# How to prove the 2nd & 3rd conditions of outer measure

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1. Feb 11, 2015

### A.Magnus

I have this question on outer measure from Richard Bass' book, supposed to be an introductory but I am lost:

Prove that $\mu^*$ is an outer measure, given a measure space $(X, \mathcal A, \mu)$ and define

$\mu^*(A) = \inf \{\mu(B) \mid A \subset B, B \in \mathcal A\}$
for all subsets $A$ of $X$.​

Here are what I have gone so far:
(1) The first condition is the easiest one:

\begin{align} \mu^*(\emptyset) &= \inf \{\mu(B) \mid \emptyset \subset B, B \in \mathcal A\}\\ &= \mu (\emptyset) \\ &= 0 \end{align}​

(2) Now the second condition. Let $D, E \in X$ and $D \subset E$,

\begin{align} \mu^*(D) &= \inf \{\mu(D') \mid D \subset D', D' \in \mathcal A\}\\ \mu^*(E) &= \inf \{\mu(E') \mid E \subset E', E' \in \mathcal A\}\\ \end{align}​

Here I need to prove $\mu^* (D) \leq \mu^*(E)$. It looks to me so intuitive especially if I draw Venn diagrams of $D, E, D'$ and $E'$, but I don't know how to say it in math-speak. I would appreciate helps on this 2nd. condition.

(3) And this 3rd. condition is my major stumbling block: Given $(A_i)_{i \in \mathbb N} \subset X$, I need to arrive at

$\mu^* (\bigcup _{i=1}^{\infty} A_i)\leq \sum_{i=1}^{\infty} \mu^* (A_i).$
Here, I know for sure I need to state this first: $\forall A_i, \exists B_i$ such that $A_i \subset B_i, B_i \in \mathcal A$, but I don't think the next step is right:

\begin{align} \mu^*(\bigcup_{i=1}^{\infty}A_i) &= \inf \{\bigcup_{i=1}^{\infty}\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}\\ &= \ldots\\ \end{align}​

I would appreciate any help on this 3rd. condition in addition to the 2nd. above. Thank you for your time and effort.

2. Feb 12, 2015

### Dick

(2) is easy. If $C$ and $D$ are sets of numbers and $C \subset D$ then $\inf(C) \ge \inf(D)$. Right? Try and use an idea like that.

For (3), convince yourself that for the sets $A_i$, for every $\epsilon_i \gt 0$ there is a set $A'_i \in \mathcal A$ such that $A_i \subset A'_i$ and $\mu^*(A_i) \le \mu(A'_i) \le \mu^*(A_i)+\epsilon_i$. It's just because of the $\inf$ in your definition. Now $\bigcup A'_i \in \mathcal A$ and it contains $\bigcup A_i$. Can you try and continue from here using countable additivity of the measure? The final step is to notice you can pick $\Sigma \epsilon_i$ to be as small as you want.

3. Feb 13, 2015

### A.Magnus

Thanks! After I posted this question couple of days ago, I researched other standard texts and learned that this is actually a classic question and yes, you are right, their solution generally involves using $\Sigma \varepsilon_i$ or $\Sigma \frac{\varepsilon}{2^i}$. At the same time, I managed to come up with my own solution which I believe makes sense although admittedly it is un-elegance, clumsy and folksy:

(1) The firs condition:

\begin{align} \mu^*(\emptyset) &= \inf \{\mu(B) \mid \emptyset \subset B, B \in \mathcal A\}\\ &= \mu (\emptyset) \\ &= 0 \end{align}
(2) The second condition: Let $D, E \subset X$ and $D \subset E$, then

\begin{align} \mu^* (D) &= \inf \{ \mu (D') \mid D \subset D', D' \in \mathcal A\},\\ \mu^* (E) &= \inf \{ \mu (E') \mid E \subset E', E' \in \mathcal A\}.\\ \end{align}​

From $D \subset E$ we need to show that

$\inf\{\mu (D') \mid D' \in \mathcal A, D \subset D'\} \leq \inf \{\mu(E') \mid E'\in \mathcal A, E \subset E'\}.$​

Suppose that by way of contraidiction, $D \subset E$ leads to $\inf \{\mu (D')\} > \inf \{\mu (E')\}$ instead. But this inequality will force $D'$ and $E'$ to merge into one, $D' = E'$, leading to $\inf \{\mu (D')\} = \inf \{\mu (E')\}$. Therefore,

\begin{align} \text{for } \inf \{\mu (D') \} &= \inf \{\mu (E') \}, \ \mu^* (D) = \mu^* (E)\\ \text{for } \inf \{\mu (D') \} &< \inf \{\mu (E') \}, \ \mu^* (D) < \mu^* (E). \\ \end{align}​

Hence in both cases $\mu^* (D) \leq \mu^* (E)$.

