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 $A_n$'s weren't empty, what would it contain?

 

[jamestrickington]

If the intersection of $A_n$'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" $\epsilon$. But to me the phrase "bn only tends to 0" means $\lim_{n\rightarrow \infty} = 0$. 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 $\lim_{n\rightarrow\infty} \mathbf{b_n} = 0$ -- 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 $\lim_{n\rightarrow\infty} = 0$ then $\bigcap_n (0,b_n] = \empty$
A proof by contradiction: if $S = \bigcap_n (0, b_n] \neq \empty$ then : $\exist \epsilon > 0 : \forall n : \epsilon\in (0,b_n]$
But by definition of the limit for $b_n$ we have $\forall \epsilon > 0 \exists N : \forall n>N: b_n < \epsilon \equiv \epsilon \notin (0,b_n]$, q.e.d.