Differentiating an Indefinite Lebesgue Integral

From Local Averages to Derivatives

An indefinite Lebesgue integral turns an integrable function into a function of the endpoint. The central question here is whether differentiating that new function recovers the original integrand. The answer is yes almost everywhere, even when the integrand is not continuous. The reason is that, at almost every point, the averages of the integrand over small intervals approach its value there.

Fix finite real numbers $a<b$. Take the measure space $([a,b],\mathcal{L}_{[a,b]},m)$, where $\mathcal{L}_{[a,b]}$ is the Lebesgue sigma-algebra on the interval and $m$ is Lebesgue measure. Let $g:[a,b]\to\mathbb{R}$ be measurable and integrable, meaning $\int_a^b |g|\,dm<\infty$. Define the real-valued function

F(x)=\int_a^x g(t)\,dt,\qquad a\le x\le b.

For $x<a$, an oriented integral is defined by $\int_a^x g=-\int_x^a g$; only points in $[a,b]$ are needed below. The earlier result on absolute continuity of indefinite integrals gives that $F$ is absolutely continuous. The result proved here identifies its derivative: $F'(x)=g(x)$ for $m$-almost every $x\in(a,b)$. Almost everywhere means that the set of exceptions is contained in a Lebesgue null set; it does not assert equality at every point.

The Local-Average Input

Theorem (Lebesgue Differentiation Theorem): If $u:\mathbb{R}\to\mathbb{R}$ is Lebesgue measurable and locally integrable, meaning $\int_K |u|\,dm<\infty$ for every compact interval $K$, then for $m$-almost every $x\in\mathbb{R}$, $$ \lim_{r\downarrow0}\frac{1}{2r}\int_{x-r}^{x+r}|u(t)-u(x)|\,dt=0. $$ The value $u(x)$ is finite at almost every such point. This is a standard result; its proof, which relies on a covering estimate, is beyond this tutorial.

The theorem says more than that the signed average of $u(t)-u(x)$ tends to zero: it controls the average absolute deviation. That stronger form is what makes the derivative argument work. Although the theorem is stated on $\mathbb{R}$, it applies to $g$ on $[a,b]$ by extending $g$ to be zero outside $[a,b]$. This extension is integrable on $\mathbb{R}$, since its integral of the absolute value is $\int_a^b|g|\,dm<\infty$. Values at the two endpoints do not affect any integral or almost-everywhere conclusion.

Theorem (Differentiating an Indefinite Lebesgue Integral): Let $a<b$, and let $g:[a,b]\to\mathbb{R}$ be Lebesgue measurable with $\int_a^b|g|\,dm<\infty$. For $F(x)=\int_a^x g(t)\,dt$, the derivative $F'(x)$ exists and equals $g(x)$ for $m$-almost every $x\in(a,b)$.

Proof. Extend $g$ by zero to a locally integrable function on $\mathbb{R}$, and choose a point $x\in(a,b)$ where the Lebesgue Differentiation Theorem holds. Such points comprise a set of full measure in $(a,b)$. For $h\ne0$ small enough that $x+h\in[a,b]$, additivity of the Lebesgue integral gives the signed-integral identity $$ \frac{F(x+h)-F(x)}{h}-g(x) =\frac{1}{h}\int_x^{x+h}\bigl(g(t)-g(x)\bigr)\,dt. $$ Taking absolute values and accounting for either sign of $h$, $$ \left|\frac{F(x+h)-F(x)}{h}-g(x)\right| \le \frac{1}{|h|}\int_{\min(x,x+h)}^{\max(x,x+h)}|g(t)-g(x)|\,dt. $$ The interval of integration is contained in $[x-|h|,x+|h|]$. Its integrand is nonnegative, so $$ \left|\frac{F(x+h)-F(x)}{h}-g(x)\right| \le 2\left(\frac{1}{2|h|}\int_{x-|h|}^{x+|h|}|g(t)-g(x)|\,dt\right). $$ As $h\to0$, the expression in parentheses tends to zero by the differentiation theorem, with $r=|h|$. Thus the difference quotient tends to $g(x)$, so $F'(x)=g(x)$. The set of points $x$ used has full measure in $(a,b)$, proving the assertion almost everywhere. $\square$

Worked Examples

Worked Example: An Integrand with a Jump

On $[0,1]$ with Lebesgue measure $m$, define $g(x)=2$ for $0\le x<\tfrac12$ and $g(x)=-1$ for $\tfrac12\le x\le1$. Find $F(x)=\int_0^xg(t)\,dt$ and determine where its derivative agrees with $g$.

The function is measurable and integrable; direct integration gives $\int_0^1|g|\,dm=2\cdot\tfrac12+1\cdot\tfrac12=\tfrac32$. If $0\le x\le\tfrac12$, the interval $[0,x]$ has integrand $2$, so $F(x)=2x$. If $\tfrac12\le x\le1$, split the integral at $\tfrac12$:

F(x)=\int_0^{1/2}2\,dt+\int_{1/2}^x(-1)\,dt =1-(x-\tfrac12)=\tfrac32-x.

For $0<x<\tfrac12$, differentiating $2x$ gives $F'(x)=2=g(x)$; for $\tfrac12<x<1$, differentiating $\tfrac32-x$ gives $F'(x)=-1=g(x)$. At $x=\tfrac12$, the left difference quotient is $2$, while the right difference quotient is $-1$, so the derivative does not exist there. Check: the two formulas for $F$ both give $1$ at $\tfrac12$, and $F(1)=\tfrac32-1=\tfrac12$, which also equals the total integral $2\cdot\tfrac12-1\cdot\tfrac12=\tfrac12$. The single exceptional point is a null set, as the theorem allows.

Worked Example: A Continuous Integrand

On $[-1,1]$ with Lebesgue measure, let $g(t)=t$. Compute its indefinite integral based at $-1$, and verify the derivative formula directly.

The function is measurable and integrable, with $\int_{-1}^1|t|\,dt=\tfrac12+\tfrac12=1$. For $x\in[-1,1]$,

F(x)=\int_{-1}^x t\,dt =\left[\frac{t^2}{2}\right]_{-1}^x =\frac{x^2-1}{2}.

For an interior point $x$ and a nonzero $h$ with $x+h\in[-1,1]$, the difference quotient is

\frac{F(x+h)-F(x)}{h} =\frac{((x+h)^2-1)-(x^2-1)}{2h} =\frac{2xh+h^2}{2h} =x+\frac{h}{2}.

As $h\to0$, this tends to $x=g(x)$, so the formula holds at every interior point. Check: at $x=0$, $F(0)=-\tfrac12$, and the quotient for any allowed $h\ne0$ is $h/2$, which tends to $0=g(0)$.

Worked Example: Changing an Integrand on a Null Set

On $[0,1]$, take $g=\chi_{\mathbb{Q}\cap[0,1]}$, where $\mathbb{Q}$ is the set of rational numbers. Determine $F(x)=\int_0^xg(t)\,dt$, and compare its derivative with the specified pointwise values of $g$.

The rational numbers in $[0,1]$ form a countable set, hence a Lebesgue null set. The function $g$ is measurable, bounded, and integrable, with $\int_0^1|g|\,dm=0$. For every $x\in[0,1]$, the integration interval meets the support of $g$ only in a null set, and therefore $F(x)=\int_0^xg(t)\,dt=0$. Consequently $F'(x)=0$ at every $x\in(0,1)$. At irrational $x$, $g(x)=0$, whereas at rational $x$, $g(x)=1$. Thus $F'=g$ at every irrational interior point and fails to equal $g$ at every rational interior point. Check: for any interior rational $x$, every difference quotient $(F(x+h)-F(x))/h$ is $0$, not $1$. The failure set is countable and null, so equality still holds almost everywhere.

What the Almost-Everywhere Qualification Means

The derivative formula concerns the values of $g$ at almost every point, not every point. The rational-indicator example shows why this distinction is necessary: changing an integrable function on a null set leaves all its integrals over intervals unchanged, and hence leaves $F$ unchanged everywhere, but can change the pointwise values on that null set. No derivative of $F$ can encode such arbitrary changes.

A related mistake is to replace the local-average condition by continuity without justification. Continuity at $x$ is sufficient for the average absolute deviation to tend to zero, but it is not required. For a merely measurable integrable $g$, the Lebesgue Differentiation Theorem supplies the needed local-average property at almost every point. The proof then turns that property into control of the difference quotient; it does not assert that every integrable function is continuous.

Takeaway: At almost every interior point, the average of $g(t)-g(x)$ over a shrinking interval around $x$ tends to zero. Every one-sided interval used in the difference quotient lies inside the corresponding centered interval, so its average deviation also tends to zero. This gives $F'(x)=g(x)$ almost everywhere.

Check Your Understanding

For each question, state the measure space and distinguish pointwise claims from almost-everywhere claims.

For the step function in the first example, compute the left and right difference quotients of $F$ at $x=\tfrac12$. Does the derivative exist there?
If $g\in L^1([a,b],m)$ is changed at one point, explain why its indefinite integral is unchanged at every $x\in[a,b]$.
In the differentiation proof, why does the bound use the absolute value of the integral and the centered average of $|g(t)-g(x)|$?
For $g=\chi_{\mathbb{Q}\cap[0,1]}$, identify the set where $F'(x)\ne g(x)$, and state its Lebesgue measure.
Which hypothesis allows the Lebesgue Differentiation Theorem to be applied to the zero extension of $g$ from $[a,b]$ to $\mathbb{R}$?

Differentiating an Indefinite Lebesgue Integral

What You'll Learn