$X_n xrightarrow{p} X Rightarrow existshspace{0.2em} X_{n_j} xrightarrow{a.s.}X$

i need a check about this demonstration

suppose that $X_n xrightarrow{p} X$ and take some decreasing ${epsilon_m}_{m in mathbb{N}}$ who satisfy $epsilon_mrightarrow0$ and define:

  • $A_{j,epsilon}= { omega in Omega : vert X_j(omega)-X(omega)vert <epsilon}$
  • $S_{n,epsilon}= bigcap_{j geq n} A_{j,epsilon}$
  • $S_{epsilon}=bigcup_{n in mathbb{N}}S_{n,epsilon}=liminf A_{j,epsilon}$
  • $S:= bigcap_{m in mathbb{N}}S_{epsilon_m}$

is easy to demonstrate that $S ={ omega in Omega: lim_{jto infty}X_j(omega) = X(omega)}  $

now fixing $m$ i can build an increasing $n : mathbb{N} to mathbb{N}$ who satisfy $$mathbb{P}(vert X_{n_j}-Xvert>epsilon_m)leq2^{-j}, forall jin mathbb{N}$$

now because $sum_{j=1}^{infty}mathbb{P}(vert X_{n_j}-Xvert>epsilon_m)leqsum_{j=1}^{infty}2^{-j}<infty$, by borel-cantelli lemma

$$mathbb{P}(limsup A_{j,epsilon_m})=0 Rightarrow mathbb{P}(S_{epsilon_m})=1$$
so this demonstate that $mathbb{P}(S_{epsilon_m})=1$ do not depend by $m $

now $mathbb{P}(bigcap_{m=1}^{infty}S_{epsilon_m})=lim_{mtoinfty}mathbb{P}(S_{epsilon_m})=1$, this is because $$epsilon < epsilon’ Rightarrow S_{epsilon} subset S_{epsilon’} Rightarrow S_{epsilon_{m+1}}subset S_{epsilon_{m}}, forall m in mathbb{N}$$

i think that all i used for demonstrate this theorem is legal