[proofplan] We verify the epsilon-delta definition of [absolute continuity](/page/Absolute%20Continuity) on $[a,b]$. The only analytic input is the absolute continuity property of the [Lebesgue integral](/page/Lebesgue%20Integral): because $|g|\in L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$, the integral of $|g|$ over a measurable set is small whenever the [Lebesgue measure](/page/Lebesgue%20Measure) of that set is small. For a finite disjoint family of subintervals, each increment $f(y_j)-f(x_j)$ is the integral of $g$ over the corresponding open interval, so the sum of the absolute increments is bounded by the integral of $|g|$ over the union. [/proofplan]

[step:Use absolute continuity of the integral to choose the measure scale]Since $g\in L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$, the function $|g|:(a,b)\to[0,\infty)$ belongs to $L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$. We use the standard absolute continuity property of the Lebesgue integral: for every $\varepsilon>0$ there exists $\delta>0$ such that, for every $E\in\mathcal B((a,b))$, \begin{align*} \mathcal L^1(E)<\delta \implies \int_E |g(t)|\,d\mathcal L^1(t)<\varepsilon. \end{align*} Fix $\varepsilon>0$, and choose such a $\delta>0$.[/step]

[guided]The goal is to prove absolute continuity of $f$, so we must start with an arbitrary tolerance $\varepsilon>0$ and produce a length tolerance $\delta>0$. The function controlling the increments of $f$ is not $g$ itself but $|g|$, because the definition of absolute continuity asks for a bound on sums of absolute values. Since $g\in L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$, its modulus $|g|:(a,b)\to[0,\infty)$ is integrable with respect to $\mathcal L^1$. The absolute continuity property of the Lebesgue integral says that integrability of $|g|$ implies the following: for every $\varepsilon>0$ there exists $\delta>0$ such that every measurable set $E\in\mathcal B((a,b))$ with sufficiently small measure satisfies \begin{align*} \mathcal L^1(E)<\delta \implies \int_E |g(t)|\,d\mathcal L^1(t)<\varepsilon. \end{align*} We fix the given $\varepsilon>0$ and choose this corresponding $\delta>0$. This is exactly the scale we will use in the definition of absolute continuity for $f$: small total interval length will produce a measurable union of small measure, and therefore a small integral of $|g|$ over that union.[/guided]

[step:Rewrite each increment of $f$ as an integral over the corresponding interval] Let $x,y\in[a,b]$ satisfy $x<y$. By the definition of $f$ and finite additivity of the Lebesgue integral over disjoint measurable sets, \begin{align*} f(y)-f(x)=\int_{(a,y)} g(t)\,d\mathcal L^1(t)-\int_{(a,x)} g(t)\,d\mathcal L^1(t). \end{align*} Since $(a,y)$ is the disjoint union of $(a,x]$ and $(x,y)$ up to endpoints, and endpoints have $\mathcal L^1$-measure zero, the difference is \begin{align*} f(y)-f(x)=\int_{(x,y)} g(t)\,d\mathcal L^1(t). \end{align*} Therefore, by the triangle inequality for the Lebesgue integral, \begin{align*} |f(y)-f(x)|\leq \int_{(x,y)} |g(t)|\,d\mathcal L^1(t). \end{align*} [/step]

[step:Estimate finite sums over disjoint subintervals]Let $n\in\mathbb N$, and let $(x_j,y_j)_{j=1}^n$ be a finite family with $a\leq x_j<y_j\leq b$ such that the open intervals $(x_j,y_j)$ are pairwise disjoint and \begin{align*} \sum_{j=1}^n (y_j-x_j)<\delta. \end{align*} Define the measurable set $E\in\mathcal B((a,b))$ by \begin{align*} E=\bigcup_{j=1}^n \bigl((x_j,y_j)\cap(a,b)\bigr). \end{align*} Because the intervals are pairwise disjoint, finite additivity of $\mathcal L^1$ gives \begin{align*} \mathcal L^1(E)\leq \sum_{j=1}^n (y_j-x_j)<\delta. \end{align*} Using the increment estimate from the previous step and finite additivity of the Lebesgue integral for nonnegative functions on disjoint measurable sets, \begin{align*} \sum_{j=1}^n |f(y_j)-f(x_j)|\leq \sum_{j=1}^n \int_{(x_j,y_j)\cap(a,b)} |g(t)|\,d\mathcal L^1(t). \end{align*} Since the sets $(x_j,y_j)\cap(a,b)$ are pairwise disjoint and their union is $E$, this becomes \begin{align*} \sum_{j=1}^n |f(y_j)-f(x_j)|\leq \int_E |g(t)|\,d\mathcal L^1(t). \end{align*} By the choice of $\delta$, \begin{align*} \int_E |g(t)|\,d\mathcal L^1(t)<\varepsilon. \end{align*} Hence \begin{align*} \sum_{j=1}^n |f(y_j)-f(x_j)|<\varepsilon. \end{align*}[/step]

