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
 A: 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).
A: The logic behind the existence of the cover $T_(n,j)$ is an implication of the order property of set of Real Numbers, $\mathbb{R}$.
We know that $\mathbb{R}$ is a complete ordered field, so for all $S \subseteq \mathbb{R}$, $\bigvee S$ and $\bigwedge S$ exist in $\mathbb {R}$.
Let us have a look at definition of $\bigwedge S$. By definition, $k =\bigwedge S$ if the two conditions hold:

*

*$k$ is a lower bound of $S$, and

*for any lower bound $v$ of $S$, $v \le k$ holds, or contrapositively, for every $k<k'$, there exists an $s \in S$ such that $k \le s <k'$.

So utilising the definition of infimum of $S$ and denoting $\epsilon = k'-k$, we obtain this basic result:
Theorem: If $k=\bigwedge S$, then for every $\epsilon >0$, there exists an $s \in S$ such that $s-k<\epsilon$, or equivalently, $s<k+\epsilon$.
Then applying this theorem to $$\mu^{*} (A_{j}) = \bigwedge \{ \sum_{n=1}^{\infty} \tau (T_{n,j}) : T_{n,j} \in \mathcal{T}, A\subseteq \bigcup_{n=1}^{\infty} T_{n,j} \}$$ we see that there exists a cover $\bigcup_{n=1}^{\infty} T_{n,j}$ from the collection of $A_{j}$ which satisfies $T_{n,j} \in \mathcal{T}$ and $$\sum_{n=1}^{\infty} \tau (T_{n,j}) < \mu^{*}(A_{n}) + \frac{\epsilon}{2^{n}}. $$
