Are the following statements equivalent?
$$a) X(t)/t\xrightarrow{a.s} c $$ $$b) X(t)\xrightarrow{a.s} t c $$
where $c$ is a constant and $X(t)$ is a sequence of random variable.
By definition, above statements are equivalent to the followings $$a) P(\lim_{t \to \infty}X(t)/t=c)=1$$ $$b) P(\lim_{t \to \infty}X(t)=tc)=1$$
which are equivalent to: $$a) P(\omega \in \Omega: \forall \epsilon>0, \,\exists T \, \mid \forall t>T,|X(t,\omega)/t-c|<\epsilon)=1$$ $$b) P(\omega \in \Omega: \forall \epsilon>0, \,\exists T \, \mid \forall t>T,|X(t,\omega)-tc|<\epsilon)=1$$
Now, they are equivalent. Right?
Let's assume without any loss of generality that $t$ is a discrete index, that is, $X_t$ is a random variable on some probability space $(\Omega, \mathcal F, \mathbb P)$ for each positive integer $t$. The statement $$\mathbb P\left(\lim_{t\to\infty}\frac{X_t}t = c \right)=1 $$ means that there exists an event $\Omega_0\in\mathcal F$ with $\mathbb P(\Omega_0)=1$ such that $$\lim_{t\to\infty}\frac{X_t(\omega)}t = c $$ for all $\omega\in\Omega_0$.
In order to make sense of the ill-defined statement $X_t \stackrel{\mathrm{a.s.}}\longrightarrow tc$, we would first have to make sense of $\lim_{t\to\infty} X_t(\omega)=tc$ for a fixed $\omega\in\Omega$ - which itself is not a well-defined statement.