Monotone convergence for the integral

By [L1], the numbers $\int f_{n}d\mu$ increase and satisfy[given, L1] $\int f_{n}d\mu \le \int fd\mu$ for every $n$. So their supremum $L$ exists in $[0,+\mathrm{\infty }]$ and $L\le \int fd\mu$.

Fix a nonnegative simple function $s\le f$ and a real $c$ with $0<c<1$.[given, L2, L3] Put $A_n:=\{f_n\ge cs\}$. Then $A_n\uparrow X$: if $s(x)=0$ then $x\in A_n$ for all $n$, while if $s(x)>0$ then $f_n(x)\uparrow f(x)\ge s(x)>cs(x)$, so eventually $x\in A_n$. Since $A\mapsto {\int }_{A}sd\mu$ is a measure by [L2], [L3] gives ${\int }_{A_n}sd\mu \uparrow {\int }_{X}sd\mu \mathrm{.}$

On $A_n$ one has $cs\le f_n$, hence $cs{\chi }_{A_n}\le f_n$. By [L1] and [L4],[step 1.2, L1, L4, algebra] $c{\int }_{A_n}sd\mu =\int cs{\chi }_{A_n}d\mu \le \int f_{n}d\mu \le L\mathrm{.}$ Letting $n\to \mathrm{\infty }$ in step 1.2 yields $c\int sd\mu \le L\mathrm{.}$ Now choose $c_m=1-2^{-m}$ and let $m\to \mathrm{\infty }$; then $\int sd\mu \le L$.

Step 2.1 holds for every simple minorant $s\le f$, so taking the supremum [step 1.1, step 2.1, given] ∎ over such $s$ gives $\int fd\mu \le L$ by the definition of the nonnegative integral. Together with step 1.1, this proves $L=\int fd\mu$, so $\int f_{n}d\mu \uparrow \int fd\mu$.