Monotone convergence for the integral

By [L1], the integrals $\int f_{n}d\mu$ increase and are bounded above by $\int fd\mu$. Write $L={\sup }_n\int f_{n}d\mu$, so $L\le \int fd\mu$ in $[0,+\mathrm{\infty }]$.

Fix a finite-valued nonnegative simple function $s\le f$ and $0<c<1$. Set $A_n=\{f_n\ge cs\}$. The sets $A_n$ increase to $X$: where $s=0$ membership is automatic, and where $s>0$, the limit $f\ge s>cs$ eventually forces $f_n\ge cs$. Since $A\mapsto {\int }_{A}sd\mu$ is a measure [L2], continuity from below [L3] gives ${\int }_{A_n}sd\mu \uparrow \int sd\mu$.

On $A_n$, $cs\le f_n$, hence $cs{\chi }_{A_n}\le f_n$ everywhere. By monotonicity [L1] and simple-integral agreement and homogeneity [L4], $c{\int }_{A_n}sd\mu \le \int f_{n}d\mu \le L$. Letting $n\to \mathrm{\infty }$ in step 1.2 yields $c\int sd\mu \le L$. Write $I=\int sd\mu$. If $L=+\mathrm{\infty }$ there is nothing to prove. If $L<+\mathrm{\infty }$ and $I=+\mathrm{\infty }$, the inequality $cI\le L$ is already impossible for $c=1\mathrm{/}2$. If $I<+\mathrm{\infty }$ and $I>L$, choose a positive integer $m>L\mathrm{/}(I-L)$ by [L5] and set $c_m=m\mathrm{/}(m+1)\in (0,1)$. Then $c_{m}I>L$, since $m(I-L)>L$, again contradicting $c_{m}I\le L$. Thus $I\le L$ in every case. This gives explicitly the elementary approximation $c_m\uparrow 1$, without invoking any additional convergence theorem.

The inequality from step 2.1 holds for every admissible simple minorant $s\le f$. Taking their supremum, as in The nonnegative Lebesgue integral, gives $\int fd\mu \le L$. Combine this with step 1.1 to obtain $\int f_{n}d\mu \uparrow \int fd\mu$.