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


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 $S$.

| cite | improve this answer | |
  • $\begingroup$ @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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.