# Does $\inf_{B \in \Lambda} \varphi(B) = \varphi\left( \inf_{B \in \Lambda} \right)$ if $\varphi$ is Borel measurable and each $B \in \Lambda$ is Borel

I have been reading the book Modern Real Analysis by Ziemer and have come to an exercise I am having trouble with in the chapter on measure theory. The exercise goes :

Let $$\varphi$$ be a Caratheodory measure. For each set $$A \subset X$$ define : $$$$\psi(A) = \inf\left\{ \varphi(B) \; : \; B \supset A \text{ , } B \text{ is a Borel set } \right\}$$$$ Prove that $$\psi$$ is an outer measure on $$X$$.

Here is a definition that appears in the chapter :

Definition 4.1 :

A function $$\varphi$$ defined for every subset $$A$$ of an arbitrary set $$X$$ is called an outer measure on $$X$$ if the following conditions are satisfied :

(i.) $$\varphi(\emptyset) = 0$$

(ii.) $$0 \leq \varphi(A) \leq \infty$$ whenever $$A \subset X$$

(iii.) $$\varphi(A_{1}) \leq \varphi(A_{2})$$ whenever $$A_{1} \subset A_{2}$$

(iv.) $$\varphi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \varphi(A_{i})$$ for every countable collection of sets $$\{A_{i}\}$$ in $$X$$

Here is my solution so far :

Let $$\mathcal{B}$$ be the set of Borel sets and : $$$$\ell(A) := \{ B \in \mathcal{B} \; \mid \; B \supset A \}$$$$ We see : $$$$\psi(A) = \inf_{B \in \ell(A)} \varphi(B)$$$$ First prove : $$$$\psi(\emptyset) = 0$$$$ We see : $$$$\ell(\emptyset) = \mathcal{B}$$$$ and we know : $$$$\emptyset \in \mathcal{B} \text{ and } \emptyset \subset B \; \forall B \in \mathcal{B}$$$$ So : \begin{align} \psi(\emptyset) & = \inf_{B \in \ell(A)} \varphi(B)\\ & = \varphi(\emptyset) = 0 \; \checkmark \end{align} Now show : $$$$\psi(A) \in [0,\infty] \; \forall A \subset X$$$$ We see : \begin{align} \varphi(B) \geq 0 \; \forall B \in \mathcal{B} & \Rightarrow \inf_{B \in \mathcal{B}} \varphi(B) = 0 \\ & \Rightarrow \inf_{B \in \ell(A) \subset \mathcal{B}} \varphi(B) \geq 0 \; \forall A \subset X\\ & \Rightarrow \psi(A) \geq 0 \; \forall A \subset X \; \checkmark \end{align} I think we can just assume that $$\psi(A) \leq \infty \; \forall A \subset X$$ since $$\mathbb{R} \cup \{\infty\}$$ is the co-domain of $$\varphi$$. So : $$$$\psi(A) \in [0,\infty] \; \forall A \subset X \; \checkmark$$$$ Now show : $$$$A_{1} \subset A_{2} \subset X \Rightarrow \psi(A_{1}) \leq \psi(A_{2})$$$$ We see : \begin{align} A_{1} \subset A_{2} & \Rightarrow \ell(A_{2}) \subset \ell(A_{1})\\ & \Rightarrow \inf_{B \in \ell(A_{2})} B \geq \inf_{B \in \ell(A_{1})} B\\ & \Rightarrow \psi(A_{2}) \geq \psi(A_{1})\\ & \Rightarrow \psi(A_{1}) \leq \psi(A_{2}) \; \checkmark \end{align} Now show : $$$$\tag{1}\label{1} \psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \psi(A_{i})$$$$ where $$A_{i} \subset X \; \forall i \in \mathbb{N}$$.

