TITLE: Why changing $\liminf$ by $\limsup$?
QUESTION [5 upvotes]: In a proof (of the dominated convergence theorem) I found the following:
$$
\liminf_{n\to\infty}\int (2g-h_n)\, d\mu=\int 2g\, d\mu-\limsup_{n\to\infty}\int h_n\, d\mu.
$$

I do not understand where the $\limsup_n$ comes from!
To my opinion it is
$$
\liminf_{n\to\infty}\int (2g-h_n)\, d\mu=\lim\inf_{n\to\infty}\int 2g\, d\mu-\liminf_{n\to\infty}\int h_n\, d\mu\\=\int 2g\, d\mu-\liminf_{n\to\infty}\int h_n\, d\mu.
$$

REPLY [13 votes]: Recall that, for any set $S\subset\mathbb{R}$,
$$\inf(-S)=-\sup(S).$$
(For example, consider $S=\{0,1\}$.)
The negative sign interchanges $\inf$'s and $\sup$'s.