# Outer measure is countably subadditive

Working on the proof that outer measure is countably subadditive in Royden

For a set $A \subset X$ we define the outer measure:

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

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

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

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

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I'm confused as to how we know that such an open cover $T_{(n,j)}$ exists, especially subject to the condition that $\sum_{j=1}^{\infty} \tau(T_{(n,j)}) \le \mu^*(A_n)+\frac{\epsilon}{2^n}$ How would such an open cover be constructed? If anyone could help me develop some intuition for this step, that would be great

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The existence of this cover follows from definition of $\mu^*$ as an infimum (since $\mu^*(A_n)<\mu^*(A_n)+\frac{\varepsilon}{2}$). Note that the set $\{\sum_{n=1}^\infty \tau(T_n): T_n\in \mathcal{T}, A\subset \bigcup_{n=1}^\infty T_n\}$ is not empty since the whole space cover $A$ and we can let $(T_n)$ be a constant sequence of sets being equal to the spacer (wich is open).

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I'm not sure that I understand this...just because the sum of the interval lengths in the cover is guaranteed to meet the specified bound, why would such a cover exist? – johyu May 27 '14 at 3:18
Could you clarify? – johyu May 27 '14 at 3:49
If there's no such cover, then every cover is such that $\sum \tau \left( T_n \right) > \mu^*\left( A \right) + \epsilon / 2^n$, which violates the definition of $\mu^*\left( A \right)$ as the infimum of such sums. – Callus May 27 '14 at 3:57