In the context of the question:
"What is the complement of $M:=\left\{x\in\mathbb{R}\mid \liminf\limits_{n\to\infty}f_n(x)\geq f(x)\right\}$?"
I ran into the following problem.
We see that
\begin{align*} &M=\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f_j(x)-f(x)>\frac{1}{m}\right\}\cup\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f(x)-f_j(x)<\frac{1}{m}\right\}. \end{align*}
So by law of De Morgan it follows directly
\begin{align*} &M^c=\left\{x\in\mathbb{R}\mid \liminf\limits_{n\to\infty}f_n(x)< f(x)\right\}\\ &=\left(\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f_j(x)-f(x)>\frac{1}{m}\right\}\cup\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f(x)-f_j(x)<\frac{1}{m}\right\}\right)^c\\ &=\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f_j(x)-f(x)\leq\frac{1}{m}\right\}\cap\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f(x)-f_j(x)\geq\frac{1}{m}\right\}. \end{align*}
If we consider the example $f:\mathbb{R}\to\{0\}$ and $f_n:\mathbb{R}\to\{-1,1\}$ with $f_n(x)=(-1)^n$, then clearly $\liminf\limits_{n\to\infty}f_n(x)=-1<0$ for all $x\in\mathbb{R}$. However, in this case $$ \bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{x\in\mathbb{R}\mid f_j(x)-f(x)\leq\frac{1}{m}\right\}=\emptyset. $$
And this makes no sense. Where is my mistake?