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
Best Answer
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).