A measure which is not continuous from above

Question

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.

Answer 1

Compute $\mu(A_n)$ what you get? Find $\cap_{n\ge 1} A_n$ what you get?

Answer 2

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 $n\in\mathbb{N}$?

Answer 3

• $μ(A_n)$ 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.