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

[Math Amateur]

TL;DR 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 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:

Hope that helps ...

Peter

 

[Dragon27]

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.

 

[member 587159]

Hint: If $\emptyset \neq A \subseteq [0,\infty]$, then $\inf(A)\geq 0$.

I leave the easy proof of this fact to you.

 

[Math Amateur]

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 ...?

PeterEDIT

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

 

[member 587159]

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 ...?

PeterEDIT

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...

 

[Math Amateur]

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

 

[member 587159]

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 $\infty \geq m^*(E)= \inf(A) \geq 0$.