Say I have a function $f(x)$ on some interval $[a,b]$. Say it is integrable such that $\displaystyle\int f(x)~dx $ is defined.

Is $\displaystyle\int f(x)~dx $ necessarily continuous? If I were to know that the integral is integrable itself such that $\displaystyle\int \int f(x)~dx $ is defined. Would that change anything?

If so why?

Thank you

  • 7
    $\begingroup$ You want $\int f(x) dx$ to denote what, exactly? $\endgroup$ Jun 26, 2013 at 7:05
  • 11
    $\begingroup$ Well that is a number then, not a function. $\endgroup$ Jun 26, 2013 at 7:06
  • 1
    $\begingroup$ @vondip You're probably looking for a general antiderivatve $F(x)$, where $F'(x)=f(x)$. This involves an indefinite integral (there is no interval $[a,b]$). $\endgroup$
    – Adriano
    Jun 26, 2013 at 7:09
  • 5
    $\begingroup$ If the integral is a Lebesgue- or Riemann-Integral, $F(y) = \int\limits_c^y f(x)\, dx$ is continuous. Don't know whether it can be discontinuous for some other integration theory. $\endgroup$ Jun 26, 2013 at 7:11
  • 4
    $\begingroup$ @Julien You can think about the Direc delta as an integrator, $d\alpha$; where $\alpha$ is the Heaviside step function. In said case, whenver $f$ is left or right continuous at the jump $\xi\in[a,b]$ we have $$\int_a^b fd\alpha=f(\xi)$$ But the Riemann Stieljes analog to my answer requieres that $f$ is $\alpha$-integrable where $\alpha$ is of bounded variation. In fact, $F(x)=\int_a^x f(t)d\alpha(t)$ will be continuous wherever $\alpha$ is. So, if the integrator is continuous and $f\in\mathscr R(\alpha)$, $F$ will be continuous. $\endgroup$
    – Pedro
    Jun 27, 2013 at 0:04

3 Answers 3


One can prove the following

THM Let $f:[a,b]\to\Bbb R$ be Riemann integrable over its domain. Define a new function $F:[a,b]\to\Bbb R$ by $$F(x)=\int_a^x f(t)dt$$ Then $F$ is continuous. That is, the map $$f\mapsto \int_a^x f$$ sends $\mathscr R[a,b]$ to $\mathscr C[a,b]$.

PROOF Let $c\in[a,b]$. Then $$F(x)-F(c)=\int_c^x f(t)dt$$

Since $f$ is integrable, we know it is bounded, say $|f(x)|\leq M$ over $[a,b]$. Then $$ -\int_c^x M\;dt\leq \int_c^x f(t)dt\leq \int_c^xM \;dt$$

which means $$-M(x-c)\leq \int_c^x f(t)dt\leq M(x-c)$$

Thus we get $$|F(x)-F(c)|\leq M|x-c|$$

Taking $x\to c$ the squeeze theorem says $\lim\limits_{x\to c}F(x)=F(c)$. $\blacktriangle$

  • $\begingroup$ If $f(x)=x^{-\frac13}$ for $x\neq0$ and $f(0)=0$, isn't $f$ unbounded but Riemann integrable? $\endgroup$
    – stewbasic
    Aug 29, 2016 at 1:36
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    $\begingroup$ @stewbasic It is improperly integrable, but not Riemann integrable in the usual sense. A necessary condition for a function to be Riemann integrable in the usual sense is that it be bounded. $\endgroup$
    – Pedro
    Aug 29, 2016 at 2:18
  • $\begingroup$ Ah you are right, sorry. I forgot that the sampling points can be chosen anywhere in the subintervals. $\endgroup$
    – stewbasic
    Aug 29, 2016 at 3:34
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    $\begingroup$ @Pedro your inequality estsblished that not only is it continuous but also Lipschitz continuos. isn't it? $\endgroup$
    – Upstart
    Jul 18, 2019 at 3:56
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    $\begingroup$ why $f$ bounded on $[a,b]$ ? The function $f(x)=0$ on $(0,\infty ]$ and $f(0)=\infty $ is not riemann integrable on $[0,1]$ ? $\endgroup$
    – Todd
    Dec 9, 2019 at 15:28

I would like to add something to Pedro Tamaroff's answer. This would hold not only for Riemann-integrable functions but even Lebesgue-integrable ones.

Let $(X,\mu)$ be any measure space and $f$ be $\mu$-integrable. One can approximate $f$ (in general $\mathscr{L}^p$ functions) by simple functions. So given $\epsilon>0$ we can find a simple function $\phi$ such that $\int|f-\phi|d\mu<\frac{\epsilon}{2}$. Let $|\phi(x)|< M$ for all $x$. Now for any measurable set $A$ we have $$|\int\limits_Afd\mu-\int \limits_A\phi d\mu|\leq\int\limits_A|f-\phi|d\mu<\frac{\epsilon}{2}$$ or $$|\int\limits_Afd\mu|<\frac{\epsilon}{2}+|\int\limits_A\phi d\mu|\leq\frac{\epsilon}{2}+M\mu(A)<\epsilon$$ whenever $\mu(A)<\frac{\epsilon}{2M}=\delta$.

Now for a Lebesgue-integrable function $f$ over $[a,b]$ we define $F(x)=\int\limits_a^x fd\lambda$. Then given $\epsilon>0$ we can find (as above) some $\delta>0$ such that $|F(x)-F(y)|=|\int\limits_y^xfd\lambda|<\epsilon$ whenever $|x-y|<\delta$. This proves the (uniform) continuity of $F$.

Moreover, this function is differentiable almost everywhere.


The following is worth pointing out:

Firstly, note that if we replace "continuous" with "differentiable", the new statement isn't true. For instance, the function $$\mathbb{R} \rightarrow \mathbb{R}$$ $$x \mapsto \begin{cases} 1, & x \geq 0 \\ -1 & x < 0\end{cases}$$ has $x \mapsto |x|$ as an indefinite integral, which is not differentiable.

We do however have the following: if $f$ is Riemann integrable over its domain and $F$ is defined as in Pedro's answer, then: $$f \mbox{ continuous at } x \rightarrow F \mbox{ differentiable at } x$$

Here's a nice article about such things. In the parlance of that article, we'd say that $|x|$ is an indefinite integral of the aforementioned step function, but not an antiderivative.

  • $\begingroup$ Can you update your link? The new link is Here @goblinGONE $\endgroup$
    – falamiw
    Nov 17, 2022 at 18:18

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