Let $\mathcal A$ be an algebra of subsets of a set $X$ and $\mathcal S (\mathcal A)$ be the $\sigma$-algebra of subsets of $X$ generated by $\mathcal A.$ Let $\mu : \mathcal A \longrightarrow [0,+ \infty]$ be a measure on $\mathcal A.$ Let $\mu^*$ be the outer measure induced by $\mu.$ Let $E \subseteq X$ and $F \in \mathcal S (\mathcal A)$ be such that $E \subseteq F$ and $\mu^* (E) = \mu^* (F) < + \infty.$ If $\mu^* (G) = 0,$ $\forall$ $G \in \mathcal S (\mathcal A)$ with $G \subseteq F \setminus E$ then show that $\mu^* (F \setminus E) = 0.$

I have proved that the outer measure $\mu^*$ on $\mathcal P(X)$ induced by $\mu$ can be equivalently defined in terms of the restriction $\bar \mu$ of $\mu^*$ to $\mathcal S (\mathcal A)$ as follows $:$ $$\mu^* (A) = \text {inf}\ \left \{\bar {\mu} (B)\ |\ B \in \mathcal S (\mathcal A), A \subseteq B \right \}.$$ Can it help anyway? I hardly believe that this result will hold. But how to find a counter-example?

Thanks in advance.

  • $\begingroup$ I can show that $\mu^* \left ( (F \setminus E)^c \right ) = \mu^* (X).$ $\endgroup$ – math maniac. May 15 '20 at 15:43
  • $\begingroup$ Can anybody give me some hint? $\endgroup$ – math maniac. May 17 '20 at 6:28
  • 1
    $\begingroup$ Shouldn't $\mu$ be a pre-measure since it's only defined on an algebra, not a $\sigma$-algebra? $\endgroup$ – md2perpe May 17 '20 at 21:17
  • $\begingroup$ @md2perpe in my book, a measure is defined as a countably additive set function over a collection of subsets of a set $X$ containing the empty set $\varnothing.$ $\endgroup$ – math maniac. May 18 '20 at 1:58
  • $\begingroup$ For more information you can go through the second edition of the e-book An Introduction to Measure and Integration by Inder K. Rana published by American Mathematical Society which is available online in pdf format. $\endgroup$ – math maniac. May 18 '20 at 2:03

The claim to prove is wrong, as shows the following example. Let $X=[0,1]$, $\mu$ be the Lebesgue measure on $X$, and $\mathcal A=\mathcal S(\mathcal A)$ be a $\sigma$-algebra of Lebesgue measurable subsets of $X$. Let $E\subset X$ be an arbitrary set such that $E\not\in\mathcal A$. For instance, $E$ can be a Vitali set. Let $F\in\mathcal A$ be a set such that $E\subseteq F$ and $\mu(F)=\mu^*(E)>0$. Clearly, for all $G \in \mathcal S(\mathcal A)$ with $G \subseteq F \setminus E$ we have $\mu^*(G) =\mu(G)=0$. But if $\mu^*(F \setminus E) = 0$ then $F\setminus E\in\mathcal A$, ans so $E\in\mathcal A$, a contradiction.

  • 1
    $\begingroup$ how did the author miss that? He claimed that in his book. $\endgroup$ – math maniac. May 20 '20 at 6:52
  • $\begingroup$ @mathmaniac. Maybe because the claim is formulated in too general and too abstract form. $\endgroup$ – Alex Ravsky May 20 '20 at 6:56

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.