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

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.

• Are you considering covers by intrrvals? – Kavi Rama Murthy Mar 16 at 5:15
• Yes I do, the sets I and J are intervals. I should have specify this – Marine Galantin Mar 16 at 5:16

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.

• 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? – Marine Galantin Mar 16 at 5:27
• If you tell me why this proof is not acceptable I can try to help you. This proof is 100% rigorous. – Kavi Rama Murthy Mar 16 at 5:31
• 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? – Marine Galantin Mar 16 at 5:57

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