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Let $E$ be a subset of $\mathbb R $, $J_n$ be a closed cover of $E$, where $\forall n, J_n = [a_n,b_n]$ and we build the open cover $I_n$ like this : $ \forall n, I_n = ]a_n - \frac {\epsilon }{2^{n+1}} , b_n + \frac {\epsilon }{2^{n+1}} [ $.

I'm struggling to prove that the outer measure on open sets is inferior to the one on closed sets. So here I'd like to show that :

$$ \inf_{ \text {open sets covering E} } ( \sum \text{length}(I_n) ) \leq \inf_{ \text {closed sets covering E} } ( \sum \text{length}(J_n) ) $$

Why am I struggling?

For now I've proved this:

$$ \sum \text{length}(I_n) = \sum \text{length}(J_n) + \epsilon$$ here I'm stuck.


Please help me to complete this proof. Moreover, I ask for not giving me a proof using "it works for all closed sets so we take the infinimum on the RHS". I'm asking for something really formal.

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  • $\begingroup$ Are you considering covers by intrrvals? $\endgroup$ – Kavi Rama Murthy Mar 16 at 5:15
  • $\begingroup$ Yes I do, the sets I and J are intervals. I should have specify this $\endgroup$ – Marine Galantin Mar 16 at 5:16
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Let $\{J_n\}$ be a covering of $E$ by closed intervals. Let $\epsilon >0$. Expand $J_n$ to an open interval $I_n$ such that length of $I_n$ equals length of $J_n +\frac {\epsilon }{2^{n}}$. Then $(I_n)$ is an open covering of $E$ so LHS $\leq \sum \text {length of } I_n \leq \sum \text {length of } J_n +\epsilon$. Taking infimum over all $(J_n)$ we get LHS $\leq RHS +\epsilon$. Since $\epsilon$ is arbitarry we get LHS $\leq $ RHS.

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  • $\begingroup$ Thank you for your answer. However, that s exactly what I tried to avoid. I read a correction of this exercice using the same trick you wrote, and I am not convinced by it... Is there no way to prove it in a another way? $\endgroup$ – Marine Galantin Mar 16 at 5:27
  • $\begingroup$ If you tell me why this proof is not acceptable I can try to help you. This proof is 100% rigorous. $\endgroup$ – Kavi Rama Murthy Mar 16 at 5:31
  • $\begingroup$ I m confused by the fact that you re taking first one infimum and then the others. It kinda feel not right... I m sure it is correct, but can isnt it possible to do it in another way? $\endgroup$ – Marine Galantin Mar 16 at 5:57
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I wrote this answer. One can also find a similar answer to this question in the Measure Theory and Integration by G. de Barra.

Let $\epsilon > 0$. Let $(J_n)$ covering of $E$ with closed sets, such that :

$$ \sum length(Jn) - \epsilon \leq \inf_{ \text {closed sets covering E} } ( \sum \text{length}( \tilde{I}_n) )$$ which exists by definition of the infimum. We then fix $(I_n)$ s.t. : $$ J_n \subset I_n $$ $$length (I_n) = (1+ \epsilon) length( J_n) $$

so $$ \sum length( J_n) = \frac 1 {1+ \epsilon} \sum length(I_n) \leq \inf_{ \text {closed sets covering E} } ( \sum \text{length}( \tilde{I}_n) ) + \epsilon =: M$$

however, $$E \subset \bigcup I_n \ \ \text{ open sets } \implies \inf_{ \text {open sets covering E} } ( \sum \text{length}( \tilde{I}_n) ) \leq \sum length(I_n) \leq (1 + \epsilon ) M $$

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