I am reading Royden's proof on Fatou's lemma.
Let ${f_n}$ be a sequence of nonnegative measurable functions on $E$ converging to $f$ pointwise almost everywhere on $E$, then $\int_E f \leq \liminf\int_E f_n$.
In the very last line of Royden's proof, it says
By the definition of the integral of $f_n$ over $E$, $\int_E h_n \leq \int_E f_n$ (where $h_n = \min (h , f_n)$ and $h$ is a bounded measurable function with finite support with $h \leq f$) Thus $$\int_E h = \lim \int_E h_n \leq \liminf \int_E f_n.$$
And I don't see how the last statement is implied, how does he apply the first line to imply the last? Just because $\int_E h_n \leq \int_E f_n$ doesn't tell me that $\inf \int_E f_n$ has to be greater than any particular $\int h_n$. How do we know that $\int_E f_{n+1}$ isn't smaller than $\int_E h_n$?
You can see this in two steps: Since $$ \int_E h_n \; \mathrm dx \leq \int_E f_n \; \mathrm dx \quad \text{for all } n \in \mathbb N $$ we get $$ \int_E h \; \mathrm dx = \lim_{n \to \infty} \int_E h_n \; \mathrm dx \leq \int_E f_n \; \mathrm dx \quad \text{for all } n \in \mathbb N \; . $$ By taking the $\liminf$ on the right-hand side, we finally get $$ \int_E h \; \mathrm dx \leq \sup_{n \geq 0} \inf_{m \geq n} \int_E f_m \; \mathrm dx = \liminf_{n \to \infty} \int_E f_n \; \mathrm dx \; . $$
Edit: The above explanation is not fully correct. Just take the $\liminf$ on both sides of line 1. Then we get $$ \int_E h \; \mathrm dx = \lim_{n \to \infty} \int_E h_n \; \mathrm dx = \liminf_{n \to \infty} \int_E h_n \; \mathrm dx \leq \liminf \int_E f_n \; \mathrm dx \; . $$