For a subset $E \subset \mathbb{R}^d$ define the outer measure
$$m_*(E)=\inf \{\Sigma \vert I_j \vert\}$$
Where the infimum is taken over all countable coverings of $E$ by closed intervals. So $E \subset \bigcup_{j=1}^\infty I_j$.
Show $m_*([0,1])=1.$
How do I go about this rigorously? What would my $I_j$ be?
How about for general $[a,b]$ how can I show its just $b-a$?
Best Answer
Assuming that $d = 1$ to keep things simple, we have by definition $|[0,1]| = 1$. (More generally, $|[a,b]| = b - a$, and for an elementary set which is a finite union of disjoint intervals we sum up their lengths.) This is the pre-measure that leads to the Lebesgue measure. Note that $| \cdot |$ can be shown to be countably subadditive for elementary sets.
Since $[0, 1]$ is itself a closed interval, we can take $I_1 = [0,1]$, $I_j = \emptyset, \; j \geq 2$ as a particular covering. Then the outer measure satisfies $m^*([0,1]) \leq |[0,1]| = 1$ (as it is an infimum).
Now for the other direction. Consider any countable covering of $[0,1]$ by closed intervals $(I_j)_{j \geq 1}$ (i.e. elementary sets with $[0,1] \subseteq \bigcup_{j \geq 1} I_j$). By countable subadditivity, $\sum_{j \geq 1} |I_j| \geq |\bigcup_{j \geq 1} I_j|$. By monotonicity, $|\bigcup_{j \geq 1} I_j| \geq |[0,1]| = 1$. Putting this together shows that $m^*([0,1]) \geq 1$ (again, because $m^*$ is an infimum).
Hence $m^*([0,1]) = 1$. For general $[a,b]$, which is an interval/elementary set, the proof is essentially identical. (More generally, this says that the Lebesgue outer measure $m^*$ agrees with the Lebesgue measure $m$ on elementary sets, which we call $|\cdot|$ here).