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Let $\mu$ be a regular "outer" measure on $\mathbb{R}^N$ (for example, the Lebesgue outer measure). By regularity I mean that for all $A\subset \mathbb{R}^N$, there is $E$ measurable with $A\subset E$ and $\mu(A)=\mu(E)$.

I have two questions which are bothering me for a long time.

1 - Let $A\subset \mathbb{R}^N$, $E$ measurable with $A\subset E$ and $\mu(A)=\mu(E)$. Let $Q$ be a cube. Is it true that $$\mu(A\cap Q)=\mu (E\cap Q)?$$

2- In the above conditions, is there any example for which $\mu(E\setminus A)\neq 0$?

Any idea or reference is appreciated.

Community
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Tomás
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1 Answers1

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Assuming the following additional conditions I give an answer to your question.

  • For each set $A\subset\mathbb R^N$ there exists a measurable set $E$ such that $A\subset E$ and $\mu(A)=\mu(E)$.
  • Q is measurable.

Since $A\subset E$ we clearly have $\mu(A\cap Q)\leq\mu(E\cap Q)$ and $\mu(A\backslash Q)\leq\mu(E\backslash Q)$. Since $Q$ and $E$ are measurable we have $\mu(A\cap Q)=\mu(A)-\mu(A\backslash Q)=\mu(E)-\mu(A\backslash Q)\geq\mu(E)-\mu(E\backslash Q)=\mu(E\cap Q)$. Consequently, $\mu(A\cap Q)=\mu(E\cap Q)$.

sranthrop
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  • If $Q$ is not measurable, do you think that equality may fail? – Tomás Aug 17 '14 at 01:25
  • I think one could find a counterexample, but I am not sure. Do you know anything else about your cube? Or about the measure? If $\mu$ were Borel regular for example, and if your cube were open/closed (=> Borel set), then you knew that it is automatically measurable. – sranthrop Aug 17 '14 at 01:34
  • In fact, in these particular cases you have solved the problem, I alread knew the solution (only for question 1), so I am more interested in the general case. – Tomás Aug 17 '14 at 01:55