# A measure which is not continuous from above

Let $\Omega= \mathbb{N}$, $F = P(\Omega)$, and $A_n = \{j \mid j \in\mathbb{N}, j \geq n\}$, $n \in\mathbb{N}$. Let $\mu$ be the counting measure on $(\Omega,F)$, so that $\mu(A) = |A|$. I need to show that $$\lim_{n\to\infty} μ(A_n) \neq \mu\bigg(\bigcap_{n\geq 1} A_n\bigg).$$

Now, for a fixed $n$, $μ(A_n)$ cannot be finite, because it is equal to $|N|-|{1,2\dots,n}|$ and then $N$ will be a finite set. So, left hand side of the identity is $\infty$. Now in the right hand side of the identity the set $\bigcap_{n\geq 1} A_n$ is the set of all natural numbers which are greater than all the natural numbers. Naturally, this set is null set. So, $\mu$ applied on it becomes zero. So, they are unequal.

The reason I posted this question is to find why is this happening. For a measure like probability this does not happen. I guess the reason is probability is a finite measure whereas here our measures are infinite. Want to know more intuition on this.

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Also, you can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Feb 19 '13 at 7:05

Compute $\mu(A_n)$ what you get? Find $\cap_{n\ge 1} A_n$ what you get?
Hint: For a fixed $n$, how many elements are in $A_n$? What elements are in $$\bigcap_{n\geq 1}A_n$$ i.e. how many natural numbers are greater than $n$ for all $n\in\mathbb{N}$?
• $μ(A_n)$ is the number of natural numbers $\ge n$, so the limit as $n\to \infty$ is $\infty$.
• $\mu\bigg(\bigcap_{n\geq 1} A_n\bigg)$ is the intersection of shrinking sets and thus goes to $\emptyset$ in the limit; its measure is zero, obviously.