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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

user190080
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vondip
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    You want $\int f(x) dx$ to denote what, exactly? – Mariano Suárez-Álvarez Jun 26 '13 at 07:05
  • the integral of the function f(x) on the interval [a,b] – vondip Jun 26 '13 at 07:05
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    Well that is a number then, not a function. – Mariano Suárez-Álvarez Jun 26 '13 at 07:06
  • no, I'm sorry. I meant it should denote the integral of a function f, no necessarily at a single point. – vondip Jun 26 '13 at 07:08
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    @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 $[a,b]$). – Adriano Jun 26 '13 at 07:09
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    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. – Daniel Fischer Jun 26 '13 at 07:11
  • @Daniel The same for the gauge integral. – Tony Piccolo Jun 26 '13 at 07:33
  • It's not the case for the Dirac delta function. That's not technically a function, though, so I don't know... – hasnohat Jun 26 '13 at 22:12
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    @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. – Pedro Jun 27 '13 at 00:04

3 Answers3

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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 $\lim\limits_{x\to c}F(x)=F(c)$. $\blacktriangle$

Pedro
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  • If $f(x)=x^{-\frac13}$ for $x\neq0$ and $f(0)=0$, isn't $f$ unbounded but Riemann integrable? – stewbasic Aug 29 '16 at 01:36
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    @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. – Pedro Aug 29 '16 at 02:18
  • Ah you are right, sorry. I forgot that the sampling points can be chosen anywhere in the subintervals. – stewbasic Aug 29 '16 at 03:34
  • @PedroTamaroff My apologies for what is likely a simply question on an old answer, but could you elaborate on why did you need to add absolute value sign around $F(x)-F(c)$ at the second to last line? – user106860 May 17 '19 at 01:09
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    @Pedro your inequality estsblished that not only is it continuous but also Lipschitz continuos. isn't it? – Upstart Jul 18 '19 at 03:56
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    why $f$ bounded on $[a,b]$ ? The function $f(x)=0$ on $(0,\infty ]$ and $f(0)=\infty $ is not riemann integrable on $[0,1]$ ? – Todd Dec 09 '19 at 15:28
  • @Todd https://math.stackexchange.com/questions/610054/if-a-function-fx-is-riemann-integrable-on-a-b-is-fx-bounded-on-a – Pedro Feb 17 '24 at 09:38
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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.

Not Euler
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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.

goblin GONE
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