(3) The third condition: Let $(A_i)_{i \in \mathbb N} \subset X$, and $(B_i)_{i \in \mathbb N} \in \mathcal A.$

\begin{align} \text{Given } \mu^* (A_i) &= \inf \{\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}, \\ \text{therefore, } \sum_{i=1}^{\infty}\mu^* (A_i) &= \sum_{i=1}^{\infty}\inf \{\mu(B_i) \mid A_i \subset B_i, B_i \in \mathcal A\}.\\ \text{On the other hand, } \ \mu^* (\bigcup_{i=1}^{\infty}A_i) &= \inf \{\mu(\bigcup_{i=1}^{\infty} B_i) \mid \bigcup_{i=1}^{\infty} A_i \subset \bigcup_{i=1}^{\infty} B_i, B_i \in \mathcal A\}.\\ \end{align}​

\CASE-1: If $B_i$'s are pairwise disjoint, that is $B_i \cap B_j = \emptyset$ for all $i, j \in \mathbb N$, then

$\sum_{i=1}^{\infty} \mu^* (A_i) = \mu^* (\bigcup_{i=1}^{\infty} A_i).$​

CASE-2: If $B_i \cap B_j \neq \emptyset$ for certain $i, j \in \mathbb N$, then

$\sum_{i=1}^{\infty} \mu^* (A_i) < \mu^* (\bigcup_{i=1}^{\infty} A_i).$​

Therefore for all cases

$\sum_{i=1}^{\infty} \mu^* (A_i) \leq \mu^* (\bigcup_{i=1}^{\infty} A_i),$​

hence $\mu^*$ is an outer measure. $\blacksquare$

4. Feb 13, 2015

### Dick

Mmm. I saw that you got a pretty good solution from stackexchange as well, which pretty much echoed what I said. I really don't understand your 'folksy' reinterpretation of them. Try to explain them in words. I'll do mine first. The proof of (2) is immediate because the inf of a subset is greater or equal to the inf of the set that its a subset of. The proof of (3) is a little less obvious but the general idea that you can approximate the outer measure of $A_i$ as well as you like with a $\mu$ measureable set $A'_i$ and then use countable additivity and shrinking epsilons to basically zero. Tell me a story about your proofs. A proof should have a narrative. That might help you to construct more coherent arguments.

Last edited: Feb 13, 2015
5. Feb 14, 2015

### A.Magnus

(1) Believe it or not, in my zeal for learning math, I have signed up 4 accounts: StackExchange (Math, Meta, LaTex & Academia,) Physics Forum, http://www.quora.com/ [Broken] for broad question and finally MyMathForum which I am not very engaged in. I am very active in MSE because it is generally the fastest forum I get feedback from, sometimes just in a matter of minutes. But for this particular question here, for some reasons I did not get feedback until another day or two. In the meantime, I had thought that I lost my question forever, therefore I decided to cross-post it here. For your info, I did not take cross-posting lightly. In fact, I did my due diligence last year by posting this question at Meta SE, asking for general consensus if cross-posting runs against any etiquette. From that posting I didn't see any backlash against cross-posting as long as the next forum is not sister to MSE.

(2) Aside from my zeal, I have an imminent reason for signing up multiple accounts: I am studying online for my MS in pure math from a local state university. Whenever I have question, it is very hard to get any help from my professors, and if at all after long wait, the answers consist of no more than one or two paragraphs. Being a HS math teacher, I understand their predicaments, they are overloaded and overwhelmed. Beside, online students are impersonal to them because of the distance. Therefore the public forums are effectively the place where I get help; the forums crowd-teach me.

(3) Now back to your request to explain my "folksy" proof in words. While I am not a prolific writer, but as you can see here I am capable of expressing my thought in words of nuances. In explaining my "folksy" proof, I think I have narrated enough that I did not see anything left to say. For this reason I decided to visualize my proof idea by putting Venn diagrams, see the attached pdf file. I hope it gets my clumsy ideas across better than just words.

Thanks again for your feedback to my question.

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6. Feb 15, 2015

### Dick

Thanks for your response! I'm not complaining about the cross posting. I encourage it. There don't seem to be many people around here that are willing to tackle measure theory problems, so go for it. My main point is simply that you got a good proof from stackexchange and your restatement of it makes no sense to me. Sorry! Start with (2). I don't see how assuming the contradiction 'forces D' and E' to merge into one'. And the pictures don't help. I'm not trying to be mean or overly critical. It's just that what you are writing, while it may make sense to you, is not a mathematical proof. Do you want to try that one again? It's even simpler than you think.

7. Feb 16, 2015

### A.Magnus

I am sorry I am coming back late, I just finished tending other classes. I took advantage of this downtime this Spring semester by taking 4 math courses!

Going back to (2), I think both of us have the same idea, but my idea looks weird because of my bad, bad choice of using the verb "merge" in the proof. I think the language should go along the math-lingo like these:

(1) Suppose that by way of contradiction, $D \subset E$ leads to $\inf \{ \mu (D′) \} > \inf \{ \mu (E′) \}$ instead.
(2) But this contradicts the definition of $D'$ being the infimum of set belong to $\mathcal A$ and containing $D$.
(3) Therefore $D \subset E$ has to lead to $\inf \{ \mu (D′) \} \leq \inf \{ \mu (E′) \}$.

I prefer this "chatty" proof because I am being extra careful in using phrases such as "obviously," "clearly" that gloss over proof. Thanks again for giving me input -- I am learning along the way!

8. Feb 16, 2015

### Dick

I don't doubt that the correct idea is in your head someplace. You just aren't expressing it clearly. Let's start with the assumption in (1). That means that there is a set $E'$ that contains $E$ such that $\mu(E')$ is less than $\mu(D')$ for any set $D'$ which contains $D$. Now the next line here is the one that you really aren't saying clearly. $E'$ is itself a set that contains $D$ since it contains $E$ and $D \subset E$. That's what I was looking for.

This goes back to what I've been saying all along. If $A$ is a set of numbers and $B$ is another set of numbers and $A \subset B$ then $inf(B) \le inf(A)$. The previous argument is just a proof of that. The set of subsets of $E$ is itself a subset of the set of subsets of $D$.

Last edited: Feb 16, 2015