TITLE: Subadditivity of the limit superior
QUESTION [5 upvotes]: $$ \limsup \left(f(h)+g(h)\right) \leq \limsup f(h)+ \limsup g(h).$$
How can we prove this? Any help would be appreciated.

REPLY [8 votes]: The general statement
$$
A\subset B\Rightarrow\sup\limits_{x\in A}f(x)\le\sup\limits_{x\in B}f(x)\tag{1}
$$
seems intuitively obvious because the proof is very simple:

Suppose that $a=\sup\limits_{x\in A}f(x)$, then for any $\epsilon>0$, there is an $x_\epsilon\in A$ so that $f(x_\epsilon)>a-\epsilon$. Since $A\subset B$, $x_\epsilon\in B$ and we have that $\sup\limits_{x\in B}f(x)>a-\epsilon$. Because $\epsilon$ was arbitrary, we get that
  $$
\sup_{x\in B}f(x)\ge a=\sup_{x\in A}f(x)
$$

Let $A_m=\{n\in\mathbb{Z}:n>m\}$, then $\sup\limits_{n>m}f(n)=\sup\limits_{n\in A_m}f(n)$.
Note that
$$
\sup_{n\in A_m}f(n)+\sup_{n\in A_m}g(n)=\sup_{(n_1,n_2)\in A_m\times A_m}f(n_1)+g(n_2)\tag{2}
$$
and
$$
\sup_{n\in A_m}f(n)+g(n)=\sup_{\substack{(n_1,n_2)\in A_m\times A_m\\n_1=n_2}}f(n_1)+g(n_2)\tag{3}
$$
and
$$
\{(n_1,n_2)\in A_m\times A_m:n_1=n_2\}\subset A_m\times A_m\tag{4}
$$
$(1)$ and $(4)$ say that $(3)\le(2)$, that is
$$
\sup_{n>m}f(n)+g(n)\le\sup_{n>m}f(n)+\sup_{n>m}g(n)\tag{5}
$$
Now take the $\lim\limits_{m\to\infty}$ of both sides of $(5)$ to get
$$
\limsup_{n\to\infty}f(n)+g(n)\le\limsup_{n\to\infty}f(n)+\limsup_{n\to\infty}g(n)\tag{6}
$$