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I was wondering if there is a set of algebraic "rules" for finding primitives of Lebesgue integrals as there is one for finding primitives of Riemann integrals. I.e. for $x^{n}$ the primitive is $\frac{x^{n+1}}{n+1} + C_{0}$ in the Riemann world first presented in school. Are there similar rules for working with Lebesgue integrable functions that would allow to compute the primitives of more functions (that is, functions that would not be Riemann integrable)?

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    $\int D(x)\,dx=C$, where $D=\chi_{\mathbb Q}$ is the Dirichlet function. Does this count?2012-05-24
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    Isn't $C$ equal to 0 in this case? Is $D$ the only function for which we have an "easy" primitive?2012-05-24
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    All of the functions I can think of with a reasonably nice antiderivative are continuous, so Riemann integrable (on intervals). What kind of functions do you have in mind?2012-05-24
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    No, I think $C$ can be any real value (but now we are getting into the question of what $\int f(x)\,dx$ means, which is controversial). More seriously, the issue with your question is that you are asking for something **"algebraic"** (whatever that means) but **not Riemann integrable**. For any reasonable value of "algebraic" (maybe "elementary" is a better word), such functions are piecewise continuous, as Qiaochu already said.2012-05-24
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    I do not have any specific function in mind. I was just comparing the way Riemann integration is introduced in class then leads to rules such as the one I mention above and the corresponding pages of exercises (and even MIT competitions), when nothing similar happens in the class on Lebesgue integration, as far as I can tell.2012-05-24
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    @Leonid: going back to your example, we could then say that the "derivative" (a new kind of derivative) of any real valued constant is the Dirichlet function? I have never seen anything like a "Lebesgue derivative". That wouldn't make sense, would it?2012-05-24
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    @Frank Yes and no, depending on the meaning of "derivative". If the derivative is understood in the classical sense, $f'(x)=\lim_{h\to 0}(f(x+h)-f(x))/h$, then the derivative of a constant function is $0$ and nothing else. If the derivative is understood in the distributional/weak sense, then $C'=D$ is correct: a distributional derivative is whatever you can use in the FTC. So it looks like derivative is not unique, but we then make it unique by passing to equivalence classes of functions: $f\sim g$ if the set $\{f\ne g\}$ has measure zero.2012-05-24
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    That being said, you should not expect a new set of calculus-type integration recipes from the Lebesgue theory. Simply put, that is not what it's for.2012-05-24
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    @Leonid Yes, I'm now realizing it's not what it's for. So, is Lebesgue integrability used... as a limit of series of functions, and the game is then to find the limit? or as a theoretical tool in proofs? My question is about how mathematicians work with that concept in their daily activity.2012-05-24
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    I'm told to "avoid extended discussions in comments", so I'll end with this comment. Lebesgue integral allows one to shift focus from individual functions to spaces of functions ($L^1$, $L^2$, and many others) in which variational problems and PDE can be efficiently treated. If we used the Riemann integral to define such spaces, they would not be complete. With the Lebesgue integral, they are. To appreciate the importance of completeness, recall that we obtain $\mathbb R$ by completing $\mathbb Q$. Working with only Riemann integral is like doing analysis with rational numbers only.2012-05-24
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    @Leonid. Thanks! That makes sense.2012-05-24

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