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Is it possible for a function to be a Lebesgue integral, but not a Riemann integral?

After the comments below I realize my question was not a good one. Thank you. This is my edited version:

Let $f$ be an integrable function on $[a,b]$. Suppose $F(x)=F(a)+\int_a^x{f(t)}dt$. and suppose $f$ is Lebesgue integrable but not Riemann integrable.

Is the following true: There is always a function $g$ which is Riemann integrable on $[a,b]$ and for which $F(x)=F(a)+\int_a^x{g(t)}dt$

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a function cannot be an integral. –  Bombyx mori Apr 17 '13 at 6:53
    
@user31953 : The definition of the Riemann integral does not use antiderivatives at all. It uses Riemann sums. The Fundamental Theorem of Calculus states the connection between the Riemann integral and antiderivatives. –  Stefan Smith Apr 17 '13 at 23:57

1 Answer 1

up vote 9 down vote accepted

You mean to be Lebesgue integrable and not Riemann integrable? The answer is yes. Classic example, let $f(x)=1$ if $x$ is a rational number and zero otherwise on the interval [0,1].

By the way, the Lebesgue integral is a generalization of the Riemann integral. Every Riemann integrable function is Lebesgue integrable. On the other hand there are plenty of functions which are Lebesgue integrable but not Riemann integrable.


Addendunm:

To the OP, looking at the comment below, I think there are several confusions here. You seem to be confusing an antiderivative with the integral. Antiderivatives and integrals are two entirely different things.

Sticking with Riemann integral, a Riemann integral is defined as the (signed) area under a curve so the Riemann integral of a function is always a number. Your textbook calls this the definite integral.

An antiderivative of $f(x)$ is a function is another function $F(x)$ such that $F'(x)=f(x)$. An antiderivative is always a function.

They are two different things but Newton and Leibniz proved that they are actually very closely related (a very useful thing by the way). The fundamental theorem of calculus says that in order to find the (definite) integral of a function, just compute its antiderivative and evaluate it at the top minus the bottom, meaning

$$\int_a^b f(x)dx=F(b)-F(a).$$

So when we say that a function is integrable, we mean that we can find the area underneath it. If you mean antiderivative then say antiderivative. Don't call the antiderivative the integral or the other way around.

Now to answer what I think you originally asked, Lebesgue integral doesn't use antiderivatives. Only the Riemann integral does. They both measure area under the curve but they use different methods. Whenever they both exist, they both agree and give you the same number but they use different methods so it doesn't even make sense to ask "if a function can be a Lebesgue integral". There is no such thing as "an indefinite Lebesgue integral". The Lebesgue integral is useful because it works on many functions that the Riemann integral can't handle BUT the Riemann integral is much easier to do/understand/develop and it also came first historically. So the Lebesgue integral is a (great) generalization of the Riemann integral so the classic example I gave you is one such function which Lebesgue can handle but Riemann can't.


Second Addendum:

Here's Martin's answer from the comments below

"In Lebesgue theory the primitive $F(x)=\int_a^x f(t)dt$ makes sense for every integrable function $f:[a,b]\rightarrow\mathbb{R}$ and by the Lebesgue differentiation theorem a function $F$ is a primitive iff it is absolutely continuous. Perhaps OP asks whether the derivative of an absolutely continuous function is always Riemann integrable. In other words, whether there is an integrable function that is not a.e. equal to a Riemann integrable function (yes: characteristic function of a fat Cantor set)."

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The statement "Every Riemann integrable function is Lebesgue integrable" is not correct. Consider $\sin[x]/x$, for example. –  Bombyx mori Apr 17 '13 at 6:58
    
No, I mean integral. –  user31953 Apr 17 '13 at 7:02
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Every proper Riemann integral exists as a Lebesgue integral. –  Random Variable Apr 17 '13 at 7:04
    
And the other way around? –  user31953 Apr 17 '13 at 7:14
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In Lebesgue theory the primitive $F(x) = \int_{a}^x f(t) \,dt$ makes sense for every integrable function $f \colon [a,b] \to \mathbb{R}$ and by the Lebesgue differentiation theorem a function $F$ is a primitive iff it is absolutely continuous. Perhaps OP asks whether the derivative of an absolutely continuous function is always Riemann integrable. In other words, whether there is an integrable function that is not a.e. equal to a Riemann integrable function (yes: characteristic function of a fat Cantor set). –  Martin Apr 17 '13 at 15:18

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