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

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

• $$A_{j,epsilon}= { omega in Omega : vert X_j(omega)-X(omega)vert
• $$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}, 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