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

Question

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.

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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.

Answer 1

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

Answer 2

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.