What Is the Derivative of an Indicator Function, and Its Integral?

An indicator function records whether a point belongs to a set: it equals 1 inside the set and 0 outside. Its derivative depends on what kind of derivative you mean. For the common case of an interval, the ordinary derivative vanishes wherever the function is locally constant, while the jumps at the interval’s edges have no ordinary derivative.

The integral is easier: integrating an interval’s indicator measures how much of the integration region lies inside that interval. A distributional derivative gives a precise way to represent the jumps as point masses.

Indicator of an interval

Let $a<b$ and define

$$\mathbf{1}_{[a,b]}(x)= \begin{cases} 1,&a\le x\le b,\\ 0,&x<a\text{ or }x>b. \end{cases}$$

Changing whether either endpoint is included does not change its integral or its distribution. A single point has length zero, so endpoint values do not affect these results.

Ordinary derivative

For every $x$ strictly between the endpoints, the indicator is locally constant at 1. For every $x$ outside the closed interval, it is locally constant at 0. Thus

$$\frac{d}{dx}\mathbf{1}_{[a,b]}(x)=0 \qquad (x\ne a,b).$$

At $a$ and $b$, the function jumps between 0 and 1. It is discontinuous at each point, so it cannot be differentiable there. The ordinary derivative is therefore zero almost everywhere, but that does not mean the function is constant.

Integral of the indicator

For any interval $[c,d]$ with $c<d$,

$$\int_c^d \mathbf{1}_{[a,b]}(x)\,dx =\text{length}\bigl([c,d]\cap[a,b]\bigr).$$

In particular, over the whole real line,

$$\int_{-\infty}^{\infty}\mathbf{1}_{[a,b]}(x)\,dx=b-a.$$

This holds as a Riemann integral on finite intervals and as a Lebesgue integral. More generally, for a measurable set $A$, the integral of its indicator over a measurable region $E$ is the measure of $A\cap E$.

Example: Take $f(x)=\mathbf{1}_{[2,5]}(x)$. On $[0,6]$, the function equals 1 from 2 to 5 and 0 elsewhere, so

$$\int_0^6 f(x)\,dx=5-2=3.$$

Its accumulated integral from the far left is

$$F(x)=\int_{-\infty}^{x}f(t)\,dt= \begin{cases} 0,&x\le 2,\\ x-2,&2<x<5,\\ 3,&x\ge 5. \end{cases}$$

The function $F$ has derivative $f$ except at the jump points, where its derivative does not exist. This is consistent with the fundamental theorem of calculus in its usual form for absolutely continuous functions; the indicator itself is not absolutely continuous because it has jumps.

Distributional derivative: the jumps as point masses

Ordinary differentiation does not record the jumps, but distribution theory does. The distributional derivative $D f$ is defined by how it acts on every smooth, compactly supported test function $\varphi$:

$$\begin{aligned} \langle Df,\varphi\rangle &=-\int_{-\infty}^{\infty}f(x)\varphi'(x)\,dx\\ &=-\int_a^b\varphi'(x)\,dx\\ &=\varphi(a)-\varphi(b). \end{aligned}$$

The Dirac delta $\delta_c$ acts on a test function by $\langle\delta_c,\varphi\rangle=\varphi(c)$. Therefore,

$$D\mathbf{1}_{[a,b]}=\delta_a-\delta_b.$$

The positive sign at $a$ represents the jump from 0 to 1; the negative sign at $b$ represents the jump from 1 to 0. For the example $[2,5]$, the distributional derivative is $\delta_2-\delta_5$. Its total mass is zero, matching the fact that the indicator starts and ends at 0 when moving from left to right across the whole real line.

Common mistakes

• Saying the derivative is simply zero. It is zero wherever the ordinary derivative exists, but the derivative does not exist at the jump points. The distributional derivative also includes point masses there.
• Using the fundamental theorem without checking its conditions. A function with jumps is not absolutely continuous, so integrating its ordinary derivative need not recover its change in value.
• Confusing the integral with the derivative. The integral of the indicator measures interval length; the distributional derivative records the locations and directions of the jumps.
• Treating endpoint conventions as consequential here. Including or excluding the endpoints changes neither the integral nor the distribution associated with the indicator.