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I have defined the lebesgue outer measure as $$\lambda(A) = \inf \{ \sum_{i=1}^\infty (b_i - a_i) : A \subset \cup_{i=1}^\infty (a_i,b_i) \}$$

I am reading the proof for $\lambda [a,b] = b-a$. So far, I have shown that $\lambda [a,b] \geq b-a$ the notes then read: since $[a,b] \subset (a-\epsilon, b + \epsilon)$ we have that $\lambda [a,b] \leq b-a + 2 \epsilon$. I don't follow how this follows from the definition:

if I let $A = [a,b]$ then the definition reads as $$\lambda(A) = \inf \{ (b+\epsilon - a + \epsilon) : [a,b] \subset (a-\epsilon, b + \epsilon) \}$$ but I get this as equal to exactly $b - a + 2 \epsilon$. I am obviously misunderstanding the definition for $\lambda$ - could someone please explain

edit: so if that definition is wrong thenI don't follow the logic of his argument. We have that $$ \lambda [a,b] = \inf \{ \sum (b_i - a_i) : [a,b] \subset \cup_i^\infty (a_i,b_i) \}$$ then I don't understand where the $[a,b] \subset (a - \epsilon, b + \epsilon)$ leads to the definition $\lambda[a,b] = b-a + 2\epsilon$?

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  • $\begingroup$ Because $\lambda[a,b]$ is the infimum taken over all possible covers with intervals, then it must be less than or equal to the total length of any particular cover, in that case the particular cover is the interval $(a-\epsilon,b+\epsilon)$ $\endgroup$ – Svetoslav Feb 22 '16 at 18:37
  • $\begingroup$ but it is the infimum of all of the sum of all possible covers, not all possible covers I thought? $\endgroup$ – FACEIT Feb 22 '16 at 18:39
  • $\begingroup$ I said it is the infimum over all possible covers (of something). What was your question again ? $\endgroup$ – Svetoslav Feb 22 '16 at 18:41
  • $\begingroup$ I don't understand what you mean by this. I read $\lambda(A)$ as the infimum of the sum $\sum (b_i - a_i)$ where $A$ is covered by $\cup_i^\infty (b_i - a_i)$ $\endgroup$ – FACEIT Feb 22 '16 at 18:42
  • $\begingroup$ Yes, what you wrote last is correct. And because $\lambda[a,b]$ is the infimum over all possible covers, then this infimum must be less than or equal to the total length of any particular cover of $[a,b]$. In your case, for a particular cover you take $(a-\epsilon,b+\epsilon)$ $\endgroup$ – Svetoslav Feb 22 '16 at 18:44
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As I said in the comments the definition for the outer measure of $A=[a,b]$ is one, and it is $$\lambda(A) = \inf \{ \sum_{i=1}^\infty (b_i - a_i) : A \subset \cup_{i=1}^\infty (a_i,b_i) \}\quad\quad (*)$$ and therefore $\lambda (A)$ is a fixed number. So after this you can not say $$\lambda(A) = \inf \{ (b+\epsilon - (a-\epsilon)) : A \subset (a-\epsilon,b+\epsilon) \}$$ because this is the infimum over only a subset of all possible covers, i.e the covers which have the particular form $(a-\epsilon,b+\epsilon)$, and therefore the above quantity is greater than or equal to $\lambda([a,b])$.

What you can say after noticing that $[a,b]\subset (a-\epsilon, b+\epsilon)$ is that $(a-\epsilon, b+\epsilon)$ is an admissible element of the set over which the infimum in $(*)$ is taken. As such, by the definition of the infimum, you have that $\lambda(A)=\lambda([a,b])\leq (b+\epsilon-(a-\epsilon))=b-a+2\epsilon$

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  • $\begingroup$ Thank you again for taking the time to explain it to me. $\endgroup$ – FACEIT Feb 22 '16 at 19:28
  • $\begingroup$ You are welcome ! $\endgroup$ – Svetoslav Feb 22 '16 at 19:29

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