Absolute Continuity of the Lebesgue Integral

Let $(X,\mathcal{F},\mu)$ be a measure space and let $g:X\to[0,\infty]$ be a measurable function with finite integral, $\int_X g\,d\mu<\infty$. Then for every real $\varepsilon>0$ there exists a real $\delta>0$ such that every $A\in\mathcal{F}$ with $\mu(A)<\delta$ satisfies

where $\mathbf{1}_{A}$ is the indicator function of $A$ as in Simple Function and Its Integral (the product $\mathbf{1}_{A}\,g$ is measurable, since $\{\mathbf{1}_{A}g>u\}=A\cap\{g>u\}$ for $u\ge0$ and $\{\mathbf{1}_{A}g>u\}=X$ for $u<0$).

In particular, let $\lambda$ be Lebesgue measure on the real line with the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$, let $a<b$ be real numbers, and let $g:\mathbb{R}\to[0,\infty]$ be Borel measurable with $\int_{\mathbb{R}}\mathbf{1}_{(a,b]}\,g\,d\lambda<\infty$. Then for every $\varepsilon>0$ there is $\delta>0$ such that all real numbers $s,t$ with $a\le s\le t\le b$ and $t-s<\delta$ satisfy $\int_{\mathbb{R}}\mathbf{1}_{(s,t]}\,g\,d\lambda<\varepsilon$.