Monotone convergence for the measure integral

For every finite real $a$, $\{f>a\}={\bigcup }_n\{f_n>a\}$, since the sequence increases to $f$. These threshold sets generate the order-topology Borel sigma-algebra in [F1], so $f$ is measurable. The simple minorants of $f_n$ are also minorants of $f_{n+1}$, so their integral suprema increase. Each is at most $\int f$; hence $L:={\lim }_n\int f_n$ exists in $[0,\mathrm{\infty }]$ and $L\le \int f$.

Fix a nonnegative simple $s\le f$ and $0<t<1$. The measurable sets $A_n:=\{f_n\ge ts\}$ increase and cover $\{s>0\}$, since $f_n(x)\to f(x)\ge s(x)>ts(x)$ there. Write $s={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{B_j}$ in disjoint level form. For each $c_j>0$, $A_n\cap B_j\uparrow B_j$; countable additivity, applied to successive disjoint differences, gives $\mu (A_n\cap B_j)\uparrow \mu (B_j)$. The finite-sum definition [F2] then gives $\int s{\mathbf{1}}_{A_n}\uparrow \int s$. As $f_n\ge ts{\mathbf{1}}_{A_n}$, monotonicity and homogeneity of the simple integral imply $L\ge t\int s$. Let $t\uparrow 1$ and take the supremum over $s\le f$; [F1, F3] give $L\ge \int f$. Together with step 1.1 this proves monotone convergence.

For nonnegative measurable $f$, define $s_n:=2^{-n}{\sum }_{k=1}^{n{2}^n}{\mathbf{1}}_{\{f\ge k{2}^{-n}\}}$. Each $s_n$ is a finite-valued measurable simple function. Every threshold counted in $s_n$ is also counted at the even-index threshold in $s_{n+1}$, so $s_n\le s_{n+1}$. If $f(x)<\mathrm{\infty }$, once $n>f(x)$ the mesh gives $0\le f(x)-s_n(x)<2^{-n}$; if $f(x)=\mathrm{\infty }$, then $s_n(x)=n$. Thus $s_n\uparrow f$. Define $t_n:=2^{-n}{\sum }_{k=1}^{n{2}^n}{\mathbf{1}}_{\{g\ge k{2}^{-n}\}}$; the same threshold and mesh argument shows $t_n\uparrow g$. By [F4], $f+g$ and $s_n+t_n$ are measurable with the defined nonnegative extended addition. Apply the monotone-convergence conclusion of step 2.1 to $s_n$, $t_n$, and $s_n+t_n$. Refine their finitely many level sets to the finite partition of intersections $B_j\cap C_k$; regrouping the resulting sums by [F2] gives $\int (s_n+t_n)=\int s_n+\int t_n$. Passing to the three monotone limits proves $\int (f+g)=\int f+\int g$.

For $c>0$, simple minorants of $cf$ are exactly $c$ times simple minorants of $f$, so the supremum definition gives $\int cf=c\int f$; for $c=0$ both sides are zero. If an integrable real function is expressed as $u-v$ with $u,v\ge 0$ integrable, the value $\int u-\int v$ is independent of the expression: $u-v=u^{\mathrm{\prime }}-v^{\mathrm{\prime }}$ implies $u+v^{\mathrm{\prime }}=u^{\mathrm{\prime }}+v$, and step 3.1 equates the finite sums of integrals. The definition using positive and negative parts is one such expression, so signed integrals are linear.