Does sequence almost sure convergence imply almost sure convergence?

This is a cross-post of this and this questions from math.stackexchange.com since I have not received any response there. I would like to seek help here.

Suppose $x(t,\omega): [0,T]\times\Omega\rightarrow \mathbf R$ is a random variable on a probability space $\Omega$.

1. Sequence $(t_k<2^{-k})_{k=1}^\infty \implies \lim_\limits{i\rightarrow\infty}x(t_i,\omega)\rightarrow0$ for almost all $\omega\in\Omega$. Does this imply $\lim_\limits{t\rightarrow0}x(t,\omega)=0$ for almost all $\omega\in\Omega$? What if we assume $\lim_\limits{t\rightarrow0}x(t,\omega)=0$ in probability?

2. For every sequence $(t_k)_{k=1}^\infty\rightarrow 0$, $\exists$ a subsequence $(t_{k_i})_{i=1}^\infty$ independent of $\omega$ $\ni\big(\lim_\limits{i\rightarrow\infty}x(t_i,\omega)\rightarrow0$ for almost all $\omega\in\Omega\big)$. Does this imply $\lim_\limits{t\rightarrow0}x(t,\omega)=0$ for almost all $\omega\in\Omega$? What if we assume $\lim_\limits{t\rightarrow0}x(t,\omega)=0$ in probability?

• This is very closely related to your questions on math.stackexchange.com: math.stackexchange.com/questions/2801027/… and math.stackexchange.com/questions/2800921/… – Anthony Quas May 29 '18 at 22:13
• @AnthonyQuas: Yes it is. I hope I have more luck getting an answer here as I have not received any response there. Do you have any idea regarding the questions? – Hans May 29 '18 at 22:20
• This isn't cool. You only posted there 4 hours ago. At the very least, you should indicate on your post that this is a cross-posting. Also, you should wait longer before cross-posting. – Anthony Quas May 29 '18 at 22:21
• @AnthonyQuas: I have now indicated as much. – Hans May 29 '18 at 22:28

No. A counterexample for all of your questions is as follows. Let $\Omega$ be $[0,1)$, with probability measure $\mathbb P$ being Lebesgue measure. Set $x(t,\omega)=1$ if the fractional part of $1/t$ is $\omega$ and 0 otherwise.