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I'm learning about Lebesgue measure and Lebesgue integration.

I know that Lebesgue integration is superior to Riemann integration in the sense that we can integrate a much larger class of function with Lebesgue integral.

I would like to ask which is the main reason that creates the superiority of Lebesgue integral?

EDIT

As for me, basically, Lebesgue integration is another way to calculate the integration (or geometrically, the area under the curve), so why just changing the way we compute can make such a huge difference between those 2 kind of integration ?

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    $\begingroup$ The main reason is I think its nice behavior with respect to limits. For instance, we have the three limit theorems: monotone convergence, Fatou's lemma and dominated convergence. These give some rather simple conditions under which limits and integrals can be exchanged. Next, the space of functions (namely the various $L^p$ spaces for $p\in[1,\infty]$) are complete normed vector spaces (i.e Banach spaces). So this first of all provides us with examples of a huge class of Banach spaces, and moreover it opens the door for applying powerful tools of functional analysis. $\endgroup$ – peek-a-boo Feb 24 at 20:44
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    $\begingroup$ The best answer so far is a comment under the question. $\qquad$ $\endgroup$ – Michael Hardy Feb 24 at 20:50
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Riemann integration has trouble with functions which take very different values very rapidly. For example, the indicator function on the rationals is not Riemann integrable, because every dissection of an interval will trap a rational and an irrational between every point of your dissection.

Lebesgue integration, however, is perfectly fine with functions taking very different values rapidly in a wide range of circumstances: all that matters is that we can somehow split up the domain into "not too horrible" (i.e. measurable) sets on which the function is constant, or that we can approximate the function by such "simple" functions. The indicator function on the rationals is fine, for example, because the rationals are a "nice" set (a measurable one), as are their complement.


I should add that I've approached this from a "conceptually why is Lebesgue integration good?" angle; peek-a-boo commented from an "in practice, what properties of Lebesgue integration make it nice to work with?" angle, and I haven't touched that aspect.

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  • $\begingroup$ "peek-a-boo commented from an "in practice, what properties of Lebesgue integration make it nice to work with?" angle". It sounds to me more like what limit properties allow abstract integrals to be well defined. $\endgroup$ – Numerically illiterate Feb 24 at 22:09
  • $\begingroup$ Thanks for your comment. Basically, Lebesgue integration is another way to calculate the integration (or geometrically, the area under the curve), so why just changing the way we look at the computation can cause such a huge difference between those 2 kind of integration ? (I have edited my question for more clarity, can you have a look at it and add some more comment in your answer please) Thank you very much for your help! $\endgroup$ – InTheSearchForKnowledge Feb 24 at 22:16
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    $\begingroup$ I'm not really sure how much clearer I can be. Riemann integration requires the function to be locally well-behaved throughout its domain. Lebesgue integration allows you to split up the domain in complicated ways if necessary so that you can glue together very different kinds of well-behavedness in many different places. $\endgroup$ – Patrick Stevens Feb 24 at 22:20
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This is not a real answer to this question, I am not aiming at this since I see there are other nice answers. What I want to highlight is that Lebesgue is better than Riemann, as long as you are talking about definite integrals over, say, an interval $[a,b]$.

As long as you have improper integrals, the situation is reversed and the Lebesgue integral is no more a generalization of the Riemann (improper) Integral, since we may have functions that are Riemann integrable in the improper sense, but they are not Lebesgue integrable. As an example pick $f(x)= \frac{\sin(x)}{x}$, then $$\int_0^{\infty}\frac{\sin(x)}{x}dx$$ is a well defined real number, while $$\int_0^{\infty}\left|\frac{\sin(x)}{x}\right|dx$$ is not since the integral diverges.

Of course I agree that the Lebsegue integral is more general if we restrict to the definite notion of Riemann integral, but this answer was just to remind we have always to keep in mind there may be instances where the two notions do not agree in any sense.

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