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The following is given in Folland: enter image description here

I don't understand why $u_0(E) \leq u^*(E)$. I thought that since $u^*(E)$ is the inf of $\sum_{n=1}^{\infty} u_0(A_j): Aj \in A, E \in \bigcup A_j$ that $u^*(E)<u_0(E)$ not the other way around. I don't understand why if : $u_0(E) \leq \sum_{n=1}^{\infty} u_0(A_j)$, must it be: $u_0(E) \leq$ inf{$\sum_{n=1}^{\infty} u_0(A_j)$}? I'm not exactly following. I know that this probably simple, but any help would be much appreciated.

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So the first part of the proof establishes that $$ \mu_0(E) \leq \sum_1^{\infty} \mu_0(A_j) \quad \forall A_j \in \mathcal A,\ E \subset \bigcup_1^{\infty} A_j$$ The point here is that this holds for any such $A_j$, hence the statement still holds true for the infinum across all such $A_j$. In other words $$ \mu_0(E) \leq \inf \left\{ \sum_1^{\infty} \mu_0(A_j): A_j \in \mathcal A,\ E \subset \bigcup_1^{\infty} A_j \right\} = \mu^*(E)$$ as required.

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  • $\begingroup$ so since it must hold for any such $A_j$ , it must hold for the smallest one so it must be less than the inf across all such $A_j$, is that correct? $\endgroup$
    – kemb
    Sep 18 '18 at 3:12
  • $\begingroup$ That's the intuitive idea, but doesn't work as a proof since inf may never be attained. The statement you're essentially trying to prove is that for some $S \subset \mathbb R$ if we have $x < y$ for all $y \in S$ then $x < \inf S$. This is done via contradiction as mentioned by @KarlPeter. $\endgroup$
    – bitesizebo
    Sep 18 '18 at 3:13
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That's because you have choosen an arbitrary family $A_j \in A$ with property $ E \in \bigcup A_j$. Since $u_0(E) \leq \sum_{n=1}^{\infty} u_0(A_j)$ holds for every such family this must also hold for the infimum. If you assump the contrary, so that $u_0(E) > inf \sum_{n=1}^{\infty} u_0(A_j)$, then there would exist a family $\bar{A}_j \in A$ with property $ E \in \bigcup \bar{A}_j$ with $u_0(E) > \sum_{n=1}^{\infty} u_0(\bar{A}_j)$, so that leads you to a contradiction to the statement above.

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