1
$\begingroup$

If $G$ is a measurable set and satisfies $m^*(G)<\infty$, then for all $\varepsilon>0$ there exists a closed set $F\subset G$ such that $m^*(F)>m^*(G)-\varepsilon$

Edit:

I know that: $\forall$ closed set $C\subset \mathbb{R}$, $\forall \varepsilon>0$, $\exists A\subset F$ a open set such that $m^*(F\backslash A)<\varepsilon$.

Using this ideas i try that:

For each open set $G$ and $\varepsilon>0$ choose $I_n'=(a_n-\frac{\varepsilon}{2^{n+2}},b_n)$ such that $G=\bigcup_{n=1}^\infty$. Now choose $F$ a closed set such that, if $I_n[a_n,b_n)$ so $F\subset \bigcup_{n=1}^\infty$ furthermore, $F$ must satisfy that $$ \sum_{i=1}^\infty c(I_n)<m^*(F)+\frac{\varepsilon}{2} $$ (I do not know if I could guarantee it that way!)

So , $$ m^*(G)<m^*(F)+\varepsilon $$

$\endgroup$
  • $\begingroup$ It is quite misleading to use outer measure ($m^{\star}$) here (there are counterexamples unless you assume $G$ is measurable). $\endgroup$ – hot_queen Feb 2 '14 at 23:40
  • $\begingroup$ @hot_queen I corrected this. $\endgroup$ – Felipe Feb 2 '14 at 23:57
  • $\begingroup$ @hot_queen can you help me in this? $\endgroup$ – Felipe Feb 3 '14 at 11:31
1
$\begingroup$

Let $A$ a measurable set and $\{I_n\}$ a family of open intervals such that $A\subset \bigcup I_n$ $$ \sum_{i=1}^\infty l(I_n) \leq m^*(A)+\varepsilon/2 $$ Let $I_n=[a_n,b_n)$ and choose intervals that $I_n'=(a_n-\varepsilon/(2^{n+2}),b_n)$, so if $O=\bigcup I_n'$ we have that $O$ is a open set. Furthermore $$ m^*(O)\leq\sum_{i=1}^\infty I_n'=\sum_{i=1}^\infty I_n+\varepsilon/2\leq m^*(A)+\varepsilon\Rightarrow m^*(O)-m^*(A)=m^*(O\backslash A)\leq \varepsilon $$ *Remember of this basic property of sets: $A\backslash B^c=B\backslash A^c$, this is usefull here!

If $A$ is a measurable set, so $A^c$ is too. The result above result that exists a open set $O'$ such that for each $\varepsilon>0$ we have $m^*(O'\backslash A^c)<\varepsilon$. Using the property above $m^*(O'\backslash A^c)=m^*(A\backslash O^c)$. Let $F=O^c$ and we have the result.

$\endgroup$

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.