A snappy proof of Fatou's lemma

Let $\lambda<1$. For any $k$, denote $$B_k := \Big\{ x \in X ~\Big|~ \forall l \geq k,~ \lambda \liminf_n f_n(x) \leq f_l(x) \Big\}.$$ Then $$ \lambda \int_{B_k} \liminf_n f_n \leq \int_{B_k} f_k \leq \int_X f_k.$$ Take $\liminf_k$ in both side : $$ \lambda \liminf_k \left( \int_{B_k} \liminf_n f_n \right) \leq \liminf_k \int_X f_k.$$

The LHS is equal to $\lambda \int_X \liminf_n f_n$ because:

• $(B_k)_k$ is increasing.

• $\bigcup_k B_k=X$. Indeed for any $x \in X$, either $\liminf_n f_n(x)>0$ in with case $\lambda \liminf_n f_n(x) < f_k(x)$ for $k \gg 1$, or $\liminf_n f_n(x)=0$ in which case it is trivial.

Some property about measure will tell you that $\lim_k \int_{B_k} g = \int_X g$.

In my view Fatou's lemma and MCT are more ore less equivalent. MCT is a particular case of Fatou (with nondecreasing sequences), and ususally Fatou is proved by applying MCT for the sequence $g_n=\inf\{f_n,f_{n+1},\ldots\}$.

If you want a direct proof, and you want to understand what is going on, try the simplest case of Fatou: if $(a_n)$ and $(b_n)$ are sequences of nonnegative numbers then $$ \liminf a_n + \liminf b_n \le \liminf(a_n+b_n). $$

The general case is the same you just have a bit more technical complication.