Taking the average of the indicator function and the positivity of measure

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Let $G$ be a Lie group and $\Gamma$ be a lattice in $G$. Let $\mu$ be the $G$-invariant Haar probability measure on $X:=G/\Gamma$. Let $f$ be a real, positive continuous function on $X$ and suppose the set $F=\{x\in X: f(x)<\delta \}$ has positive measure.

Let $(g_t)$ be a one-parameter subgroup of $G$ and assume that for $\mu$-a.e. $x\in X$, $$\lim_{T\to \infty}\frac{1}{T} \int_0^T \textbf{1}_F(g_t.x) dt >0.\tag{1}$$

Can we conclude from here that $$\mu\{x\in X:\liminf_{t\to \infty} f(g_t.x) < \delta\}>0 \tag{*}.$$

The hard part about this problem is how to relate the set $F_{g_t}:=\{x\in X:\liminf_{t\to \infty}f(g_t.x) < \delta\}$ to the set $F=\{x\in X: f(x)<\delta \}$ (the latter has positive measure by assumption).


Although $(*)$ is what I need, I believe the following stronger statement might be true:

$$\mu\{x\in X:\limsup_{t\to \infty} f(g_t.x) < \delta\}>0 \tag{**}.$$

The intuition comes from that (1) implies for a.e. $x\in X$, its $g_t$ orbit visits the set $F$ "infinitely often" and stays there for a "positive proportion of time" in large time scale.

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This is a silly question, from (1) one has for any $T_0>0$,

$$\lim_{T\to \infty}\frac{1}{T} \int_{T_0}^T \textbf{1}_F(g_t.x) dt >0.\tag{2}$$

which implies that in particular for a.e. $x\in X$, there exists $t>t_0$ such that $g_t.x$ falls into $F$ and therefore $$\limsup_{t\to \infty} f(g_t.x)<\delta.$$

Intersection this full-measure set with the set $F$, we obtain $(**)$.