Take a sequence of binary random variables $(Y_t)_t$ such that
$$ (A) \quad \Pr\Big(\lim_{T\rightarrow +\infty} \frac{1}{T}\sum_{t=1}^T Y_t=\nu\Big)=1. $$
Consider another sequence of discrete random variables $(X_t)_t$ such that, if $Y_t=1$, then $X_t=x$, where $x$ is some element of the support of $X_t$. That is, $$ Y_t\leq 1(X_t=x) \quad \forall t=1,2,\dots $$ where $1(...)$ takes value 1 if the condition inside is satisfied.
Can you help me to show that, by (A), $$ \Pr\bigg(\liminf_{T\rightarrow +\infty} \frac{1}{T}\sum_{t=1}^T 1(X_t=x)\geq \nu\bigg)=1 \quad ? $$
Let $a_n\leq b_n,\forall n$ and assume $a_n\to a$. We know that there exists an $N$ s.t. $a_n\in(a-\varepsilon,a+\varepsilon)$ for all $n\geq N$. So for any $\varepsilon>0$, there exists an $N$ s.t for all $n\geq N$ we have $b_n>a-\varepsilon$. So $\liminf_nb_n\geq a$. Indicate with $\overline{Y}_n,\overline{X}_n$ the two running averages. We have $\overline{X}_n\geq \overline{Y}_n,\forall n$. Now let $\omega\in \{\overline{Y}_n\to \nu\}$. Then by the above we have $\liminf_n\overline{X}_n(\omega)\geq \nu$. So $\{\overline{Y}_n\to \nu\}\subseteq \{\liminf_n\overline{X}_n\geq \nu\}$ and since $P(\overline{Y}_n\to \nu)=1$ we conclude.