Define : $$$$W = \left\{ \{B_{i}\}_{i=1}^{\infty} \subset \mathcal{B} \; \mid \; B_{i} \in \ell(A_{i}) \; \forall i \in \mathbb{N} \right\}$$$$ Let $$\alpha \in W$$ s.t. $$\alpha = \{B_{i}\}_{i=1}^{\infty}$$. Define : $$$$f(\alpha) = \sum_{i=1}^{\infty} \varphi(B_{i})$$$$ and : $$$$g(\alpha) = \varphi\left( \bigcup_{i=1}^{\infty} B_{i} \right)$$$$ We know since $$\mathcal{B}$$ is a $$\sigma$$-algebra : $$$$\bigcup_{i=1}^{\infty} B_{i} \in \mathcal{B}$$$$ and by definition 4.1 (iv) : $$$$\tag{2}\label{2} g(\alpha) \leq f(\alpha) \; \forall \alpha \in W$$$$ Also : \begin{align} B_{i} \supset A_{i} \; \forall i \in \mathbb{N} & \Rightarrow \bigcup_{i=1}^{\infty} B_{i} \supset \bigcup_{i=1}^{\infty} A_{i}\\ & \Rightarrow \bigcup_{i=1}^{\infty} B_{i} \in \ell\left( \bigcup_{i=1}^{\infty} A_{i} \right) \end{align} So : $$$$\tag{3}\label{3} \psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) = \inf_{B \in \ell\left( \bigcup_{i=1}^{\infty} A_{i} \right)} \varphi(B) \leq g(\alpha) \; \forall \alpha \in W$$$$ and : $$$$\tag{4}\label{4} \sum_{i=1}^{\infty} \psi(A_{i}) = \sum_{i=1}^{\infty} \inf_{B \in \ell(A_{i})} \varphi(B) \leq f(\alpha) \; \forall \alpha \in W$$$$ So : $$$$\ref{2} \text{ and } \ref{3} \Rightarrow \psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq f(\alpha) \; \forall \alpha \in W$$$$ Now define a partial order relation $$\prec$$ on $$W$$. Let $$\alpha,\beta \in W$$ s.t. $$\alpha = \{B_{i}\}_{i=1}^{\infty}$$ and $$\beta = \{E_{i}\}_{i=1}^{\infty}$$. We define : $$$$\alpha \prec \beta \Leftrightarrow B_{i} \subset E_{i} \; \forall i \in \mathbb{N}$$$$ We see : \begin{align} \alpha \prec \alpha \; \forall \alpha \in W \; \text{(reflexive)}\\ \alpha \prec \beta \text{ and } \beta \prec \alpha \Rightarrow \alpha = \beta \; \text{(antisymmetric)}\\ \alpha \prec \beta \text{ and } \beta \prec \gamma \Rightarrow \alpha \prec \gamma \; \text{(transitive)} \end{align} So $$\prec$$ is a valid order relation.

Clearly due to definition 4.1 (iii) : $$$$\alpha \prec \beta \Rightarrow f(\alpha) \leq f(\beta)$$$$ We see : $$$$\sum_{i=1}^{\infty} \inf_{B \in \ell(A_{i})} \varphi(B) = f\left( \inf(W) \right) \Rightarrow \psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \psi(A_{i})$$$$ So we can prove : $$$$\sum_{i=1}^{\infty} \inf_{B \in \ell(A_{i})} \varphi(B) = f\left( \inf(W) \right)$$$$ First prove : $$$$\color{red}{\inf_{B \in \ell(A_{i})} \varphi(B) = \varphi\left( \inf_{B \in \ell(A_{i})} B \right)}$$$$ (... NEED TO FINISH THIS PART ...)

So : $$$$\inf_{B \in \ell(A_{i})} \varphi(B) = \varphi\left( \inf_{B \in \ell(A_{i})} B \right) \; \checkmark$$$$ So we can prove : $$$$\tag{5}\label{5} \sum_{i=1}^{\infty} \varphi\left( \inf_{B \in \ell(A_{i})} B \right) = f\left( \inf(W) \right)$$$$ Now let $$\alpha = \inf(W)$$ with $$\alpha = \{ D_{i} \}_{i=1}^{\infty}$$. We see : $$$$\inf_{B \in \ell(A_{i})} B = D_{i} \; \forall i \in \mathbb{N}$$$$ So \ref{5} is true. This means : $$$$\psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \psi(A_{i}) \; \checkmark$$$$

I'm not sure if the answer needs to be this long, but I think that I may be missing something here. The part that I need assistance with is proving : $$$$\color{red}{\inf_{B \in \ell(A_{i})} \varphi(B) = \varphi\left( \inf_{B \in \ell(A_{i})} B \right)}$$$$ Is this a true statement ? Is there a way to prove that it is true ?

I think you're making things overly complicated. Let $$\{A_k\}$$ be a sequence of subsets of $$X$$. If $$\psi(A_k) = \infty$$ for some $$k$$ then subadditivity is automatic. Otherwise, if $$\psi(A_k) < \infty$$ for each $$k$$, then for a fixed $$\epsilon > 0$$ you can by definition find a set $$B_k \in \cal B$$ for which $$A_k \subset B_k$$ and $$\varphi(B_k) \le \psi(A_k) + \frac \epsilon{2^k}.$$ Since $$\bigcup_k A_k \subset \bigcup_k B_k$$ and $$\bigcup_k B_k \in \cal B$$, you have by definition $$\psi \left( \bigcup_k A_k \right) \le \varphi \left( \bigcup_k B_k \right).$$ Since $$\varphi$$ is a measure it is countably subadditive and you get $$\varphi \left( \bigcup_k B_k \right) \le \sum_k \varphi(B_k) \le \sum_k \left( \psi(A_k) + \frac \epsilon{2^k} \right) = \epsilon + \sum_k \psi(A_k).$$ Now let $$\epsilon \to 0^+$$ to reach the conclusion.