(I_n)\leq m^*(I)+2[\sum_{n=1}^{\infty}l(I_n)-m^*(I)]$$
$$\sum_{n=1}^{\infty}l(I_n)\leq m^*(I)+\frac{\epsilon}{2}$$ with $\epsilon :=4[\sum_{n=1}^{\infty}l(I_n)-m^*(I)]$, but that does not sound like an arbitrary $\epsilon$ to me…
Any hints ? thanks.
Following hints to exploit the infinum definition for $m^*$, I come to the following argument.
Argue the opposite, namely that
$$ m^*(I)+\frac{\epsilon}{2}<\sum_{n=1}^{\infty}l(I_n)
\tag{1}$$ so $m^*(I)+\frac{\epsilon}{2}$ is a lower bound for $\sum_{n=1}^{\infty}l(I_n)$, also $m^*(I)+\frac{\epsilon}{2} > m^*(I)$ another lower bound for the sum. So it remains to find a cover of $I$ with length $m^*(I)+\frac{\epsilon}{2}$, and it would then be the greatest lower bound contradicting the definition of $m^*(I)$ as the g.l.b.
Let $\{I_n^*\}$ with endpoints $a_n^*,b_n^*$ be the cover that achieves $m^*$, define the cover $\{I_n^{\epsilon}\}$ $$I_1^{\epsilon}:=(a_1-\frac{\epsilon}{4},b_1+\frac{\epsilon}{4})\\I_i^{\epsilon}=I_i^*,i>1$$ then $\sum_{n=1}^{\infty}l(I_n^{\epsilon})=\sum_{n=1}^{\infty}l(I_n^*)+\frac{\epsilon}{2} = m^*(I)+\frac{\epsilon}{2}$
So we found a cover that acheives l.h.s of $(1)$, contradicting the definition of $m^*(I)$ as the g.l.b.
Is that correct ?
Best Answer
go through the definition of outer measure. that'll help you out. it's infimum over the covers that covers $I$. now as that is infimum hence there'll be a cover such that $\sum_{n=1}^{\infty}l(I_n) \leqslant m^*(I)+\frac{\epsilon}{2}$