# comparison between Lebesgue and Riemann Integration

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Major Section:
Reference
Type of Math Object:
Example

## Mathematics Subject Classification

### Improber Lebesque Integrals

I wonder why improper "Lebesque" Integrals are not defined, but when it comes to define improper integrals they are almost always defined as limits of "Riemann" integrals. Functions like Sin(x)/x multiplied by the characteristic function of the set of irrational numbers would have been integrable in the first sense.

### Re: Improber Lebesque Integrals

Here are the reasons I can think of:

(i) improper integrals fail translation invariance:

\int_{-\infty}^\infty f(x) dx \neq \int_{-\infty}^\infty f(x+t) dx

(ii) it doesn't have an obvious generalization to higher dimensions and other measure spaces (in what way do you take the domain of integration out to infinity?)

(iii) Fubini and limit theorems fail

(iv) In the context of Fourier analysis, where need to perform "improper integrals" when the function f is to be transformed is not in L^1, one can always take _any_ sequence
f_n in L^1 \cap L^2 that converges to f in L^2.
It's common to use the symmetric limit \int_{-T}^T f(x) e^{2\pi i \xi x} \, dx,
T \to \infty, but this restriction is not necessary.

### Re: Improber Lebesque Integrals

> Here are the reasons I can think of:
[why improper Lebesgue integrals are not defined]
>
> (i) improper integrals fail translation invariance:
>
> \int_{-\infty}^\infty f(x) dx \neq
> \int_{-\infty}^\infty f(x+t) dx
>
> (ii) it doesn't have an obvious generalization to higher
> dimensions and other measure spaces (in what way do you take
> the domain of integration out to infinity?)

The same is true for Riemann improper integrals. The reason they are called improper is that their value is defined by the limit of properly defined integrals. On the real line, the limit involves taking each of the integration limits progressively closer to a singularity or to infinity, one by one. There also exist other limiting procedures, e.g. Cauchy principal value. In higher dimensions, there is no limiting procedure quite as natural, but some can still be concocted. For example, on R^2 one could take the limit of integrals over successively larger discs centered at the origin. This would make the functions sin(sqrt(x^2+y^2))/(x^2+y^2)) improper integrable.

The same limiting procedures can be applied when Riemann integrals are replaced by Lebesgue ones. Obviously, the standard theorems of Lebesge integration will not apply to the result, but they would apply to each step in the limit. So the function sin(x)/x times the characteristic function of the irrationals is "improper Lebesgue" integrable as the original poster suggested.

### Re: Improber Lebesque Integrals

I just want to clarify my original comments here;
otherwise there is nothing new here.

> The same is true for Riemann improper integrals.
...
> There also exist other limiting procedures, e.g. Cauchy principal value.
...
> The same limiting procedures can be applied when Riemann integrals are replaced by
> Lebesgue ones.

All the more reason why Lebesgue integrals are usually not defined; there's no coherent and *useful* theory (i.e. equipped with Fubini-type and limit theorems)
that can applied to all (or perhaps even most) instances of "improper integrals".

> Obviously, the standard theorems of Lebesge integration will not apply
> to the result, but they would apply to each step in the limit.

So again there is no real loss in not considering theories of improper integrals
(actually, really just "conditionally convergent" integals), since everything we do with
improper integrals have to be justified step-by-step with proper integrals anyway. Improper integrals are defined in the Riemann case only because the original class of Riemann-integrable functions is too small --- arguably, this is not a problem with the Lebesgue integral.