Lebesgue Outer Measure ... Carothers, Proposition 16.2 (i) ...

  • #1
Math Amateur
Gold Member
1,067
47

Summary:

I need help in order to construct and express a valid, convincing, formal and rigorous proof to Carothers Proposition 16.2 (i) ...
I am reading N. L. Carothers' book: "Real Analysis". ... ...

I am focused on Chapter 16: Lebesgue Measure ... ...

I need help with the proof of Proposition 16.2 part (i) ...

Proposition 16.2 and its proof read as follows:


Carothers - Proposition 16.2 ... .png




Carothers does not prove Proposition 16.2 (i) above ...

Although it seems intuitively obvious, I am unable to construct and express a valid, convincing, formal and rigorous proof of the result ...

Can someone please demonstrate a formal and rigorous proof of Proposition 16.2 (i) above ...

Peter



========================================================================================================


It may help readers of the above post to have access to Carothers introduction to Lebesgue outer measure ... so I am providing the same as follows:



Carothers - Proposition 16.1 ... .png





Hope that helps ...

Peter
 

Answers and Replies

  • #2
86
47
I am unable to construct and express a valid, convincing, formal and rigorous proof of the result
But you should at least make an attempt.
 
  • #3
Math_QED
Science Advisor
Homework Helper
2019 Award
1,692
718
Hint: If ##\emptyset \neq A \subseteq [0,\infty]##, then ##\inf(A)\geq 0##.

I leave the easy proof of this fact to you.
 
  • Like
Likes Math Amateur
  • #4
Math Amateur
Gold Member
1,067
47
Hint: If ##\emptyset \neq A \subseteq [0,\infty]##, then ##\inf(A)\geq 0##.

I leave the easy proof of this fact to you.

Thanks for the hint Math_QED ...

We need to prove the following ...

If ##\emptyset \neq A\subseteq [0,\infty]##, then ##\inf(A)\geq 0##

Proof

Assume ##\text{inf} (A) \lt 0##

Then there exists ##x \in A## such that ##\text{inf} (A) \lt x \lt 0## ...

... contradiction as there are no negative numbers in ##A## ...

Therefore ##\inf(A)\geq 0## ...


Could also prove this in the same way for ##m^*(E)## ...

This would establish that ##m^*(E) \geq 0## ...

But that would not necessarily mean that ##m^*(E)## ranges from ##0## to ##\infty## ...

So ... how do we proceed from here ...?

Peter


EDIT

To prove that ##m^*(E)## ranges from ##0## to ##\infty## ... would it be sufficient to note that

##m^*([a, b]) = m^*((a, b)) = b - a## for ##a, b \in \mathbb{R}## where ##a \lt b##

... and that ##m^*( \emptyset ) = 0## ... and ##m^*( ( 0, \infty ) ) = \infty - 0 = \infty##

Does that complete the proof?

Peter
 
Last edited:
  • #5
Math_QED
Science Advisor
Homework Helper
2019 Award
1,692
718
Thanks for the hint Math_QED ...

We need to prove the following ...

If ##\emptyset \neq A\subseteq [0,\infty]##, then ##\inf(A)\geq 0##

Proof

Assume ##\text{inf} (A) \lt 0##

Then there exists ##x \in A## such that ##\text{inf} (A) \lt x \lt 0## ...

... contradiction as there are no negative numbers in ##A## ...

Therefore ##\inf(A)\geq 0## ...


Could also prove this in the same way for ##m^*(E)## ...

This would establish that ##m^*(E) \geq 0## ...

But that would not necessarily mean that ##m^*(E)## ranges from ##0## to ##\infty## ...

So ... how do we proceed from here ...?

Peter


EDIT

To prove that ##m^*(E)## ranges from ##0## to ##\infty## ... would it be sufficient to note that

##m^*([a, b]) = m^*((a, b)) = b - a## for ##a, b \in \mathbb{R}## where ##a \lt b##

... and that ##m^*( \emptyset ) = 0## ... and ##m^*( ( 0, \infty ) ) = \infty - 0 = \infty##

Does that complete the proof?

Peter
If ##A\subseteq [0,\infty]##, then ##0## is a lower bound of ##A## hence...
 
  • Like
Likes Math Amateur
  • #6
Math Amateur
Gold Member
1,067
47
If ##A\subseteq [0,\infty]##, then ##0## is a lower bound of ##A## hence...

Oh ... OK ... then ##m^* (A) \geq 0## ...

But ... does that necessarily prove that ##m^* (A)## ranges from ##0## to ##\infty##?

Peter
 
  • #7
Math_QED
Science Advisor
Homework Helper
2019 Award
1,692
718
I'll be very explicit now.

Lemma: Let ##\emptyset \neq A \subseteq [0, \infty]##. Then ##\inf A \in [0, \infty]##.
Proof: ##0## is a lower bound for ##A##, hence by definition of infinum as greatest lower bound ##\inf(A) \geq 0##. ##\quad \square##

Apply the lemma with ##A:= \{\sum_{n=1}^\infty l(I_n): E \subseteq \bigcup_{n=1}^\infty I_n\}##. Then you obtain ##\infty \geq m^*(E)= \inf(A) \geq 0##.
 
  • Like
Likes Math Amateur

Related Threads on Lebesgue Outer Measure ... Carothers, Proposition 16.2 (i) ...

Replies
5
Views
216
  • Last Post
Replies
2
Views
1K
Replies
6
Views
828
Replies
12
Views
3K
Replies
4
Views
974
Replies
14
Views
343
Replies
8
Views
2K
Replies
16
Views
2K
Top