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Sometimes definite integral is defined using antiderivatives:

$$\int_{a}^b{f(t)dt}=F(b)-F(a)$$ where $F$ is any continuous function such that: $$(\forall t\in[a,b]\setminus C)(F'(t) \space exists \space and\space F'(t)=f(t))$$ where $C$ is a countable set. Then (if I have written the definition correctly) it can be proved, the integral is well-defined.

some not lebesgue integrable functions are integrable with this definition.

1) are all lebesgue integrable functions, integrable with above definition?

2) can this definition be extended to measure spaces?

3) will this definition be valid if we require $C$ have zero measure only (not necessarily countable)?

4) Has this integral a formal name? (eg Dieudonné integral)

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This maybe not pertinent, but there is a very general version if one is prepared to use the framework of distribution theory. Every distribution has a primitive and one then can use the concept of the limit of a distribution at a point or at infinity (which can be defined in an elementary fashion, i.e., without functional analysis) to define definite integrals---the fundamental theorem then holds. For example, the integral of the $\delta$-distribution is $1$ as one would expect, more surprisingly that of $\cos x$ is zero. Details are in an article by J. Sebastião e Silva (MR0216289). –  jbc Jun 9 '13 at 13:40

2 Answers 2

up vote 1 down vote accepted

To add to Gerald Edgar’s answer:

1) Kurzweil–Henstock integral satisfies $\int_a^bf(x)\\,dx=F(b)-F(a)$ even under the weaker assumptions that $F$ is continuous and $F'(x)=f(x)$ for all but countably many $x\in[a,b]$, hence it fully subsumes the integral you want to define. It is, however, strictly more general: if $A\subseteq[a,b]$ is a null set and $f$ its characteristic function, then $\int_a^bf(x)=0$ (as a Kurzweil–Henstock or Lebesgue integral), so the only choice would be $F$ constant, but then $0=F'(x)=f(x)$ only holds outside $A$. Thus, in general, the set of exceptions may be an arbitrary null set. (It cannot be any worse: the Lebesgue differentiation theorem mentioned by Gerald Edgar holds for Kurzweil–Henstock integral as well.) In particular, your integral (unlike Kurzweil–Henstock integral) does not extend Lebesgue integral. It does not even extend Riemann integral, as we can take for $A$ a closed set.

3) If we allow $C$ to be an arbitrary null set, then the definition no longer makes sense: if we take for $F$ the Cantor function, then $F$ is continuous and $F'(x)=0$ almost everywhere, so we would be forced to put $0=\int_0^10\\,dx=F(1)-F(0)=1$.

One can salvage the definition by making stronger requirements on the function $F$. That is, we can put $\int_a^bf(x)\\,dx=F(b)-F(a)$ if $F'(x)=f(x)$ for all $x\in[a,b]$ except a null set, and $F$ is an admissible indefinite integral function. One can obtain a characterization of Lebesgue integral in this way by taking $F$ absolutely continuous. For Kurzweil–Henstock integral, one needs a broader class of functions, whose description is a bit more delicate, see e.g. http://link.springer.com/article/10.1007%2Fs10587-008-0081-0 .

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In case the exceptional set $C$ is empty, I have seen this called the Newton integral.

It can exist in cases where the Riemann integral does not exist, also in cases where the Lebesgue integral does not exist. There is an integral (the generalized Riemann integral or gauge integral or Kurzweil integral or Henstock integral) that does have that theorem: if $f = F'$ everywhere, then the integral exists and satisfies $\int_a^b f(x)dx = F(b)-F(a)$. An elementary reference: Generalized Riemann Integral by R. M. McLeod.

If $f$ is Riemann integrable, then it is true that there is a null set $C$ (but not, in general countable) such that $F'(x)=f(x)$ for all $x \in [a,b]\setminus C$ where $$ F(x) = \int_a^x f $$
and this $F$ is of course continuous. The same for Lebesgue integral.

In measure spaces $(X, \mathcal F,\mu)$, the usual thing is not to connect function $f$ to another function $F$ where $F'=f$, but to connect a function $f$ and a signed measure $\nu$ where $$ \int_A f d\mu = \nu(A)\qquad\text{for all $A \in \mathcal F$} $$

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So you answered (1) completely, and (2) and (4) partially. How about (3)? Is the integral well-defined if $C$ is a null set. to be more precise; Let the continuous functions $F$ and $G$ have equal derivatives outside a null set. Are they equal? –  user31968 Mar 5 '13 at 19:54
    
The Cantor function (more picturesquely, the devil's staircae) is a continuous, non-zero function whose derivative vanishes almost everywhere---see the corresponding Wikipedia article. –  jbc Jun 9 '13 at 13:29

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