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I received a very helpful answer from David Mitra concerning the following question:

Let $X$ be a set, $S$ = {$\emptyset$, X} and define $\mu(\emptyset) = 0$, $\mu(X) = 1$. Determine the outer measure $\mu^*$ induced by the set function $\mu: S \rightarrow [0, \infty)$ and the $\sigma$-algebra of measurable sets.

Solution:

$$ \mu^*(E)=\text{inf}\Bigl\{\ \textstyle\sum\limits_{k=1}^\infty \mu(E_k) : E\subset\bigcup_{k=1}^\infty E_k, E_k\in S\Bigr\}. $$

If $E_k\in S$, then either $E_k=\emptyset$ or $E_k=X$.

Let $E\subset X$ be non-empty. Take the covering $E_1=X$ and $E_n=\emptyset$ for $n>1$, we have $E\subset \bigcup_{k=1}^\infty E_k$ and $\sum\limits_{k=1}^\infty u(E_k)=1$. Then, for all coverings, $1=u^*(E)=\inf\sum\limits_{k=1}^\infty u(E_k) \leq \sum\limits_{k=1}^\infty u(E_k)$.

So $\mu^*(E)=\cases{1, & if $E\ne\emptyset$\cr 0, & if $E=\emptyset$}$ This defines the outer measure.

A set $E$ is measurable if $$\mu^*(T)= \mu^*(T\cap E)+\mu^*(T\cap E')$$ for all $T\subset X$.

In particular, if $E$ is measurable, we must have $$ 1=\mu^*(X)= \mu^*(X\cap E)+\mu^*(X\cap E'). $$

So $\mu^*(E)=\cases{1, & if $E\ne\emptyset$\cr 0, & if $E=\emptyset$}$ This defines the outer measure.

My question is: How would this change if $X = [1,2]$, so that $S$ = {$\emptyset$, $[1,2]$} of subsets of $R$ and $\mu(\emptyset) = 0$, $\mu([1,2])=1$? Wouldn't it be the same?

What if we are talking about S being all subsets of $R$, where $\mu: S \rightarrow R$ and $\mu(E)$ is the number of integers in E? This one is different, but wouldn't it end up being something like:

$\mu^*(E) = \inf\sum_{k=1}^\infty \mu(E_k) = \mu(E_1) + \mu(E_2) + ... + \mu(E_n) = 1 + 2 + ... + n + ...$ where each k is the number of integers. So, as $n \rightarrow \infty, \mu^*(E) \rightarrow \infty$.

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For the first question, the answer is "yes", nothing would change (you've only specified what $X$ is). Your computation for the second question is off. How do you get, e.g., $\mu E_2=2$? If $E$ is a subset of $\Bbb R$, to find $\mu^*(E)$ you need to compute the infimum defining $\mu^*$. In this case, note $E$ is an element of $S$; so the infimum is achieved by taking $E_1=E$ and $E_n=\emptyset$ for $n>1$. Thus $\mu^*(E)=\mu(E)$. –  David Mitra Feb 16 '13 at 1:58
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Incidentally, you did not provide the correct answer for the second part of the earlier post: which sets are measurable? Which sets satisfy the equality $1=\mu^*(X\cap E)+\mu^*(X\cap E')$? –  David Mitra Feb 16 '13 at 1:59
    
OK, so then, the original solution holds for $X = [a, b]$ where $a, b \in R$. For the last one my thinking was, since we want $\bigcup_{k=1}^\infty E_k$ to cover $S$, then we'll be taking all intervals and singletons in $R$, then wouldn't there be covers around all single integers so $lim_{n \rightarrow \infty} n\mu(E_1) = n*1 \rightarrow \infty$, covers around all intervals containing two integers $lim_{n \rightarrow \infty} n\mu(E_2) = n*2 \rightarrow \infty$, so on...(which will have a countably infinite measure, since $\mu(E)$ is the number of integers in E). –  Jake Casey Feb 16 '13 at 17:41
    
But you're saying that we always need to just look for the smallest one that works and it is the infimum? Since $\mu^*(E) = \mu(E)$ which is the number of integers in E, the answer is $\mu^*(E) = \infty$. –  Jake Casey Feb 16 '13 at 17:42
    
Since $ E \subset X$, $1=\mu^*(X)= \mu^*(X\cap E)+\mu^*(X\cap E')$ holds for any E. –  Jake Casey Feb 16 '13 at 17:43

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