Proof of Linearity and Monotonicity of the Lebesgue Integral

Step 0 (auxiliary facts). (a) Sums and scalar multiples of measurable functions are measurable. By claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line, a function $h:X\to\mathbb{R}$ is measurable if and only if $\{h>a\}\in\mathcal{F}$ for every real $a$; we use this criterion in both directions. 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 rational $q$, which are countably many by The Integers and the Rational Numbers are Countable (if $f(x)+g(x)>a$, choose rational $q$ with $f(x)>q>a-g(x)$ by claim 1 of The Rational Numbers are Dense 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 $\{f\ge b\}=\bigcap_{j\in\mathbb{N}}\{f>b-1/j\}\in\mathcal{F}$, the inclusion $\supseteq$ by The Archimedean Property of the Real Numbers; 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 $c\ge0$ and the conventions of Measure, Measure Space, and Probability Measure. The same intersection gives $\{f\ge b\}\in\mathcal{F}$ for such $f$ and real $b$. In the sum identity the choice of $q$ needs two extra cases. If $f(x)+g(x)>a$ and $g(x)=\infty$, take $q=-1$, which is less than $f(x)\ge0$. If $f(x)=\infty$ and $g(x)$ is real, take a rational $q$ with $a-g(x)<q<a-g(x)+1$, by claim 1 of The Rational Numbers are Dense in the Real Numbers. Conversely, $f(x)>q$ and $g(x)>a-q$ give $f(x)+g(x)>a$ in every case.

(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 its level sets are the sets $\{j2^{-m}\le f<(j+1)2^{-m}\}$ for $0\le j<m2^{m}$ and $\{f\ge m\}$. Each of these lies in $\mathcal{F}$, because $\{f\ge b\}\in\mathcal{F}$ for real $b$ and $\{f<b\}=X\setminus\{f\ge b\}$. Hence $s_m^{-1}(B)$, a finite union of level sets, lies in $\mathcal{F}$ for every Borel set $B$, so $s_m$ is measurable. Further, $s_m\le s_{m+1}$, because $2^{-m}\lfloor2^{m}t\rfloor\le2^{-m-1}\lfloor2^{m+1}t\rfloor\le t$ for real $t\ge0$. And $\sup_m s_m=f$ pointwise: if $f(x)$ is real, then $f(x)-2^{-m}<s_m(x)\le f(x)$ once $m>f(x)$; if $f(x)=\infty$, then $s_m(x)=m$.

(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). For monotonicity, suppose $f\le g$. Every nonnegative simple function $s$ with $s\le f$ also satisfies $s\le g$. So the set of values whose supremum defines $\int_X f\,d\mu$ in Lebesgue Integral of a Nonnegative Measurable Function is contained in the set defining $\int_X g\,d\mu$, and hence $\int_X f\,d\mu\le\int_X g\,d\mu$. 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). Since $|f|=f^{+}+f^{-}$, claim 1 gives $\int|f|\,d\mu=\int f^{+}d\mu+\int f^{-}d\mu<\infty$, and likewise for $g$. Pointwise $|af+bg|\le|a|\,|f|+|b|\,|g|$, so claim 1 (monotonicity, additivity and homogeneity) gives $\int|af+bg|\,d\mu<\infty$. Writing $h=af+bg$, claim 1 again gives $\int|h|\,d\mu=\int h^{+}d\mu+\int h^{-}d\mu$, so both integrals on the right are finite, and $h$ 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. $\bigl|\int_X f\,d\mu\bigr|\le\int_X|f|\,d\mu$. $\blacksquare$