For independent r.v. the $\liminf X_n$ is almost surely constant I need to show that for independent real-valued r.v. $(X_n)_{n\in \mathbb{N}}$ the $X_*:= \liminf_\limits{n\rightarrow \infty}X_n$ is almost surely constant with $X_*\in\mathbb{R}\cup\{-\infty,+\infty\}$.
A r.v. $X$ is constant iff $X(\omega)=c$ for $\omega \in \Omega$ and $c\in\mathbb{R}\cup\{-\infty,+\infty\}$. This means that I need to show that
$$
\exists c\in\mathbb{R}\cup\{-\infty,+\infty\}: P(\liminf_\limits{n\rightarrow \infty}X_n=c)=1
$$
We know that for the r.v. on $(\Omega,\mathcal{F})$ and $ \sigma(X_k:k\geq n):=\sigma((X^{-1}B_k: B_k\in \mathcal{F}_k, k\geq n)$ the Terminal $\sigma$-Algebra is defined as
$$
\mathcal{A}(X_k:k\geq 1):=\bigcap_{n\geq 1}\sigma(X_k:k\geq n)
$$
I further know that for Events $A_n$
$$
\liminf_\limits{n\rightarrow \infty}A_n:=\bigcup_{m\geq1}\bigcap_{k\geq m}A_k
$$
I am not quite sure from where to "attack" this problem. If $c\in\mathbb{R}$ than $\{X_n=c\} = \{\omega \in \Omega:X_n(\omega)=c\}:=A_n$ and since $X_n$ are independent $A_n$ are also independent. It is also clear to me that $\liminf_\limits{n\rightarrow \infty}A_n \in \sigma(1_{A_n}:k\geq n).$ Where do I go from here?
Since $A_n$ are independent I know that
$$
\liminf_\limits{n\rightarrow \infty}A_n:=\bigcup_{m\geq1}\bigcap_{k\geq m}A_k= \bigcup_{m\geq1}\sum_{k\geq m}A_k
$$
I don't know how to go further. Doesn't Kolmogorov's $0$-$1$-law tell me that $P(\liminf_\limits{n\rightarrow \infty}X_n=c)=0$ or $P(\liminf_\limits{n\rightarrow \infty}X_n=c)=1$? But how do I show the latter?

The event $\{X_{*} \leq x\}$ belongs to $\sigma (X_k,X_{k+1},...)$ for each $k$. By 0-1 law $P(X_{*} \leq x) =0$ or $1$ for each $x$. The distribution function $F$  take only the values $0$ and $1$ and so it jumps from $0$ to $1$ at some point $c$. ($c=\sup \{t: F(t) =0\}$). It follows that $X_{*}=c$ with probability $1$. 
