Suppose function $f:[a,b] \to \mathbb{R}$ has an antiderivative. This means there is some differentiable function $F$ so that $f(x) = F'(x)$ in all of $[a,b]$.

Must the absolute value $|f|$ have an antiderivative?

What if we also specify that $f$ is Riemann integrable? I know that if $f$ is Riemann integrable, then $|f|$ must be also, but not every integrable function has an antiderivative.

  • $\begingroup$ Even if $f$ is continuous it is a little hard to construct the antiderivative of $|f|$ explicitly from $F$: First we need take a countable family of disjoint intervals $(a_n,b_n)$ of $[a,b]$ with dense union (i.e each $a_n$ is also $b_m$ and vice-versa, except possibly at the boundary points $a,b$) such that $f$ is either positive, negative, or zero on each interval. Take an antiderivative of $|f|$ on each $(a_n,b_n)$ ($F$ or $-F$). Then an antiderivative of $|f|$ needs to be constructed by translates of $F$ and $-F$ in a step-by-step manner. $\endgroup$ May 3, 2019 at 17:24
  • $\begingroup$ @SAS: For clarification are you not asking if the existence of $F$ such that $F'(x) = f(x)$ for all $x \in [a,b]$ implies that there exists $G$ such that $G'(x) = |f(x)|$ for all $x \in [a,b]$. $\endgroup$
    – RRL
    May 5, 2019 at 8:17
  • $\begingroup$ @RRL: Yes that is correct. Was something not clear. $\endgroup$
    – SAS
    May 6, 2019 at 23:41

2 Answers 2


I believe your understanding of the term antiderivative is correct. A function $F:[a,b] \to \mathbb{R}$ is an antiderivative of $f:[a,b] \to \mathbb{R}$, if it is differentiable and $F'(x) = f(x)$ for all $x \in [a,b]$.

It seems then the question is if $f$ has an antiderivative, then is it necessary that the function $x \mapsto |f(x)|$ has an antiderivative. Specifically, does there exist differentiable $G:[a,b] \to \mathbb{R}$ such that $G'(x) = |f(x)|$ for all $x \in [a,b]$?

The answer is no -- it is not necessary.

A counterexample is

$$f(x) = \begin{cases}\cos(x^{-1}), & 0 < x \leqslant 1 \\ 0, & x= 0 \end{cases}$$

Here $f$ has an antiderivative

$$F(x) = \int_0^x \cos (t^{-1}) \, dt$$

Since $f$ is continuous on $(0,1]$ we have, directly by the FTC, that $F'(x) = \cos(x^{-1})$ for $0 < x \leqslant 1$. We can also verify that $F(0) = 0$ and

$$\tag{*}F'(0) = \lim_{x \to 0} \frac{F(x) - F(0)}{x} = \lim_{x \to 0}\frac{1}{x}\int_0^x \cos (t^{-1}) \, dt = 0 = f(0) $$

The details of deriving the limit (*) are given here.

On the other hand, we have

$$g(x) = |f(x)| = \begin{cases}|\cos(x^{-1})|, & 0 < x \leqslant 1 \\ 0, & x= 0 \end{cases}$$

Suppose there exists an antiderivative $G$ such that $G'(x) = g(x)$ for all $x \in [0,1]$. Since $g$ is Riemann integrable (continuous except at one endpoint), it follows by the FTC that

$$G(x) - G(0) = \int_0^x g(t) \, dt$$

By continuity of $g$, clearly have $G'(x) = |\cos(x^{-1}|$ for $0 < x \leqslant 1$.


$$\tag{**}G'(0) = \lim_{x \to 0} \frac{G(x) - G(0)}{x} = \lim_{x \to 0}\frac{1}{x}\int_0^x |\cos (t^{-1})| \, dt = \frac{2}{\pi} \neq 0 = g(0) $$

The details of deriving the limit (**) are given here in robjohn's answer.

This proves that no such antiderivative $G$ exists.


The issue of integrability of $f$ is peripheral as there are Riemann integrable functions without antiderivatives. Furthermore, if we have

$$F(x) = \int_{[a,x]} f,$$

as a Lebesgue integral and $F'(x) = f(x)$ almost everywhere but not for every $x \in [a,b]$, then $F$ does not qualify as an antiderivative.

  • $\begingroup$ Thank you. This makes sense. How did you arrive at such counterexample? $\endgroup$
    – SAS
    May 6, 2019 at 23:42

Note the following quote from the wikipedia article on the Henstock–Kurzweil integral (under Properties)

In general, every Henstock–Kurzweil integrable function is measurable, and f is Lebesgue integrable if and only if both f and |f| are Henstock–Kurzweil integrable. This means that the Henstock–Kurzweil integral can be thought of as a "non-absolutely convergent version of Lebesgue integral". It also implies that the Henstock–Kurzweil integral satisfies appropriate versions of the monotone convergence theorem (without requiring the functions to be nonnegative) and dominated convergence theorem (where the condition of dominance is loosened to g(x) ≤ fn(x) ≤ h(x) for some integrable g, h).

If F is differentiable everywhere (or with countable many exceptions), the derivative F′ is Henstock–Kurzweil integrable, and its indefinite Henstock–Kurzweil integral is F.

From the highlighted sentences (and the Henstock–Kurzweil differentiation theorem), it follows that if $f$ has an anti-derivative, then $|f|$ will also have an anti-derivative if and only if $f$ is Lesbegue integrable.

  • $\begingroup$ Thank you for your contribution. I don't think it is correct though, if it only provides diff. everywhere with countable exceptions. $\endgroup$
    – SAS
    May 6, 2019 at 23:46
  • $\begingroup$ Perhaps it falls short in providing only an almost-everywhere anti-derivative for $|f|$, but it does show that there are $f$ with anti-derivatives where $|f|$ does not have have one. Any non-Lebesgue Integrable function cannot have anti-derivatives for both $f$ and $|f|$, That is all you got out of the other answer as well. RRL gave you a specific example, which is always nice. But I gave you a necessary condition, which is also sufficient if one relaxes the concept of anti-derivative only slightly. $\endgroup$ May 7, 2019 at 0:13

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