[guided]Now we check the actual definition of absolute continuity. Let $n\in\mathbb N$, and let $(x_j,y_j)_{j=1}^n$ be a finite family of pairwise disjoint subintervals with endpoints in $[a,b]$, where $x_j<y_j$ for each $j$, and suppose that their total length satisfies \begin{align*} \sum_{j=1}^n (y_j-x_j)<\delta. \end{align*} The function $g$ is defined as an $L^1$ function on the open interval $(a,b)$, so endpoint values at $a$ and $b$ are irrelevant. To express all increments using sets inside the [measure space](/page/Measure%20Space) $(a,b)$, define \begin{align*} E=\bigcup_{j=1}^n \bigl((x_j,y_j)\cap(a,b)\bigr). \end{align*} This set belongs to $\mathcal B((a,b))$ because it is a finite union of Borel subsets of $(a,b)$. Since the intervals are pairwise disjoint, the sets $(x_j,y_j)\cap(a,b)$ are also pairwise disjoint. Therefore finite additivity of Lebesgue measure gives \begin{align*} \mathcal L^1(E)=\sum_{j=1}^n \mathcal L^1\bigl((x_j,y_j)\cap(a,b)\bigr). \end{align*} For each $j$, removing or adding endpoints does not change one-dimensional Lebesgue measure, so \begin{align*} \mathcal L^1\bigl((x_j,y_j)\cap(a,b)\bigr)\leq y_j-x_j. \end{align*} Thus \begin{align*} \mathcal L^1(E)\leq \sum_{j=1}^n (y_j-x_j)<\delta. \end{align*} For each interval, the preceding increment computation gives \begin{align*} |f(y_j)-f(x_j)|\leq \int_{(x_j,y_j)\cap(a,b)} |g(t)|\,d\mathcal L^1(t). \end{align*} Summing this inequality over $j\in\{1,\dots,n\}$ gives \begin{align*} \sum_{j=1}^n |f(y_j)-f(x_j)|\leq \sum_{j=1}^n \int_{(x_j,y_j)\cap(a,b)} |g(t)|\,d\mathcal L^1(t). \end{align*} Because $|g|$ is nonnegative and the measurable sets $(x_j,y_j)\cap(a,b)$ are pairwise disjoint, finite additivity of the Lebesgue integral over disjoint measurable sets yields \begin{align*} \sum_{j=1}^n \int_{(x_j,y_j)\cap(a,b)} |g(t)|\,d\mathcal L^1(t)=\int_E |g(t)|\,d\mathcal L^1(t). \end{align*} Finally, the construction of $\delta$ applies because $\mathcal L^1(E)<\delta$, so \begin{align*} \int_E |g(t)|\,d\mathcal L^1(t)<\varepsilon. \end{align*} Combining the last three displays gives \begin{align*} \sum_{j=1}^n |f(y_j)-f(x_j)|<\varepsilon. \end{align*} This is precisely the required estimate in the definition of absolute continuity.[/guided]

[step:Conclude absolute continuity on $[a,b]$] We have shown that for every $\varepsilon>0$ there exists $\delta>0$ such that every finite pairwise disjoint family of intervals with endpoints in $[a,b]$ and total length less than $\delta$ satisfies \begin{align*} \sum_{j=1}^n |f(y_j)-f(x_j)|<\varepsilon. \end{align*} This is the epsilon-delta definition of absolute continuity on $[a,b]$. Therefore $f$ is absolutely continuous on $[a,b]$. [/step]