Theorem: Let A, B $\in \mathcal{P}(\mathbb{R})$. If $A \subseteq B$ then $m^*(A) \leq m^*(B)$ with $m^*$ being the outer measure of Lebesgue. The question is about the conclusion of the proof. First they say: Let $A,B \in \mathcal{P}(\mathbb{R})$ be such that $A \subseteq B$. If $\{I_n = (a_n,b_n)\}_{n=1}^{\infty}$ is such that $B\subseteq \bigcup_{n=1}^{\infty}I_n$ then $A\subseteq \bigcup_{n=1}^{\infty} I_n$. So far, so good, nothing to ask there. The conclusion of that is where I have the doubt:
$$
\left\{\sum_{n=1}^{\infty}l(I_n): B\subseteq \bigcup_{n=1}^{\infty} \{I_n = (a_n, b_n)\}_{n=1}^{\infty} \right\} \subseteq \left\{\sum_{n=1}^{\infty}l(I_n): A\subseteq \bigcup_{n=1}^{\infty} \{I_n = (a_n, b_n)\}_{n=1}^{\infty} \right\}
$$
(From there one can easily conclude the proof because of properties of the infimum). What I don't understand is the centered part. Intuitively if the $I_n$ covers B, the sum of the lengths should be greater than the sum lengths of the covers of A and in such case, the $\subseteq$ symbol should go the other way around. Why that isn't true then?
Best Answer
I think you are confusing $\leq$ with $\subseteq$.
What the proof demonstrated is that if $x \in \left\{\sum_{n=1}^{\infty}l(I_n): B\subseteq \bigcup_{n=1}^{\infty} \{I_n = (a_n, b_n)\}_{n=1}^{\infty} \right\} =: S_B$, then $x \in \left\{\sum_{n=1}^{\infty}l(I_n): A\subseteq \bigcup_{n=1}^{\infty} \{I_n = (a_n, b_n)\}_{n=1}^{\infty} \right\} =: S_A$.
Then since each of these sets actually has the form $[r,\infty]$ for some $r \in \overline{\mathbb R}$, we have $S_B = [b,\infty] \subseteq [a, \infty] = S_A$, as a result, the smallest element of $S_A$ is actually smaller than the smallest element of $S_B$. Thus the relation $\subseteq$ actually implies the elements of $S_B$ are $\geq$ those of $S_A$ (this language is imprecise, the precise meaning is as I stated above).