# Outer measure: countable subadditivity proof

We have a space $X, \mathcal{T} \subset \mathcal{P}(X)$ (system of all subsets of $X$) and a set function $\tau: \mathcal{T} \to [0,\infty)$. For a set $A \subset X$ we define the outer measure:

$$\mu^{*}(A) = \inf\{ \sum_{n=1}^{\infty} \tau(T_n): \space T_n \in \mathcal{T}, \space A \subset \bigcup_{n=1}^{\infty} T_n \}$$

We want to prove that $\mu^{*}$ has the property of subadditivity: for a sequence $\{ A_n\}_{n=1}^{\infty}$ the following is true:

$$\mu^{*}( \bigcup_{n=1}^{\infty} A_n ) \le \sum_{n=1}^{\infty} \mu^{*}(A_n)$$

$\mathbf{Proof}$: Exists $(n,j)$ such that $A_n \subset \bigcup_{j=1}^{\infty} T_{(n,j)}$ and $\sum_{j=1}^{\infty} \tau(T_{(n,j)}) \le \mu^*(A_n)+\frac{\epsilon}{2^n}$

We now choose a simple mapping $\phi: \mathbb{N} \to \mathbb{N}\times \mathbb{N}$ and $k = \phi^{-1}( n,j)$. We claim that $A \subset \bigcup_{k=1}^{\infty} T_{\phi(k)}$.

We'll choose $m \in \mathbb{N}$, then exists $p \in \mathbb{N}$ such that $\forall k \le m$: $$\phi(k)\in Q =\{(r,s) \in \mathbb{N} \times \mathbb{N}: r\le p, s \le p\}$$

Hence, \begin{align} \sum_{k=1}^m \tau(T_{\phi(k)}) &\le \sum_{n=1}^p \sum_{j=1}^p \tau(T_{(n,j)}) \\ & \le \sum_{n=1}^{\infty} \sum_{j=1}^{\infty} \tau(T_{(n,j)}) \\ & \le \sum_{n=1}^{\infty} \Bigg(\mu^*(A_n)+\frac{\epsilon}{2^n} \Bigg)\\ \end{align} Therefore $$\mu^*(A) \le \sum_{n=1}^{\infty} \Bigg(\mu^*(A_n)+\frac{\epsilon}{2^n} \Bigg)$$

$\mathbf{Q}$: I'm just a little surprised the proof is that complicated. Why couldn't we just say:

Exists $(n,j)$ such that $$A_n \subset \bigcup_{j=1}^{\infty} T_{(n,j)} \text{ and } \sum_{j=1}^{\infty} \tau(T_{(n,j)}) \le \mu^*(A_n)+\frac{\epsilon}{2^n}$$ Then,

$$A=\bigcup_{n=1}^{\infty}A_n = \bigcup_{n=1}^{\infty}\bigcup_{j=1}^{\infty} T_{(n,j)} \Rightarrow \sum_{n=1}^{\infty} \sum_{j=1}^{\infty} \tau(T_{(n,j)}) \le \sum_{n=1}^{\infty}\Bigg( \mu^*(A_n)+\frac{\epsilon}{2^n} \Bigg)$$

Q.E.D?

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We could, it'll be ok – Norbert May 30 '13 at 16:16
actually my textbook says that we can't do that because we haven't defined the double summation – shimee May 30 '13 at 16:34
so you answered your question – Norbert May 30 '13 at 17:42
yeah, but i don't know what "defining the double summation" means – shimee May 30 '13 at 20:05
nothing, the problem is in the proof that double infinite sum can be substituted by unary infinite sum. – Norbert May 30 '13 at 20:24