What @brom writes is essentially true:
Take some sequence $t_n \to \infty$ such that $\lim_n \int f_{t_n} d\mu \to \liminf_{t \to \infty} f_t d\mu$.
We then have $\liminf_{ t \to \infty} f_t \leq \liminf_n f_{t_n}$ and hence we have that
$$
\int \liminf_{t \to \infty} f_t d\mu \leq \int \liminf_n f_{t_n} \leq \liminf_n \int f_{t_n} d\mu = \liminf_{t \to \infty}\int f_t d\mu,
$$
where we used the "ordinary" Fatou. The above is true if $\liminf_t f_t$ is measurable.
But to ensure this, we have to place additional assumptions. For example, let $V \subset \Bbb{R}$ be a non-measurable set. Define
$$
f_t (x) := -\chi_{V \times \Bbb{N}} (x,t).
$$
Then $(x,t) \mapsto f_t (x)$ is Lebesgue-measurable (because $V \times \Bbb{N}$ is a null-set), but we easily see that
$$
\liminf_{t \to \infty} f_t(x) = -\chi_V (x),
$$
which is not measurable. I am not completely certain wether one can construct a similar counterexample in which $(x,t) \mapsto f_t (x)$ is even Borel-measurable.
For the dominated convergence theorem, the situation is easier. Let us assume that $\lim_{t \to \infty} f_t(x) \to f(x)$ for almost every $x$ and where each $f_t$ is measurable and that $|f_t (x)| \leq g(x)$ with $g \in L^1$.
Then let $t_n \to \infty$ be any sequence converging to $\infty$. This implies that $f(x) = \lim_n f_{t_n}(x)$ is measurable and ordinary dominated convergence yields
$$
\lim_n \int f_{t_n} d\mu = \int f d\mu.
$$
As this holds for arbitrary $t_n \to \infty$, we derive
$$
\lim_{t \to \infty} \int f_{t}d\mu = \int f d\mu.
$$
For monotone convergence, the argument is essentially the same (again without the measurability problem).
EDIT: I have now figured out the behaviour for Borel-measurable functions.
Note that the following uses some facts from descriptive set theory,
in particular about analytic sets. Also, I only prove my claims in the setting of $\Bbb{R}^2$, but the proof should generalize easily to general polish spaces.
In general, $\liminf_{t\to\infty}f_{t}\left(x\right)$ will not be
Borel measurable, even if $\left(x,t\right)\mapsto f_{t}\left(x\right)$
is. To see this, it suffices to show the claim for $\limsup$ instead
of $\liminf$. To this end, let $A\subset\left[0,\frac{1}{2}\right]^{2}$
be a Borel-set for which the projection $\pi_{1}\left(A\right)\subset\left[0,\frac{1}{2}\right]$
is not a Borel-set. Here, $\pi_{1}\left(x,y\right):=x$. The existence
of such a set is a well-known fact in descriptive set theory. Then
$$
B:=\bigcup_{n\in\mathbb{N}}A+\left(0,n\right)\subset\left[0,\frac{1}{2}\right]\times\mathbb{R}
$$
is a Borel-set. Define $f_{t}\left(x\right):=\chi_{B}\left(x,t\right)$.
For $x\in\left[0,\frac{1}{2}\right]$, there are now two possibilities:
We have $x\in\pi_{1}\left(A\right)$, i.e. there is some $y\in\left[0,\frac{1}{2}\right]$
such that $\left(x,y\right)\in A$. This implies $\left(x,y+n\right)\in B$
for all $n\in\mathbb{N}$ and hence
$$
\limsup_{t\to\infty}f_{t}\left(x\right)\geq\lim_{n}f_{y+n}\left(x\right)=1.
$$
We have $x\notin\pi_{1}\left(A\right)$, i.e. $\left(x,y\right)\notin A$
for all $y\in\mathbb{R}$. If we have $f_{t}\left(x\right)>0$ for
some $t\in\mathbb{R}$, this would imply $\left(x,t\right)\in A+\left(0,n\right)$
for some $n$ and hence $\left(x,t-n\right)\in A$, a contradiction.
Hence,
$$
\limsup_{t\to\infty}f_{t}\left(x\right)\leq0.
$$
Now define $g\left(x\right):=\limsup_{t\to\infty}f_{t}\left(x\right)$
for $x\in\mathbb{R}$. The above implies that
$$
g^{-1}\left(\left(\frac{1}{2},\infty\right)\right)=\pi_{1}\left(A\right)
$$
is not Borel-measurable.
But $g=\limsup_{t\to\infty}f_{t}$ will be Lebesgue-measurable. To
see this, first note
$$
g\left(x\right)=\lim_{n\to\infty}\sup_{t\geq n}f_{t}\left(x\right),
$$
so that it suffices to show that $g_{n}:=\sup_{t\geq n}f_{t}$ is
Lebesgue-measurable for each $n\in\mathbb{N}$. To see this, let $\alpha\in\mathbb{R}$
be arbitrary and $f\left(x,t\right):=f_{t}\left(x\right)$. We have
\begin{eqnarray*}
g_{n}^{-1}\left(\left(\alpha,\infty\right]\right) & = & \left\{ x\in\mathbb{R}\,\mid\,\exists t\geq n:\: f_{t}\left(x\right)>\alpha\right\} \\
& = & \left\{ x\in\mathbb{R}\,\mid\,\exists t\in\mathbb{R}:\:\left(x,t\right)\in f^{-1}\left(\left(\alpha,\infty\right]\right)\cap\left(\mathbb{R}\times\left[n,\infty\right)\right)\right\} \\
& = & \pi_{1}\left(f^{-1}\left(\left(\alpha,\infty\right]\right)\cap\left(\mathbb{R}\times\left[n,\infty\right)\right)\right).
\end{eqnarray*}
But $f$ is Borel-measurable by assumption, so that the set above
is an analytic set. But from descriptive set theory, we know that
all analytic sets are Lebesgue measurable, so that each $g_{n}$ is
Lebesgue measurable
