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 $m^*(F)>m^*(G)-\varepsilon$


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

  • $\begingroup$ 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

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.


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