# Outer Measure ... Axler, Result 2.5 ...

#### Peter

##### Well-known member
MHB Site Helper
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:

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

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

But how do we rigorously prove this ...

Can someone please demonstrate a formal and rigorous proof that:

$$\displaystyle \mid A \mid \leq \sum_{ k = 1 }^{ \infty } l(I_k) \Longrightarrow \mid A \mid \leq \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k )$$ ...

Help will be much appreciated ...

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My thoughts ...

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

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

Hope that someone can help ...

Peter

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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:

Hope that helps ...

Peter

Last edited:

#### GJA

##### Well-known member
MHB Math Scholar
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?

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle \mid A \mid \ \gt \text{ inf } ( \sum_{ k = 1 }^{ \infty } l(I_k) \text{ where } B \subset \cup_{ k = 1 }^{ \infty } I_k )$$

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

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

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

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

Is the above correct?

Can someone please critique the above argument ...

Help will be appreciated ...

Peter

GJA

#### GJA

##### Well-known member
MHB Math Scholar
Looks great, Peter . Nicely done!

#### Peter

##### Well-known member
MHB Site Helper
Thanks GJA ... appreciate your help as usual...

Peter

GJA