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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.)

inZugzwang
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    The doubly indexed collection $(E_{n,k}){n\geq 1, k\geq 1}$ is a countable cover of $\cup{n=1}^\infty A_n$. That's all you need. –  Nov 01 '15 at 21:09
  • I understand this, but I don't see how I can use the definition of $m^$ for doubly indexed collections. We have $\cup_{n\geqslant 1}{A_n} \subset \cup_{n,k\geqslant 1} E_{n,k}$, but how do I use this in the definition of $m^$? – inZugzwang Nov 01 '15 at 21:20
  • Single, double, or triple index is irrelevant. What is relevant is that it is a countable index. –  Nov 01 '15 at 21:22
  • So the definition of $m^$ generally works like this: if $A\subset \cup_{n,k,j\geqslant 1} E_{n,k,j}$ then $m^=\inf{\sum_{n,k,j}\psi(E_{n,k,j})}$, etc. Is this what you are saying? If yes, why is this so? – inZugzwang Nov 01 '15 at 21:25

2 Answers2

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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.

Mark Perlman
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  • What do you mean by "reindex" them. How can I do that? – inZugzwang Nov 01 '15 at 21:21
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    For example, let $j=1$ correspond to $n=1,k=1$. Let $j=2$ correspond to $n=2,k=1$. Let $j=2$ correspond to $n=1,k=2$. Take your bijection of choice from $\mathbb{N}\times\mathbb{N} \to \mathbb{N}$ to turn the double indices $n,k$ into a single index $j$. – Mark Perlman Nov 01 '15 at 21:45
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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.

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.

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