Assuming MCT holds we need to prove the NIP
Monotone Convergence theorem: A monotone and bounded sequence is convergent.
Nested Interval Property: If for all $n\in \mathbb{N}, I_n=[a_n, b_n]$ and $I_{n+1} \subset I_n$ then $\bigcap_{n=1}^\infty I_n\neq \emptyset $
Proof: $(a_n)$ is a monotone a bounded sequence and increasing then it converges to some $a$. We say that $I_n=[a_n, a]$, then $I_{n+1}\subset I_n$, as $a_n<a_{n+1}$. Now the intersection $\bigcap_{n=1}^k I_n=[a_k,a]$, now we can take the limit $$\lim_{k\rightarrow \infty}\bigcap_{n=1}^k I_n=a\neq\emptyset$$This proves the NIP.
Does this proof look ok?
Best Answer
Your proof is not O.K. You wrote $I_n=[a_n, a]$. But we have $I_n=[a_n, b_n]$ !
We have:
$a_1 \le .... \le a_n \le a_{n+1} \le b_{n+1} \le b_b \le .... \le b_1$. Hence the sequences $(a_n)$ and $(b_n)$ are bounded and monotonic, hence convergent.
Let $a= \lim a_n$ and $b= \lim b_n$. Then we have $a \le b$. Now show that $[a,b ] \subseteq \bigcap_{n=1}^\infty I_n.$