Doubt in Dominated Convergence theorem

Question

From Browder's Mathematical Analysis

On applying Fatou's Lemma to sequence ${(g \pm f_n)}$ we get $\int \liminf (g \pm f_n) \leq \liminf \int(g \pm f_n)$.

My question is how they got as $\int (g \pm f) \leq \liminf \int (g \pm f_n)$.

Is $\int \liminf (g \pm f_n)=\int g \pm f$,How?

Thanks in advance!

Answer 1

Recall (or try to prove) some basic facts about numerical sequences:

• for any numeric convergent sequences $a_n$ and $b_n$, $\lim_n(a_n+b_n)=\lim_na_n+\lim_nb_n$
• if $c=\liminf_nc_n:=\sup_n\inf_{m\geq n}c_m$, then there is a subsequence $n_k$ along which $\lim_k c_{n_k} = c$.
• For any nonempty numeric subset $-\sup A=\inf(-A)$, where $-A=\{-a:\in A\}$. From this, it follows that $-\limsup_m d_n=\liminf_n(-d_n)$
• If $a_n$ converges, then $\liminf_n(a_n+b_n)=\lim_na_n +\liminf_nb_n$ and $\limsup_n(a_n+b_n)=\lim_na_n+\limsup_n b_n$.
• In the case that concerns you, for each $x$, there is a subsequence $n^{(x)}_n$ such that $\lim_k \big(-f_{n^{(k)}_k}(x)\big)=\liminf_n\big(-f_n(x)\big)$.