Let X be a random variable such that $0<X<2 \ \mathbb{P}$-a.s and $\mathbb{E}[X] = 1$. Let $C_{n} := \{X \geq 2-\frac{1}{n} \} $ for all $n \in \mathbb{N}$.
Is it true that $\lim_{n\to\infty} \mathbb{P}\left(C_{n}\right) = 0$?
I am pretty sure (and I hope) it is true. I tried using the Markov inequality but that did not really help. Does anyone have any idea if this is actually true and why\why not?
If $D_n$ denotes the complement of $C_n$ then $D_1\subseteq D_2\subseteq\cdots$ together with $\bigcup_{n=1}^{\infty}D_n=\{X<2\}$ so that $P(D_n)\uparrow P(X<2)=1$.
This justifies the conclusion that $P(C_n)=1-P(D_n)\downarrow 0$.
It is not needed here that $\mathbb EX=1$.