MHB Outer Measure .... Axler, Result 2.5 ....

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I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 1: Measures ...

I need help with the proof of Result 2.5 ...

Result 2.5 and its proof read as follows:
Axler - Result 2.5 .png

Now $$ \mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k)$$ follows from Axler's definition of outer measure ( is that correct?) ... see definition below ...

Then essentially we have to prove that $$ \mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k) \Longrightarrow \mid A \mid \leq \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) $$ ...

But how do we rigorously prove this ...

Can someone please demonstrate a formal and rigorous proof that:

$$\mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k) \Longrightarrow \mid A \mid \leq \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) $$ ...Help will be much appreciated ...===============================================My thoughts ...

Perhaps we can assume that $$\mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) $$ ... and obtain a contradiction ...

We have that $$ \mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) \Longrightarrow \ \exists \ \sum_{ k = 1 }^{ \infty } l(I_k)$$ such that $$\mid A \mid \ \gt \sum_{ k = 1 }^{ \infty } l(I_k)$$ where $$\sum_{ k = 1 }^{ \infty } I_k$$ covers $$B$$ ... Is this a contradiction ...? why exactly? How would you explain the contradiction clearly and rigorously ...

Hope that someone can help ...

Peter=============================================================================================================

Readers of the above post may be assisted by access to Axler's definition of the length of an open interval and his definition of outer measure ... so I am providing access to the relevant text ... as follows:
Axler - Length of Interval & Outer Messure ... .png

Hope that helps ...

Peter
 
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Hi Peter,

You're almost there. Note that the collection of intervals is also a cover of $A$. Can you arrive at your contradiction using this fact?
 
GJA said:
Hi Peter,

You're almost there. Note that the collection of intervals is also a cover of $A$. Can you arrive at your contradiction using this fact?
Thanks for the help GJA ...

I think I can proceed now ... as follows ...

Assume that $$\mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k ) $$

... then $$\ \exists \ $$ a particular covering $$I_1, I_2, I_3,$$ ... of $$B$$ such that $$\mid A \mid \ \gt \sum_{ k = 1 }^{ \infty } l(I_k)$$ where $$B \subset \cup_{ k = 1 }^{ \infty } I_k $$

But this particular covering also covers $$A$$ since $$A \subset B$$ ...

... so that $$\mid A \mid \ \leq \sum_{ k = 1 }^{ \infty } l(I_k)$$ ... since $$\mid A \mid$$ is a lower bound on $$\sum_{ k = 1 }^{ \infty } l(I_k)$$ for all coverings $$I_1, I_2, I_3,$$ ... of $$A$$

... BUT ... $$\mid A \mid \ \leq \sum_{ k = 1 }^{ \infty } l(I_k)$$ is a contradiction of our assumption ...
Is the above correct?

Can someone please critique the above argument ...

Help will be appreciated ...

Peter
 
Looks great, Peter. Nicely done!
 
Thanks GJA ... appreciate your help as usual...

Peter
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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