Proof of Linearity and Monotonicity of the Lebesgue Integral

Step 0 (auxiliary facts). (a) Sums and scalar multiples of measurable functions are measurable. For real-valued measurable $f,g$ and $a\in\mathbb{R}$: $\{f+g>a\}=\bigcup_{q}\bigl(\{f>q\}\cap\{g>a-q\}\bigr)$, the union over the countably many rational $q$ (if $f(x)+g(x)>a$, choose rational $q$ with $f(x)>q>a-g(x)$ by Density of the rational numbers in the real numbers); each member is in $\mathcal{F}$, so the union is. For $c>0$, $\{cf>a\}=\{f>a/c\}$; for $c<0$, $\{cf>a\}=\{f<a/c\}$, where $\{f<b\}=X\setminus\{f\ge b\}$ and $\{f\ge b\}=\bigcap_{j\in\mathbb{N}}\{f>b-1/j\}\in\mathcal{F}$; and for $c=0$ the function $cf$ is constant, hence measurable. The same identities give these facts for $[0,\infty]$-valued measurable functions in the sense of Lebesgue Integral of a Nonnegative Measurable Function, with the conventions of Measure, Measure Space, and Probability Measure.

(b) Simple approximation. Every measurable $f:X\to[0,\infty]$ is the pointwise supremum of a nondecreasing sequence of nonnegative simple functions: put

$$s_m(x)=\min\bigl\{m,\;2^{-m}\lfloor 2^{m}f(x)\rfloor\bigr\}$$

where $\lfloor t\rfloor$ denotes the greatest integer $\le t$ and $s_m(x)=m$ when $f(x)=\infty$. Each $s_m$ takes finitely many values and is measurable (its level sets are finite unions of sets of the form $\{a<f\le b\}$ or $\{f>m\}$, all in $\mathcal{F}$ by the identities above), $s_m\le s_{m+1}$, and $\sup_m s_m=f$ pointwise.

(c) Additivity for simple functions. If $s,t$ are nonnegative simple with standard representations on partitions $(A_i)$, $(B_j)$, then on the common refinement $(A_i\cap B_j)$ both are constant, and computing the integral of $s+t$ over the refinement gives, by finite additivity of $\mu$ (Measure, Measure Space, and Probability Measure) and rearrangement of finite sums,

$$\int_X(s+t)\,d\mu=\int_X s\,d\mu+\int_X t\,d\mu,\qquad \int_X cs\,d\mu=c\int_X s\,d\mu\ (c\ge 0),$$

using that the integral of a nonnegative simple function may be computed from any representation over a finite measurable partition on which it is constant (grouping equal values and adding measures).

Claim 1. Measurability of $f+g$ and $cf$ is Step 0(a). Monotonicity was shown in Step 1 of the proof of Monotone Convergence Theorem. For additivity: take simple approximations $s_m\uparrow f$, $t_m\uparrow g$ from Step 0(b); then $s_m+t_m\uparrow f+g$, and by Monotone Convergence Theorem and Step 0(c),

$$\int_X(f+g)\,d\mu=\sup_m\int_X(s_m+t_m)\,d\mu=\sup_m\Bigl(\int_X s_m\,d\mu+\int_X t_m\,d\mu\Bigr)=\int_X f\,d\mu+\int_X g\,d\mu,$$

the last step because both inner sequences are nondecreasing, so the supremum of the sum is the sum of the suprema in $[0,\infty]$. The scalar case is analogous via $cs_m\uparrow cf$ (for $c>0$; $c=0$ is trivial by the convention $0\cdot\infty=0$).

Claim 2. Let $f,g$ be integrable and $a,b\in\mathbb{R}$. $af+bg$ is measurable by Step 0(a), and $|af+bg|\le|a|\,|f|+|b|\,|g|$ pointwise, so by claim 1 and monotonicity $\int|af+bg|\,d\mu<\infty$; hence $af+bg$ is integrable by Integrable Function and the Lebesgue Integral. For additivity with $a=b=1$: write $h=f+g$; from $h^{+}-h^{-}=f^{+}-f^{-}+g^{+}-g^{-}$ (pointwise identity of real numbers) we get

$$h^{+}+f^{-}+g^{-}=h^{-}+f^{+}+g^{+}$$

pointwise, with all six functions nonnegative measurable; claim 1 gives equality of the integrals of the two sides, all terms finite, and rearranging real numbers yields $\int h\,d\mu=\int f\,d\mu+\int g\,d\mu$. For scalars: if $a\ge 0$ then $(af)^{\pm}=af^{\pm}$ and claim 1 applies; if $a<0$ then $(af)^{+}=(-a)f^{-}$ and $(af)^{-}=(-a)f^{+}$, and the definition gives $\int af\,d\mu=a\int f\,d\mu$. Combining, $\int(af+bg)\,d\mu=a\int f\,d\mu+b\int g\,d\mu$.

Finally, $\pm f\le|f|$ pointwise, so by the integrable-case monotonicity — which follows since $g-f\ge 0$ implies $\int g\,d\mu-\int f\,d\mu=\int(g-f)\,d\mu\ge 0$ by additivity and nonnegativity of the integral of a nonnegative function — we get $\pm\int_X f\,d\mu\le\int_X|f|\,d\mu$, i.e. $\bigl|\int_X f\,d\mu\bigr|\le\int_X|f|\,d\mu$. $\blacksquare$