3
$\begingroup$

I want to show that if $E$ is Jordan measurable then for every $\epsilon > 0$ there exists $A \subset E \subset B$ such that $m(A-B) \leq \epsilon$.

I think I have the right ideas but feel I am missing some details. I'd like some feedback for the following proof:

Let $\epsilon>0$ be given. Suppose $E$ is Jordan measurable. Then there exists $A$ and $B$ elementary with $A \subset E \subset B$ such that

$$ \sup m(A) = \inf m(B) $$

and thus

$$ m(A) \leq m(B) $$

which implies

$$ 0 \leq m(B) - m(A) $$

Now since $A \subset B$ we know

$$ 0 \leq m(B) - m(A) = m(B-A) $$

and we can always choose $A,B$ such that

$$ m(B-A) \leq \epsilon $$

$\endgroup$
1
  • $\begingroup$ Can this be reopened? What other clarifications are needed? $\endgroup$
    – Chris
    Jul 26 '16 at 14:46
2
$\begingroup$

There's maybe a typo in the problem statement? I'm assuming that what you mean is "if $E$ is Jordan measurable, then for every $\epsilon > 0$ there exist elementary sets $A \subseteq E \subseteq B$ such that $m(B-A) \leq \epsilon$" -- is that right?

Assuming so --

You say that because $E$ is measurable, there exist $A \subseteq E \subseteq B$ elementary such that $\sup m(A) = \inf m(B)$. I'm not sure what that means. Since $A$ is elementary, $m(A)$ is a number, not something you can take sup or inf of.

From the definition of Jordan measurable, since $E$ is measurable you have that $\sup_{A \subseteq E} m(A) = \inf_{E \subseteq B} m(B) = m(E)$ where the inf and sup are taken over all elementary subsets (resp. supersets).

Since the inf and sup are taken over all such subsets, you can now use properties of inf and sup to choose $A$ and $B$ such that the differences $m(E) - m(A)$ and $m(B) - m(E)$ are small.

EDIT: okay, so once you have $\sup_{A \subseteq E} m(A) = \inf_{E \subseteq B} m(B) = m(E)$, which is just the definition of Jordan measurable, what you want to do is take some $\epsilon > 0$. Use definition of sup/inf to choose $A \subseteq E$ elementary such that $m(A) > m(E) - \epsilon/2$ and $B \supseteq E$ elementary such that $m(B) < m(E) + \epsilon/2$. combining these yields $m(B) - m(A) < \epsilon$. Now use additivity of measure over disjoint unions to get $m(B-A) = m(B) - m(A)$.

$\endgroup$
3
  • $\begingroup$ Yes that was misstated! I've made edits to the question. $\endgroup$
    – Chris
    Jul 29 '16 at 13:53
  • $\begingroup$ Indeed I left out the the important detail that $A \subset E$ and $E \subset B$ in my sup and info notations respectively. Your suggestion is exactly what I was getting at though. I understand this idea but I am not sure I know how to write it properly as a proof. $\endgroup$
    – Chris
    Jul 29 '16 at 13:59
  • $\begingroup$ I added some more details on how to go from the definition of Jordan measurability to a formal (ish) proof. hope that helps? $\endgroup$
    – Ada Morse
    Jul 29 '16 at 15:27

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.