Measure Theory: Continuity from below for set of intervals

[jamestrickington]

I have a question from Measure Theory / Probability Theory. This is the question:



I have established that the measure v is not sigma-additive.
I have also checked the value of An for most cases.

I am only interested in the case where An = (0, bn] where as n -> inf, b -> 0. In this case, does the infinite countable intersection of An become the empty set? If so then the statement does not hold.

 

[blamocur]

I am only interested in the case where An = (0, bn] where as n -> inf, b -> 0. In this case, does the infinite countable intersection of An become the empty set? If so then the statement does not hold.

If the intersection of [imath]A_n[/imath]'s weren't empty, what would it contain?

 

[jamestrickington]

If the intersection of [imath]A_n[/imath]'s weren't empty, what would it contain?

The smallest bn which is just greater than 0. Since bn only tends to 0, it will never be equal to 0 and always greater than 0. At least that is what I think.

 

[blamocur]

The smallest bn which is just greater than 0. Since bn only tends to 0, it will never be equal to 0 and always greater than 0. At least that is what I think.

Let's call this "smallest bn" [imath]\epsilon[/imath]. But to me the phrase "bn only tends to 0" means [imath]\lim_{n\rightarrow \infty} = 0[/imath]. Do you remember the definition of limits for sequences?

 

[blamocur]

But to me the phrase "bn only tends to 0" means lim⁡n→∞=0\lim_{n\rightarrow \infty} = 0limn→∞=0.

I meant [imath]\lim_{n\rightarrow\infty} \mathbf{b_n} = 0[/imath] -- sorry for this omission.

 

[jamestrickington]

No, I have not come across the definition of limits for sequences. Can you please help?

 

[blamocur]

No, I have not come across the definition of limits for sequences. Can you please help?

Limit of a sequence - Wikipedia

en.wikipedia.org

When you say "bn tends to zero" you do need to use a formal definition.

 

[blamocur]

Here is a closing summary:
If [imath]\lim_{n\rightarrow\infty} = 0[/imath] then [imath]\bigcap_n (0,b_n] = \empty[/imath]
A proof by contradiction: if [imath]S = \bigcap_n (0, b_n] \neq \empty[/imath] then : [imath]\exist \epsilon > 0 : \forall n : \epsilon\in (0,b_n][/imath]
But by definition of the limit for [imath]b_n[/imath] we have [imath]\forall \epsilon > 0 \exists N : \forall n>N: b_n < \epsilon \equiv \epsilon \notin (0,b_n][/imath], q.e.d.