Let $ (\Omega, \mathcal{A}, P) $ be a probability space. Suppose that $ \{ X_n := 1_{A_n} \}_{n \in \mathbb{N}} $ is a sequence of simple random variables and $ X \equiv 0$. It is claimed that $ X_{n} \rightarrow_{P} X $ in probability is equivalent to the statement $ P(A_n) \rightarrow 0. $ Sadly I can't seem to grasp how this is so. I have the following argument:
By definition $ X_n \rightarrow_{P} X $ implies for every $ \epsilon > 0 $, $$ \lim_{n \rightarrow \infty}P[ \omega \in \Omega: |X_{n}-X|\geq \epsilon ] = \lim_{n \rightarrow \infty}P[w \in \Omega: 1_{A_n}\geq \epsilon]=0.$$ Then for every $ n \in \mathbb{N} $ and every $ \omega \in [1_{A_n}\geq \epsilon] $ we have that $\omega$ satisfies $$1_{A_n}(\omega) \geq \epsilon $$ for every $\epsilon \geq 0$. In particular, if we take $ \epsilon > 1 $ then for no $\omega $ can the above inequality be correct, thus $[1_{A_n}\geq \epsilon ] \neq A_n $ since this would require that $1_{A_{n}}=1$.
What is going on here? Much appreciate for any help!
For $0<\epsilon\leq1$ we get: $$1_{A_n}(\omega)\geq\epsilon\iff\omega\in A_n$$ so that: $$\{\omega\in\Omega\mid 1_{A_n}(\omega)\geq\epsilon\}=A_n$$and $$P(\{\omega\in\Omega\mid 1_{A_n}(\omega)\geq\epsilon\})=P(A_n)$$
For $\epsilon>1$ we get:$$\{\omega\in\Omega\mid 1_{A_n}(\omega)\geq\epsilon\}=\varnothing$$so that $$P(\{\omega\in\Omega\mid 1_{A_n}(\omega)\geq\epsilon\})=P(\varnothing)=0$$
So apparantly the following statements are equivalent: