The Integral Representation Theorem for Absolutely Continuous Functions

From Absolute Continuity to an Integral Formula

Absolute continuity is the condition that lets a function be recovered from its derivative by integration. The statement is stronger than saying that a derivative exists almost everywhere: it identifies the original function at every point, including points where the derivative does not exist. The proof has two essential parts. First, the almost-everywhere derivative must be Lebesgue integrable. Then the difference between the function and the indefinite integral of that derivative has derivative zero almost everywhere, so the zero-derivative lemma makes that difference constant.

Throughout, $a<b$, and $f:[a,b]\to\mathbb{R}$ is real-valued. The measure is Lebesgue measure $m$ on $[a,b]$, with its Lebesgue sigma-algebra; it is finite and complete. A function $f$ is absolutely continuous on $[a,b]$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that for every finite collection of pairwise disjoint intervals $(u_j,v_j)\subseteq[a,b]$, the condition $\sum_j(v_j-u_j)<\delta$ implies $\sum_j|f(v_j)-f(u_j)|<\varepsilon$.

We use the earlier results that an absolutely continuous function on a compact interval has bounded variation and is differentiable almost everywhere. We also use two results from the earlier material on indefinite Lebesgue integrals and zero derivatives. Precisely, if $g\in L^1([a,b],m)$ is real-valued, then $G(x)=\int_a^x g\,dm$ is absolutely continuous and $G'(x)=g(x)$ for $m$-almost every $x$. Also, an absolutely continuous real-valued function on $[a,b]$ whose derivative is zero almost everywhere is constant on all of $[a,b]$. These are results about absolute continuity; absolute continuity of a measure is a different notion.

Theorem (Integral Representation): If $f:[a,b]\to\mathbb{R}$ is absolutely continuous, then its derivative exists and is finite almost everywhere, has a Lebesgue measurable representative in $L^1([a,b],m)$, and $$ f(x)=f(a)+\int_a^x f'(t)\,dm(t)\qquad\text{for every }x\in[a,b]. $$ The derivative is specified only almost everywhere; changing its values on a Lebesgue null set does not change the integral. The displayed identity, by contrast, holds at every point of the interval.

Why the Derivative Is Integrable

Write $V=V_a^b(f)$, the total variation of $f$ on $[a,b]$. By the earlier result that absolute continuity implies bounded variation, $V<\infty$. We first control difference quotients by $V$. Fix $h$ with $0<h\le b-a$, and set $q_h(x)=|f(x+h)-f(x)|/h$ for $a\le x\le b-h$. Then $$ \int_a^{b-h}q_h(x)\,dm(x)\le V. $$ To verify this estimate, write each $x\in[a,b-h]$, apart from choices of endpoints that do not affect the integral, as $x=a+r+kh$, where $0\le r<h$ and $k$ is a nonnegative integer. For each fixed $r$, the eligible terms are consecutive increments along the increasing list $a+r,a+r+h,a+r+2h,\ldots$, stopping when the next point would exceed $b$. Their absolute differences have sum at most $V$: inserting $a$ and $b$, when needed, gives a partition of $[a,b]$, and the added variation terms are nonnegative. Changing variables on each interval of length $h$ therefore gives $$ \int_a^{b-h}|f(x+h)-f(x)|\,dm(x) =\int_0^h\sum_{\substack{k\ge0\\a+r+(k+1)h\le b}} |f(a+r+(k+1)h)-f(a+r+kh)|\,dm(r) \le hV. $$ Dividing by $h$ proves the estimate.

Here are the details that turn this bound into integrability. Choose $h_n=(b-a)/n$ for $n\ge2$, and define the nonnegative measurable function $g_n$ on $[a,b]$ by $g_n(x)=|f(x+h_n)-f(x)|/h_n$ when $x\le b-h_n$, and $g_n(x)=0$ otherwise. Measurability follows from continuity of $f$. The estimate just proved gives $\int_a^b g_n\,dm\le V$ for each $n$. At every interior point where $f'$ exists, $g_n(x)\to|f'(x)|$, since $h_n\downarrow0$ and eventually $x\le b-h_n$. Differentiability holds almost everywhere. Fatou's lemma for the nonnegative measurable functions $g_n$ now yields $$ \int_a^b |f'|\,dm =\int_a^b\liminf_{n\to\infty}g_n\,dm \le\liminf_{n\to\infty}\int_a^b g_n\,dm \le V<\infty. $$ In this display, $|f'|$ may be assigned any measurable value on the null set where the derivative is not defined. To be precise about measurability, the signed forward difference quotients, set equal to zero when $x>b-h_n$, are measurable and converge to $f'$ wherever the derivative exists in the interior. Their limit on the measurable set where it exists, extended by zero elsewhere, is a measurable representative. The exceptional set is Lebesgue measurable because the quotients are measurable, and completeness ensures its null subsets are measurable. Thus the representative of $f'$ is in $L^1([a,b],m)$.

Proof of the Representation

Proof. If $a=b$, the claimed identity is $f(a)=f(a)+0$, so assume $a<b$. By the preceding estimate, choose a measurable representative of the almost-everywhere derivative $f'$ that belongs to $L^1([a,b],m)$. Define $G(x)=\int_a^x f'(t)\,dm(t)$. By the earlier theorem on indefinite Lebesgue integrals, $G$ is absolutely continuous and $G'=f'$ almost everywhere. The difference $H=f-G$ is absolutely continuous, because sums and differences of absolutely continuous functions are absolutely continuous. At almost every point where both derivatives exist, $H'=f'-G'=0$. The zero-derivative lemma therefore implies that $H$ is constant everywhere on $[a,b]$, not merely almost everywhere. Since $G(a)=\int_a^a f'\,dm=0$, we have $H(a)=f(a)$. Hence $H(x)=f(a)$ for every $x$, or equivalently $f(x)=f(a)+\int_a^x f'(t)\,dm(t)$ for every $x\in[a,b]$. $\square$

The pointwise conclusion depends on the zero-derivative lemma: knowing only that $f'=G'$ almost everywhere is not itself an equality of functions everywhere. Absolute continuity rules out a nonconstant difference with derivative zero almost everywhere, and therefore fixes the additive constant through the value at $a$.

Worked Example: A Polynomial on a Compact Interval

Let $f(x)=x^3-2x$ on $[0,2]$. This function is continuously differentiable and hence absolutely continuous. Its derivative is $f'(t)=3t^2-2$, which is integrable on the finite interval. For $0\le x\le2$, compute the right-hand side of the representation:

f(0)+\int_0^x(3t^2-2)\,dt =0+\bigl[t^3-2t\bigr]_0^x =x^3-2x.

This is $f(x)$ for every $x$ in the interval. Check at $x=2$: $f(2)=2^3-2\cdot2=8-4=4$, while the integral is $[t^3-2t]_0^2=(8-4)-(0-0)=4$, and $f(0)=0$. The representation gives $0+4=4$, as required.

Worked Example: A Corner Where the Derivative Does Not Exist

Let $f(x)=|x|$ on $[-1,1]$. This piecewise linear function is absolutely continuous. Its derivative is $-1$ on $(-1,0)$, $1$ on $(0,1)$, and is undefined at $0$. Assign any value at $0$, for example $f'(0)=0$; this choice does not affect the Lebesgue integral. For $-1\le x\le0$,

f(-1)+\int_{-1}^x f'(t)\,dm(t) =1+\int_{-1}^x(-1)\,dt =1-(x+1)=-x=|x|.

For $0\le x\le1$, split the integral at $0$. The value assigned to the derivative at $0$ contributes nothing, since a singleton has Lebesgue measure zero:

f(-1)+\int_{-1}^x f'(t)\,dm(t) =1+\int_{-1}^0(-1)\,dt+\int_0^x1\,dt =1-1+x=x=|x|.

Check at $x=0$: the integral from $-1$ to $0$ is $-1$, so the expression is $1-1=0=f(0)$. Thus a failure of differentiability at one point does not prevent the representation.

Worked Example: An Integrable Unbounded Derivative

On $[0,1]$, take $f(x)=\sqrt{x}$. The power-function result from the earlier material on absolute continuity gives that $f$ is absolutely continuous. For $t>0$, $f'(t)=1/(2\sqrt{t})$. The derivative is unbounded near zero, but it is integrable: for $0\le x\le1$, using the improper-integral limit at zero,

\int_0^x\frac{1}{2\sqrt{t}}\,dm(t) =\lim_{s\downarrow0}\int_s^x\frac{1}{2\sqrt{t}}\,dt =\lim_{s\downarrow0}(\sqrt{x}-\sqrt{s}) =\sqrt{x}.

The limit is valid also at $x=0$, where the integral over the degenerate interval is zero. Since $f(0)=0$, this gives $f(0)+\int_0^x f'(t)\,dm(t)=\sqrt{x}=f(x)$ throughout $[0,1]$. Check at $x=1/4$: the integral is $\sqrt{1/4}=1/2$, and $f(1/4)=\sqrt{1/4}=1/2$. An unbounded derivative is therefore compatible with the theorem; what matters is its integrability.

Common Pitfalls and the Role of the Hypotheses

A frequent error is to read “$f'=G'$ almost everywhere” as “$f=G$ everywhere.” Derivative information does not determine an additive constant. For instance, if $f(x)=3+x$ on $[0,1]$, then $f'(x)=1$, but $\int_0^x f'(t)\,dt=x$, not $3+x$. The correction is the initial value: $f(0)+\int_0^x f'(t)\,dt=3+x$.

Another error is to require the derivative to exist at every point. In the absolute-value example, $f'$ is undefined at zero, yet its values almost everywhere determine the integral representation. The theorem asserts existence and integrability of the derivative almost everywhere, not everywhere. Its integral is a Lebesgue integral, and changes on null sets do not alter it.

Absolute continuity is essential to this conclusion. The proof uses it both to ensure finite variation, which bounds the integral of the derivative, and through the zero-derivative lemma, which turns an almost-everywhere derivative statement into a global conclusion. Continuity or differentiability almost everywhere alone does not provide these steps. The integral representation is useful because it converts a global property of the function into a concrete formula involving an $L^1$ function, while retaining validity at every point of the interval.

Check Your Understanding

For each question, distinguish what is asserted almost everywhere from what holds at every point.

Which estimate involving total variation is used to prove that the derivative of an absolutely continuous function is integrable?
Why does the proof define forward difference quotients to be zero near the right endpoint?
For $f(x)=|x|$ on $[-1,1]$, why does the value assigned to $f'(0)$ not affect the integral representation?
If $G(x)=\int_a^x f'(t)\,dm(t)$, what hypotheses ensure that $G$ is absolutely continuous and $G'=f'$ almost everywhere?
Why does the theorem specify $f(a)$ in addition to the integral of $f'$?

The Integral Representation Theorem for Absolutely Continuous Functions

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