Inequality for Limit inferior I was wondering if anyone can give a  mathematical explanation for the claim below. 
If $F(x-\epsilon)\le F_n(x)+p_n$ where  $\lim_{n\to\infty}p_n=0.$  (1)
Then $F(x-\epsilon)\le \liminf\limits_{n\to\infty}F_n(x)$. (2)
My intuition says it should be based on the definition of $\lim\inf$ of a  sequence which is defined as $\lim\limits_{n\to\infty}\inf\limits_{k\ge n}x_k$.
As my first attempt, I took the limit inferior on both sides of eq. (1). The left hand side doesn't change as it is a constant, and since $\lim_{n\to\infty}p_n=0$, therefore the limit inferior is also $0$. Hence we can justify the claim. But I am not sure if this is the right approach. Could anyone provide any help/suggestion?
 A: use the fact that
$$
a_n\leqslant b_n\implies \liminf a_n\leqslant \liminf b_n \quad\text{and}\quad\liminf c= c
$$
We prove that if $\lim b_n$ exists, then
$$
\liminf (a_n+b_n)= \liminf a_n+\lim b_n \tag 1
$$
Let $a_{n_k}$ be a sequence that 
$$
\liminf_{n\to\infty} a_n=\lim_{k\to\infty} a_{n_k}\tag 2
$$
Since $\lim b_n$ exists
$$
\liminf_{n\to\infty} b_n=\lim_{n\to\infty} b_n=\lim_{k\to\infty} b_{n_k}\tag 3
$$
So $a_{n_k}+b_{n_k}\to\liminf (a_n+b_n)$.
And by $(2)$ and $(3)$
$$
\liminf_{n\to\infty} (a_n+b_n)=\lim_{k\to\infty} (a_{n_k}+b_{n_k})=\lim_{k\to\infty} a_{n_k}+\lim_{k\to\infty} b_{n_k}=\liminf_{n\to\infty} a_n+\lim_{n\to\infty} b_{n}
$$
So we have $(1)$ hold. For each $x$ we have 
\begin{align}
F(x-\epsilon)&=\liminf_{n\to\infty} F(x-\epsilon) 
\\
&\leqslant \liminf\limits_{n\to\infty}(F_n(x)+p_n)
\\
&= \liminf\limits_{n\to\infty}F_n(x)+ \lim\limits_{n\to\infty}p_n
\\
&=\liminf\limits_{n\to\infty}F_n(x)
\end{align}
A: In general, we have only:
$$\liminf\bigl( F_n(x)+p_n(x)\bigr)\ge \liminf F_n(x)+\liminf p_n(x),$$
 which would not allow to conclude. However, if one of the sequences has a limit, the inequality becomes an equality.
