Question: Let $K$ be collection of sets $(k_1, \hat{k}_1] \cup (k_2, \hat{k}_2] \cup \cdots\cup (k_n, \hat{k}_n]$ where, $n\in\mathbb{N}$ and every $k_1 <\hat{k}_1 < k_2 < \hat{k}_2 < \cdots< k_n < \hat{k}_n$ in sample space $Ω$ $(0,1]$. I have to verify that $K$ is an algebra but not a sigma algebra.
Attempted Answer: $\varnothing, Ω ∈ K $, $(k_i, \hat{k}_i] ∈ K$ and then $\Omega\setminus(k_i, \hat{k}_i] \in K$.
And, $(k_1, \hat{k}_1] \cup (k_2, \hat{k}_2] \cup \cdots\cup (k_n, \hat{k}_n]$ since $K = (k_1, \hat{k}_1] \cup (k_2, \hat{k}_2] \cup \cdots\cup (k_n, \hat{k}_n]$ Therefore, $K$ is an algebra.
My guess, K is not closed under countable finite union that is why it is not sigma-algebra. But my question is how do I show it?
Any hints, examples would be greatly appreciated. Thanks in advance.
I guess you are working with an algebra $K$ over $(0,1]$ generated by left-open intervals $(a,b]$. If $K$ is a $\sigma$-algebra, then $(0,1) = \bigcup_{n=0}^\infty (1-1/2^n,1-1/2^{n+1})$ is also a member of $K$. Hence $\{1\}\in K$.
Now observe that every element of $K$ can be represented as a finite union of half-open intervals, which are all infinite sets. What could you conclude from this?