Proof of Absolute Continuity of the Lebesgue Integral

Step 1 (Truncation). For each natural number $n\ge1$ define $g_n:X\to[0,\infty]$ by $g_n(x)=\min(g(x),n)$. Each $g_n$ is measurable in the sense of that definition: for real $a$, $\{g_n>a\}=\{g>a\}$ if $a<n$ and $\{g_n>a\}=\emptyset$ if $a\ge n$. The sequence $(g_n)_n$ is nondecreasing pointwise with $\sup_n g_n(x)=g(x)$ for every $x$ (if $g(x)<\infty$ then $g_n(x)=g(x)$ for all $n\ge g(x)$; if $g(x)=\infty$ then $g_n(x)=n\to\infty$). By the monotone convergence theorem, $\int_X g_n\,d\mu\to\int_X g\,d\mu<\infty$.

Step 2 (Small tail). Define $h_n:X\to[0,\infty]$ by $h_n(x)=g(x)-g_n(x)$, with the convention $\infty-n=\infty$; then $h_n$ is measurable in the same sense, since for $a\ge0$ we have $\{h_n>a\}=\{g>n+a\}$ and for $a<0$ we have $\{h_n>a\}=X$. Pointwise $g=g_n+h_n$, so by additivity of the integral for nonnegative measurable functions (Linearity and Monotonicity of the Lebesgue Integral),

All three integrals are finite (the first is finite by hypothesis and the others are dominated by it, using monotonicity from Linearity and Monotonicity of the Lebesgue Integral), so $\int_X h_n\,d\mu=\int_X g\,d\mu-\int_X g_n\,d\mu\to0$ by Step 1.

Step 3 (Conclusion of the main claim). Let $\varepsilon>0$. Choose $n\ge1$ with $\int_X h_n\,d\mu<\varepsilon/2$ and set $\delta=\varepsilon/(2n)$. Let $A\in\mathcal{F}$ with $\mu(A)<\delta$; the product $\mathbf{1}_{A}\,g$ is measurable by the computation recorded in the statement. We claim the pointwise bound

Indeed, if $g(x)\le n$ then $\mathbf{1}_A(x)g(x)\le n\mathbf{1}_A(x)$ and $h_n(x)=0$; if $g(x)>n$ then $g_n(x)=n$, so $\mathbf{1}_A(x)g(x)\le g(x)=n+h_n(x)\le n\mathbf{1}_A(x)+h_n(x)$ when $x\in A$, while for $x\notin A$ the left side is $0$. By monotonicity and additivity (Linearity and Monotonicity of the Lebesgue Integral), and since $n\mathbf{1}_A$ is a simple function with integral $n\,\mu(A)$,

Step 4 (Interval form). Now let $\lambda$, $a<b$, and $g$ be as in the second claim. The function $g'=\mathbf{1}_{(a,b]}\,g$ is measurable ($\{g'>u\}=(a,b]\cap\{g>u\}$ for $u\ge0$, and $\{g'>u\}=\mathbb{R}$ for $u<0$; the interval $(a,b]$ is a Borel set) and has finite integral by hypothesis. Apply the main claim on the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ to $g'$ and $\varepsilon$, obtaining $\delta>0$. For $a\le s\le t\le b$ with $t-s<\delta$, the set $A=(s,t]$ is Borel with $\lambda(A)=t-s<\delta$ by Existence of Lebesgue Measure on the Real Line (Lebesgue measure assigns to an interval its length). Since $(s,t]\subseteq(a,b]$, we have $\mathbf{1}_{(s,t]}\,g=\mathbf{1}_{A}\,g'$ pointwise, and therefore