If $A \subset B \subset \mathbb{R}$ such that $A$ is Lebesgue measurable and $m^*(B)=m(A)< \infty$. Show that $B$ is Lebesgue measurable.

I am doing this by showing that there exist a closed and open sets that are subset and superset of given set respectively. But I can't show the superset part.How to do this?

  • $\begingroup$ Are you only allowed to use subset/superset idea? Or are you allowed to use the fact that $m^*(E)=0\implies E$ is measurable and $m(E)=0$? $\endgroup$ – Clayton Apr 26 '18 at 19:40
  • $\begingroup$ this is also acceptable but I don't know how to apply that $\endgroup$ – Ashish Ranjan Apr 26 '18 at 19:41
  • $\begingroup$ Do you define measurable like Caratheodory? If yes: try to prove that $m^*(B\setminus A) = 0$ and thus $B\setminus A$ is measurable. $\endgroup$ – user251257 Apr 26 '18 at 19:43

Since $A$ is measurable and $A \subset B$, we have $m^*(B-A) = m^*(B) - m^*(A)=0$. Therefore, $B-A$ is measurable.

Since $B = A \cup (B-A)$, it follows that $B$ is measurable.

  • $\begingroup$ for what sets $A$ and $B$ we can write $m^*(B \setminus A)= m^*(B)-m^*(A)$ . I read a lemma that A should be closed and bounded and A should be open and bounded. $\endgroup$ – Ashish Ranjan Apr 26 '18 at 19:50
  • $\begingroup$ @AshishRanjan Since $A$ is measurable, $m^*(E) = m^*(E \cap A) + m^*(E - A)$ for all $E \subset \mathbb{R}$ (this is the usual definition of a measurable subset of $\mathbb{R}$). Setting $E=B$, using the fact that $A \subset B \implies B \cap A = A$, and rearranging gets the result. So we can write the equality in question for any sets $A,B$ such that $A$ is measurable and $A \subset B$. $\endgroup$ – grndl Apr 26 '18 at 19:53

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