0
$\begingroup$

I am trying to prove that for any bounded set $A$ in the borel $\sigma$ algebra, that for the Lebesgue measure $m$ $$ m(A)=\inf\{m(U)|U\text{ is open and }A\subseteq U\} $$ here is my attempt.


Let $U$ be an open set containing $A$. Then $m(A)\leq m(U)$ by monotonicity. Then $m(A)$ is a lower bound on the set $\{m(U)|U\text{ is open and }A\subseteq U\}$. Denote the infimum by $M$. By definition of infimum, there exists a sequence of open sets $(U_n)_{n\in\mathbb{N}}$ such that $A\subset U_n$ for each $n$, and for all $\epsilon>0$ there exists an $N\in\mathbb{N}$ such that for all $n\geq N$, we have $m(U_n)<M+\epsilon$.

Letting $\epsilon>0$ be arbitrary, then there exists $N\in\mathbb{N}$ such that $ m(U_n)<M+\epsilon $ whenever $n\geq N$. Since $A\subseteq U_n$ for all $n$, then we have $$ m(A)\leq m(U_n)<M+\epsilon $$ which shows that $M-m(A)<\epsilon$, and since $\epsilon$ was arbitrary, then we must have equality.


Just looking to see if there are any mistakes in my proof. Thanks!

$\endgroup$
  • $\begingroup$ How do you get $M-m(A) <\epsilon$? You see to have $m(A)-M <\epsilon$ which is of no use. $\endgroup$ – Kavi Rama Murthy Apr 18 at 8:49
  • $\begingroup$ Should I have $m(U_n)-\epsilon<M$ rather than what I have said? That way I can say $M-m(A)< \epsilon$? @KaviRamaMurthy $\endgroup$ – JB071098 Apr 18 at 8:54
  • $\begingroup$ No. For proving regularity you need the fact that $m(A)=M$. You have only stated that $m(A)$ is a lower bound. This is not enough to prove regularity. $\endgroup$ – Kavi Rama Murthy Apr 18 at 8:59
  • $\begingroup$ @KaviRamaMurthy Could you show me how? Even a hint would be appreciated. $\endgroup$ – JB071098 Apr 18 at 9:02
1
$\begingroup$

For any Lebesgue measurable set $A$ we have $m(A)=\inf \{\sum m(I_n): A \subset \cup_n I_n, I_n\, \text {is an open interval}\,\}$. This comes from the construction of Lebesgue measure. Once you have this regularity is easy. Let $\epsilon >0$ and choose open intervals $I_n$ such that $A \subset \cup_n I_n$ and $\sum m(I_n) <m(A)+\epsilon$. Let $U=\cup_n I_n$. Then $U$ is open, $A \subset U$ and $m(A) \leq m(U) \leq m(A)+\epsilon$ from which regularity follows easily.

$\endgroup$

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.