Proof of Absolute Continuity of the Integral

Proof of (i). Let $\varepsilon>0$. Since $\int_{X}f\,d\mu<\infty$, this integral is a real number, and so is every integral of an element of $\mathcal{M}^{+}$ below, each being at most $\int_{X}f\,d\mu$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii) and Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (iii), with Basic Properties of the Integral of Nonnegative Measurable Maps (iii), and with Integrals over Measurable Subsets (viii) and Integrals over Measurable Subsets (v) for $\int_{A}f\,d\mu\le\int_{X}f\,d\mu$ when $A\in\mathcal{F}$).

Step 1: a bounded minorant carrying almost all of the integral. For $m\in\mathbb{N}$ let $s_{m}=s_{m}(f)$ be the dyadic approximation of $f$ of order $m$. By Approximation of Measurable Functions by Dyadic Simple Functions (i), $s_{m}\in\mathcal{S}^{+}\subseteq\mathcal{M}^{+}$ (Basic Properties of the Integral of Nonnegative Measurable Maps (i)), $s_{m}(x)\le s_{m+1}(x)\le f(x)$ and $f(x)=\sup_{m}s_{m}(x)$ for every $x\in X$; moreover $s_{m}(x)\le m$ for every $x\in X$, by Dyadic Approximations of a Map into [0,∞] (i). By the monotone convergence theorem,

$$\int_{X}f\,d\mu=\sup_{m}\int_{X}s_{m}\,d\mu .$$

Since $\int_{X}f\,d\mu-\varepsilon/2$ is less than this supremum, there is $N\in\mathbb{N}$ with

$$\int_{X}f\,d\mu-\int_{X}s_{N}\,d\mu<\frac{\varepsilon}{2}$$

(Order and Suprema in the Extended Real Line (x)). First fix such an $N$, which depends only on $f$ and $\varepsilon$; then put

$$\delta=\frac{\varepsilon}{2N}>0$$

($N\ge1$ in $\mathbb{R}$ by Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (ii), so $\delta$ is defined and positive, and $N\ge0$).

Step 2: the estimate. Let $A\in\mathcal{F}$ with $\mu(A)<\delta$, so that $\mu(A)<\infty$ and $\mu(A)$ is a real number (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii)), and let $B=X\setminus A\in\mathcal{F}$. As $A$ and $B$ are disjoint with union $X$, additivity over disjoint sets gives, for $h=f$ and for $h=s_{N}$,

$$\int_{X}h\,d\mu=\int_{A}h\,d\mu+\int_{B}h\,d\mu$$

(Integrals over Measurable Subsets (vii) and Integrals over Measurable Subsets (v)). Writing integrals over a set as integrals of products with its indicator (Integrals over Measurable Subsets (i)), the pointwise bounds $s_{N}\mathbf{1}_{B}\le f\mathbf{1}_{B}$ and $s_{N}\mathbf{1}_{A}\le N\mathbf{1}_{A}$ give

$$\int_{B}s_{N}\,d\mu\le\int_{B}f\,d\mu,\qquad\int_{A}s_{N}\,d\mu\le\int_{X}N\mathbf{1}_{A}\,d\mu=N\mu(A)$$

(Basic Properties of the Integral of Nonnegative Measurable Maps (iii), Basic Properties of the Integral of Nonnegative Measurable Maps (iv) and Basic Properties of the Integral of Nonnegative Measurable Maps (ii)). Hence

$$\int_{A}f\,d\mu=\int_{X}f\,d\mu-\int_{B}f\,d\mu\le\int_{X}f\,d\mu-\int_{B}s_{N}\,d\mu=\Bigl(\int_{X}f\,d\mu-\int_{X}s_{N}\,d\mu\Bigr)+\int_{A}s_{N}\,d\mu<\frac{\varepsilon}{2}+N\delta=\varepsilon .$$

Proof of (ii). Here $|f|\in\mathcal{L}^{1}(\mu)$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x), and $f$ and $|f|$ are integrable on every $A\in\mathcal{F}$ by Integrals over Measurable Subsets (vi), so the integrals in (ii) are defined. Let $\varepsilon>0$. By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv), $|f|\in\mathcal{M}^{+}$ and $\int_{X}|f|\,d\mu<\infty$, so (i), applied to $|f|$ and $\varepsilon$, gives $\delta>0$ such that $\int_{A}|f|\,d\mu<\varepsilon$ for every $A\in\mathcal{F}$ with $\mu(A)<\delta$; this integral of $|f|\in\mathcal{M}^{+}$ over $A$ equals that of $|f|$ as a function integrable on $A$ (Integrals over Measurable Subsets (iii)). For such $A$, let $(A,\mathcal{F}_{A},\mu_{A})$ be the restriction of $(X,\mathcal{F},\mu)$ to $A$, a measure space (Integrals of Functions over a Measurable Subset (i)). Since $\lvert f|_{A}\rvert=\lvert f\rvert|_{A}$, the definition of integrals over $A$ (Integrals of Functions over a Measurable Subset (iv)) and the absolute-value bound in $(A,\mathcal{F}_{A},\mu_{A})$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x)) give

$$\Bigl|\int_{A}f\,d\mu\Bigr|=\Bigl|\int f|_{A}\,d\mu_{A}\Bigr|\le\int\lvert f|_{A}\rvert\,d\mu_{A}=\int_{A}|f|\,d\mu<\varepsilon .$$