Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space.
Define an rv such that $X_n=1$ with $\mathbb{P}(X_n=1)=1$ and 0 with $\mathbb{P}(X_n=0)=0$ for all n.
Then, $X_n\rightarrow 1$ in $\mathbb{P}$
I am confused as to how to prove this. I am aware that we must show that
$\mathbb{P}(|X_n-1|>\epsilon)\rightarrow 0$ and $n\rightarrow \infty$ but i'm having issues in constructing suitable cases.
Attempt:
Let $0<\epsilon<1$ then
$\{$ $X_n=0$ $\}$ $=$ $\{$ $|X_n-1|>\epsilon$ $\}$
thus, $\mathbb{P}(X_n=0 )=\mathbb{P}(|X_n-1|>\epsilon)=0$
is this correct?
If you ignore $\mathbb{P}(X_n=0)$ then your proof could become:
For all $\epsilon >0$ you have
$0 \le \mathbb{P}(|X_n-1|>\epsilon) \le \mathbb{P}(|X_n -1| \not=0) = 1-\mathbb{P}(X_n=1)=1-1=0$
so $\mathbb{P}(|X_n-1|>\epsilon) =0$