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I posted another question of the same proof but I found out that I do not understand either subadditivity property intuiton either its proof. Maybe I lack some knowledge on coverings that might show up on this proof, but I need so badly to understand it. $$\mu^{*}(\bigcup_{n=1}A_{n})\leq\sum_{n=1}^\infty \mu^{*}(A_n) \tag{$\forall \ \{A_n\}_{n\ge1} \subset \mathcal{P}(\Omega)$}$$

$\mu^{*}(A)$ is defined as follows:

$$\mu^{*}(A)\equiv \mathrm{inf}\left\{\sum_{n=1}^\infty \mu(A_n):\{A_n\}_{n\ge1} \subset \mathcal{C}, A \subset \bigcup_{n=1}A_{n}\right\}$$

Here $\mu$ is a measure. $\mathcal{C}\subset \mathcal{P}(\Omega)$ is a semialgebra. $\Omega$ is non-empty.

Proof: Fix $\epsilon > 0$. By definition of $\mu^*$, for each $n$ we may find a sequence $B_{n,k} \in \mathcal{C}$ such that $A_n \subset \bigcup_k B_{n,k}$ and $\sum_k \mu(B_{n,k}) \le \mu^*(A_n) + 2^{-n} \epsilon$.

Now $A \subset \bigcup_{n,k} B_{n,k}$ and $$\sum_{n} \sum_k \mu(B_{n,k}) \le \sum_n (\mu^*(A_n) + \epsilon 2^{-n}) = \epsilon + \sum_n \mu^*(A_n).$$ Hence $\mu^*(A) \le \epsilon + \sum_n \mu^*(A_n)$. Since $\epsilon$ was arbitrary we must have $\mu^*(A) \le \sum_n \mu^*(A_n)$.

I do not understand the proof, I mean the role of epsilon. It may be to my lack of knowledge about intervals. But for me subaddititvity property does not make sense. I would be grateful if someone could clarify me on this. Thanks.

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I think you need to grasp the definition of $\mu^*$ in terms of the infimum.

For each fixed $n$, the outer measure $\mu^*(A_n)$ is given in terms of an infimum. Since $\mu^*(A_n)+\frac{\epsilon}{2^n}$ is strictly greater than $\mu^*(A_n)$, the infimum property guarantees that there is some countable covering $B_{n,k}$ of $A_n$ with

\begin{align}\tag{1}\label{eq1}\mu^*(A_n)\leq \sum_k\mu(B_{n,k}) \leq \mu^*(A_n)+\frac{\epsilon}{2^n}.\end{align}

Now, $\bigcup_{n,k}B_{n,k}$ is itself a countable covering of $\bigcup_n A_n$, so that, by the infimum the definition of $\mu^*\big(\bigcup_n A_n\big)$, we have that

\begin{align}\tag{2}\label{eq2}\mu^*\left(\bigcup_n A_n\right)\leq \sum_{n,k}\mu(B_{n,k}).\end{align}

Using $\eqref{eq1}$ in $\eqref{eq2}$, we find that

\begin{align}\tag{3}\label{eq3}\mu^*\left(\bigcup_n A_n\right)\leq\sum_n \Bigg[\mu^*(A_n)+\frac{\epsilon}{2^n}\Bigg].\end{align}

And noticing that

\begin{align}\tag{*}\label{eq*}\sum_{n=1}^{\infty}\frac{\epsilon}{2^n}=\epsilon,\end{align}

we may rewrite $\eqref{eq3}$ as

\begin{align}\tag{4}\label{eq4}\mu^*\left(\bigcup_n A_n\right)\leq\epsilon+\sum_n \mu^*(A_n).\end{align}

Because $\eqref{eq4}$ holds for every $\epsilon>0$, in the limit when $\epsilon\to0$ we find the desired inequality:

\begin{align}\tag{5}\label{eq5}\mu^*\left(\bigcup_n A_n\right)\leq \sum_n \mu^*(A_n).\end{align}

$\epsilon$ here functions as a sort of 'buffer'. It guarantees, via the infimum property, that we may find the coverings in $\eqref{eq1}$, which are slightly larger than $A_n$. In fact, taking $\epsilon$ as small as we want, the coverings can be as 'tight' we wish. We need only take care to choose these buffers in such a way that they can be summed, as in $\eqref{eq*}$, while still yielding a quantity that can be made as small as needed.

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  • $\begingroup$ Thanks for the answer. But what is the intuition of having the measure of the union lower than the measure of the sum. Is there an intersection between the An s or is the inequality only due to epsilon? $\endgroup$ – Pedro Gomes Dec 25 '16 at 19:21
  • $\begingroup$ Oh, think about a Venn diagram. When you sum the measure of the $A_n$'s, you may be overcounting the measure of intersections. $\endgroup$ – Fimpellizieri Dec 25 '16 at 19:57
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Fixing $\varepsilon$ allows us to find the appropriate sequence $\{ B_{n,k} \} \subset \mathcal C$. Note that we want to find this sequence such that

$$ \sum_k \mu \left( B_{n,k} \right) $$

is very close to $\mu^* (A)$. Fixing $\varepsilon$ allows us to fix exactly how close 'very close' is. If we don't do this, then there is no way to determine how close

$$ \sum_n \sum_k \mu \left( B_{n,k} \right) $$

is to $\sum_n \mu^* (A_n)$.

The last part of the proof argues that we can take $\varepsilon$ to $0$ while preserving the inequality, as shown in the answers to your previous question.

This question might also be of interest.

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  • $\begingroup$ Thanks again for answering. How can you take epsilon and keep the inequality? Is it the sum that brings epsilon to zero? What is the intuition behind the subadditivity property? Why could I have it done the other way around making the measure of the union higher than the sum by adding epsilon? $\endgroup$ – Pedro Gomes Dec 25 '16 at 19:29
  • $\begingroup$ We can take $\varepsilon$ to $0$ while maintaining the inequality for the reason given in the answer to your previous question. If you don't tell me what exactly it is about the previous answer you don't understand, I can't really help you. See Fimpellizieri's answer for intuition on subadditivity. I'm not sure what you mean by the last question. $\endgroup$ – Theoretical Economist Dec 26 '16 at 8:45
  • $\begingroup$ But, one last thing re subadditivity -- it's just a property of outer measures. I'm not sure you can demand much in the way of intuition from it. However, given that constructing an outer measure is one step toward generalising our intuitive notion of length, area and volume, one should at least expect an outer measure to be subadditive. Think about the length of two line segments (that may potentially overlap) versus the length of the combination of the two segments. $\endgroup$ – Theoretical Economist Dec 26 '16 at 8:48
  • $\begingroup$ Sorry for not answering earlier. I have been studying your answers. I would like to ask wether the sum is higher due to intersection of the An s or due to the fact there is a covering that surpasses the interval in the sum of each outer measure? $\endgroup$ – Pedro Gomes Dec 26 '16 at 12:54
  • $\begingroup$ Sorry, I'm not clear on what you mean. Which sum is higher than what? $\endgroup$ – Theoretical Economist Dec 26 '16 at 13:22

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