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Let $S$ be a collection of subsets of $X$ and $\mu : S \to [0, \infty]$ a set function. Is every set in $S$ measurable with respect to the outer measure induced by $\mu$

Here is how we defined outer measure induced by $\mu$

Let $S$ be a collection of subsets of set $X$ and $\mu: S \to [0,\infty]$ a set function. Define $\mu^*(\emptyset)=0$ and for $E\subset X, E\neq \emptyset$ define $\mu^*(E)=\inf \sum\limits_{k=1}^{\infty} \mu(E_k)$ where the infimum is taken over all countable collections $\{E_k\}_{k=1}^{\infty}$ of sets in $S$ that cover $E$. Then the set function $\mu^*: 2^X \to [0, \infty]$ is an outer measure called the outer measure induced by $\mu$

I thought the answer is Yes, Outer measure is countable monotone (for any countable collection $\{E_k\}_{k=1}^{\infty}$ of measurable sets that covers a measurable set $E$ $\mu(E)\leq \sum\limits_{k=1}^{\infty} \mu(E_k)$)

Any help, with a formal detailed answers

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No. Let $S=\{2^\mathbb{[1,2]}\}$. And define $\mu^*$ as: $$ \mu^*(E)= \begin{cases} 0, & \text{if $E=\emptyset$}\\ 1, & \text{if $E=[1,2]$}\\ 0.75, & \text{if $E\subset [1,2]$}\\ \infty, & \text otherwise\\ \end{cases} $$ We see that for $A$, $B$, $(A_n)$ $\in S$: $$\mu^*(\emptyset)=0$$

$$ A\subseteq B \implies \mu^*(A) \leq \mu^*(B)$$ $$\mu^*(\bigcup_{n=1}^{\infty}A_n) \leq\sum_{n=1}^{\infty} \mu^*(A_n)$$ So $\mu^*$ is an outer measure.

We call a set E measurable[ref:Real analysis page 347] with respect to $\mu^*$ $\iff$ for every $A\subseteq[1,2], \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$

Then if $E=[1,1.5]\in S$, we can choose $A=[1,2]\in 2^\mathbb{R}$, then: $$ \mu^*(A)=1 $$ $$ \mu^*(E\cap A)=\mu^*(E)=0.75 $$ $$ \mu^*(A\cap E^c) = \mu^*([1.5,2])=0.75 $$

By our definition of measurable, we conclude that $E$ is not measurable, so there exists an non-measurable set in $S$.

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  • $\begingroup$ @Daniel Erickson Maybe our conception about outer measure $\mu^*$ induced by $\mu$ are different. According to Royden's real analysis book, 0.75 should be $\infty$ , see link $\endgroup$ – DuFong Feb 9 '16 at 0:38
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If we assume $\mu$ is a measure, then the answer is yes.

Let $S\subseteq \mathscr{P}(X)$ be a sigma algebra, with $\mu:S\to\mathbb{R}$ a measure on $X$.

We define our outer measure $\mu^*:X\to\mathbb{R}$ as $\mu^*(E):=inf\sum_{k=1}^{\infty}E_k$ over countable sequences $(E_k)\in S$ that cover $E$.

For any $A\in S$, define $(E_k)$ as $E_1:=A$, $E_n:=\emptyset$ for all $n\ge 2$. Then clearly $\mu^*(A)=\mu(A)$ as any other sequence of sets will have a greater than or equal value to $\mu(A)$ by countable additivity of $\mu$.

Now, given $E\subseteq X$ and rational $\varepsilon_q>0$, there exists a sequence $(B_{q_n})$ in $S$ such that $\sum_{n=1}^{\infty}\mu^*(B_{q_n})<\mu^*(E)+\varepsilon_q$ and $E\subseteq\bigcup_{n=1}^{\infty}B_{q_n}$. Let $B_q:=\bigcup_{n=1}^{\infty}B_{q_n}$. It's easy to see that $B_q\in S$, and $B:=\bigcap_{q}B_q\in S$ has the property that $E\subseteq B$, $\mu^*(B)=\mu^*(E)$. Furthermore, $A\cap B\in S$ and $A^c\cap B\in S$.

Then, by the subadditivity and monotonicity of $\mu^*$ and since $C\in S\implies \mu^*(C)=\mu(C)$, $$\mu^*(E)\leq\mu^*(A\cap E)+\mu^*(A^c\cap E)\leq \mu^*(A\cap B)+\mu^*(A^c\cap B)=\mu(A\cap B)+\mu(A^c\cap B)=\mu(B)=\mu^*(B)=\mu^*(E)$$ This implies $\mu^*(E)=\mu^*(A\cap E)+\mu^*(A^c\cap E)$ for any $E\subseteq X$, and therefore A is $\mu^*$ measurable.

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  • $\begingroup$ actually, if $\mu$ is a measure, the outer measure induced by $\mu$ is a measure, which is trivial. $\endgroup$ – DuFong Feb 10 '16 at 21:25
  • $\begingroup$ That is incorrect. $\mu^*$ does not necessarily have additivity (it does, however, have subadditivity). If the domain of $\mu^*$ is restricted to $S$, then yes that is trivially true. $\endgroup$ – Daniel Erickson Feb 11 '16 at 0:27

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