Can anyone give a counterexample to prove the following statement?
The existence of a further a.s. convergent subsequence of $(_n)_{n\in \mathbb N}$ 's every subsequence, do not imply a.s convergence of $(_n)_{n\in \mathbb N}$.
While there is no equivalence for a.s. convergence, there are for convergence in probability or $L^p$.
On
If a sequence converges in probability then any subsequence has a further subsequence which is a.s. convergent. But there are simple examples of sequences which converge in probability but not a.s. One can be obtained by linearly indexing the "moving block", given by $X_{n,k}(\omega)=\begin{cases} 1 & \omega \in [k2^{-n},(k+1)2^{-n}] \\ 0 & \text{otherwise} \end{cases}$,where $n=1,2,\dots$ and $k=0,1,\dots,2^n-1$, and the underlying probability space is $[0,1]$ with the Lebesgue measure.
Suppose Y is a uniform distribution on $\Omega=[0,1]$. We note $S_n=\sum_{i=2}^{n}\frac{1}{i}$.
$A_n=[S_n mod 1,S_{n+1} mod 1]$ if $S_n mod 1<S_{n+1} mod 1$
or
$A_n=[S_n mod 1,1]\cup [0,S_{n+1} mod 1]$ if $S_n mod 1>S_{n+1} mod 1$
We define $X_n=1-(1-\frac{1}{n})\chi_{A_n}$.
We find that $X_n$ converge in probability to 1, while it does not converge a.s. to 1. Since with measure 1, $X_n(\omega)$ do not converge.