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 : \begin{equation} \psi(A) = \inf\left\{ \varphi(B) \; : \; B \supset A \text{ , } B \text{ is a Borel set } \right\} \end{equation} 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 : \begin{equation} \ell(A) := \{ B \in \mathcal{B} \; \mid \; B \supset A \} \end{equation} We see : \begin{equation} \psi(A) = \inf_{B \in \ell(A)} \varphi(B) \end{equation} First prove : \begin{equation} \psi(\emptyset) = 0 \end{equation} We see : \begin{equation} \ell(\emptyset) = \mathcal{B} \end{equation} and we know : \begin{equation} \emptyset \in \mathcal{B} \text{ and } \emptyset \subset B \; \forall B \in \mathcal{B} \end{equation} So : \begin{align} \psi(\emptyset) & = \inf_{B \in \ell(A)} \varphi(B)\\ & = \varphi(\emptyset) = 0 \; \checkmark \end{align} Now show : \begin{equation} \psi(A) \in [0,\infty] \; \forall A \subset X \end{equation} 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 : \begin{equation} \psi(A) \in [0,\infty] \; \forall A \subset X \; \checkmark \end{equation} Now show : \begin{equation} A_{1} \subset A_{2} \subset X \Rightarrow \psi(A_{1}) \leq \psi(A_{2}) \end{equation} 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 : \begin{equation} \tag{1}\label{1} \psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \psi(A_{i}) \end{equation} where $A_{i} \subset X \; \forall i \in \mathbb{N}$.

Define : \begin{equation} W = \left\{ \{B_{i}\}_{i=1}^{\infty} \subset \mathcal{B} \; \mid \; B_{i} \in \ell(A_{i}) \; \forall i \in \mathbb{N} \right\} \end{equation} Let $\alpha \in W$ s.t. $\alpha = \{B_{i}\}_{i=1}^{\infty}$. Define : \begin{equation} f(\alpha) = \sum_{i=1}^{\infty} \varphi(B_{i}) \end{equation} and : \begin{equation} g(\alpha) = \varphi\left( \bigcup_{i=1}^{\infty} B_{i} \right) \end{equation} We know since $\mathcal{B}$ is a $\sigma$-algebra : \begin{equation} \bigcup_{i=1}^{\infty} B_{i} \in \mathcal{B} \end{equation} and by definition 4.1 (iv) : \begin{equation} \tag{2}\label{2} g(\alpha) \leq f(\alpha) \; \forall \alpha \in W \end{equation} 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 : \begin{equation} \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 \end{equation} and : \begin{equation} \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 \end{equation} So : \begin{equation} \ref{2} \text{ and } \ref{3} \Rightarrow \psi\left( \bigcup_{i=1}^{\infty} A_{i} \right) \leq f(\alpha) \; \forall \alpha \in W \end{equation} 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 : \begin{equation} \alpha \prec \beta \Leftrightarrow B_{i} \subset E_{i} \; \forall i \in \mathbb{N} \end{equation} 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) : \begin{equation} \alpha \prec \beta \Rightarrow f(\alpha) \leq f(\beta) \end{equation} We see : \begin{equation} \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}) \end{equation} So we can prove : \begin{equation} \sum_{i=1}^{\infty} \inf_{B \in \ell(A_{i})} \varphi(B) = f\left( \inf(W) \right) \end{equation} First prove : \begin{equation} \color{red}{\inf_{B \in \ell(A_{i})} \varphi(B) = \varphi\left( \inf_{B \in \ell(A_{i})} B \right)} \end{equation} (... NEED TO FINISH THIS PART ...)

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

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 : \begin{equation} \color{red}{\inf_{B \in \ell(A_{i})} \varphi(B) = \varphi\left( \inf_{B \in \ell(A_{i})} B \right)} \end{equation} Is this a true statement ? Is there a way to prove that it is true ?


1 Answer 1


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.


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