Hi I was reading Cohn's book and I have problem with the following exercises (only the return of b is what I don't know), I'd appreciate any help and suggestion, if necessary, for a):

a) Show that a subset $B$ of $\bf{R}$ is Lebesgue measurable iff $\lambda^*(I)=\lambda^*(I\cap B)+\lambda^*(I\cap B^c)$ for any open interval.

b) Let $I$ a bounded subinterval of $\bf{R}$. Show that a subset $B$ of $I$ is Lebesgue measurable iff it satisfies $\lambda^*(I)=\lambda^*(B)+\lambda^*(I\cap B^c)$

a) We only show the sufficiently. Let $A\subset \bf{R}$ and we may assume that $\lambda^* (A)<+\infty$. Let $\{(a_n,b_n)\}$ a sequence of open intervals such that $A\subset \bigcup_n(a_n,b_n)$ and $\sum_nb_n-a_n<\lambda^* (A)+\varepsilon$, where $\varepsilon$ is an arbitrary positive number. Thus

\begin{align}\lambda^*(A\cap B)+\lambda^*(A\cap B^c)\le \sum _n\lambda^*((a_n,b_n)\cap B)+\sum_n\lambda^*((a_n,b_n)\cap B^c)\\ = \sum _n\lambda^*((a_n,b_n)\cap B)+\lambda^*((a_n,b_n)\cap B^c)\\=\sum _n\lambda^*((a_n,b_n))=\sum_nb_n-a_n\\<\lambda^* (A)+\varepsilon\end{align}

Letting $\varepsilon \downarrow0$, $\lambda^*(A\cap B)+\lambda^*(A\cap B^c)\le \lambda^* (A)$. Hence $B$ is Lebesgue measurable.

(b) One side is obvious, but the problem is with the return...

  • $\begingroup$ You made a typo at the ende of proof of (a). You have shown that $B$ is measurable, not $A$. $\endgroup$ – user251257 Jul 16 '15 at 23:56
  • $\begingroup$ @user251257 I don't see the mistake. In the proposition we need to show that $B$ is measurable not $A$. $\endgroup$ – Jose Antonio Jul 17 '15 at 0:11
  • $\begingroup$ the last sentence reads: Hence A is Lebesgue measurable $\endgroup$ – user251257 Jul 17 '15 at 0:12
  • $\begingroup$ @user251257 I see, thanks ;) $\endgroup$ – Jose Antonio Jul 17 '15 at 0:13

First, show that if $B$ is a subset of a bounded closed interval $I$, then $\lambda_*(B)+\lambda^*(I\backslash B)=\lambda(I)$. Then, by assumption, we have that $\lambda_*(B)+\lambda^*(I \backslash B)=\lambda^*(B)+\lambda^*(I\backslash B)$, hence, $\lambda_*(B)=\lambda^*(B)<\infty$ and $B$ is measurable.

  • $\begingroup$ I really appreciate your help, but the inner measure is not defined yet in the book. I only know that a set is measurable if divides each subset in such a way that the sizes (as measured by the lebesgue outer measure in this case) of the pieces add properly. $\endgroup$ – Jose Antonio Jul 18 '15 at 2:38

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.