Construction of Outer Measures

Question

I am trying to understand the following result. I know that there are a lot of questions about this here, but none addresses the point I am stuck on.

Let $\mathcal E\subset\mathcal P(X)$ and take any function $\psi\colon \mathcal E\to[0,\infty]$. If we define $$m^*(A)=\inf\left\{\sum_{n=1}^\infty \psi(E_n)~\middle|~ E_n\in\mathcal E\text{ and } A\subset \bigcup_{n=1}^\infty E_n\right\}$$ for any $A\subset X$, then $m^*$ is an outer measure on $X$.

Now the proof for properties (1) and (2) of outer measures is straightforward. For (3), my professor wrote down the following.

Take any sequence of subsets $A_n\subset X$ and fix $\varepsilon>0$. For each $n$, there are $E_{n,k}\in \mathcal E$ such that $A_n\subset\bigcup_{k=1}^\infty E_{n,k}$ and $\sum_{k=1}^\infty \psi\left(E_{n,k}\right)\leqslant m^*(A_n)+{\varepsilon}/{2^n}$. Now, $\bigcup_{n=1}^\infty A_n\subset\bigcup_{n=1}^\infty\bigcup_{k=1}^\infty E_{n,k}$ and so $$m^*\left(\bigcup_{n=1}^\infty A_n\right)\color{RED}{\leqslant} \sum_{n=1}^\infty\sum_{k=1}^\infty\psi(E_{n,k})\leqslant \sum_{n=1}^\infty m^*(A_n) + \varepsilon.$$The result follows.

Now, what I fail to understand is how he got the first (red) inequality. I know that, since $\bigcup_{n=1}^\infty A_n\subset\bigcup_{n=1}^\infty\bigcup_{k=1}^\infty E_{n,k}$, the definition of $m^*$ implies that $m^*\left(\bigcup_{n=1}^\infty A_n\right)\leqslant\sum_{n=1}^\infty \psi\left(\bigcup_{k=1}^\infty E_{n,k}\right)$ but I don't understand why this is less than the double sum, (or indeed less than $\sum_{n,k=1}^\infty \psi(E_{n,k})$).

Any help/explanation will be appreciated.

(I've added the Lebesgue Measure tag because the proof of the similar result for that measure has an identical step.)

Answer 1

An alternative definition of outer measure does not require that the collection $\Gamma$ is indexed by $\mathbb{N}$. We could just as well define $m^*$ as $$m^*(A)=\inf\left\{\sum_{E\in \Gamma} \psi(E)~\middle|~\Gamma\subseteq {\cal E} \text{ and } A\subset \bigcup_{E\in\Gamma} E\right\}.$$ The collection $\Gamma$ can be indexed by $\mathbb{N}$ or $\mathbb{N}\times \mathbb{N}$ or any other way. The indexing is irrelevant for both the sum and for the union, and so really plays no role at all.

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Whether you take Mark's approach or mine, there is still the non-trivial task of showing that the sum over a doubly indexed set is equal to the iterated sum. See Fubini's Theorem for Infinite series for more information on this.

Answer 2

This seems like notational trouble. Once you've chosen your $E_{n,k}$ you could reindex them under the single countable index $\{E_{j}\}_{j=1}^{\infty}$, so that $\bigcup_{n=1}^{\infty}A_{n} \subset \bigcup_{j=1}^{\infty} E_{j}$ and the inequality follows from the definition of outer measure.