0
$\begingroup$

We know

Let $\epsilon\subset\mathscr{P}(X)$ and $\rho:\epsilon\rightarrow[0,\infty]$ be such that $\emptyset\in\epsilon, X\in\epsilon,\rho(\emptyset)=0$, then the set function defined on $\mathscr{P}(X)$: $$\mu^*(A):=\inf\{\sum_1^\infty\mu(E_j):E_j\in\epsilon\ and\ A\subset\bigcup_1^\infty E_j\} $$ is an outer measure.

My question is: are sets in $\epsilon$ always $\mu^*-$measurable? If not, could anyone give some examples?

$\endgroup$
  • $\begingroup$ Presumably in the def'n of $\mu^*(A)$ you mean to write $\rho (E_j)$, not $\mu (E_j).$ Note that then $\mu^*(A)$ is defined for every $A\subset X$. But what is your def'n of "measurable"? $\endgroup$ – DanielWainfleet Dec 12 '17 at 9:01
0
$\begingroup$

No. One-dimensional Hausdorff measure in the real line (= Lebesgue measure in the real line) may be defined using all subsets with $\rho(A)$ the diameter of $A$. But of course some sets are not Lebesgue measurable.

$\endgroup$

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.