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I am fairly sure this question will sound rather naive, but I do have a problem with applying the Lebesgue Integral. Actually this question can be divide in two sub-question, related to two examples I have in mind.

  1. I know that the integral of $f(x) = x^{-\frac{1}{2}}$ over $[0,1]$ cannot be obtained through Riemann Integration. Fair enough, but how do we actually compute it through Lebesgue Integration?
    [Here I mean, from the very basics of the theory, without using calculus shortcuts, with the addition that we are allowed to do this from some theorem]

  2. I (sort of know) that we cannot compute the integral of a gaussian function $e^{-x^{2}}$ by using standard Riemann techniques. Can the Lebesgue integral accomplish it? If so, how?


PS: I do hope what I wrote make sense. If it is not the case, please feel free to point me out any conceptual mistake.

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    $\begingroup$ The improper Riemann Integral exists. But I take it that you mean strictly Riemann integrable and not in the improper sense that all of us know, love, then take for granted. $\endgroup$ – Mark Viola Aug 31 '15 at 21:52
  • $\begingroup$ Indeed, you are right. :) $\endgroup$ – Kolmin Aug 31 '15 at 21:58
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Computing integral of Gaussian on whole line is possible using standard techniques. The problem is that no elementary antiderivative exists so you can't easily obtain integrals over arbitrary intervals. The function is integrable in the Lebesgue sense, just as it is integrable in "improper" Riemann sense. So Lebesgue theory doesn't introduce anything new here. In general it doesn't equip us with many new computational tools. Instead of that significance relies on its good properties: working well with limits and nice spaces of integrable functions. But on the side of computations it gives you nothing new, except that it allows you to justify some useful procedures involving limits which would be difficult to justify in Riemann setting. The computation itself is unchanged though.

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The "usual" Riemann integral is not defined, but the Lebesgue integral is and is equal to the "improper" Riemann integral:

let $f(x) = x^{-\frac{1}{2}}$ on $(0,1]$ and $f(0)=0$. The $f$ is defined on $[0,1]$.

Now define $f_n(x)=f(x)\cdot\chi _{(\frac{1}{n},1]}(x)$, Then,

$f_n\nearrow f$ a.e. and so the Monotone Convergence Theorem applies to say that

$ \int_{[0,1]}fdt=\lim_{n\to \infty}\int _{[0,1]}f_ndt$ and now just observe that the RHS of this is the usual (convergent) Improper Riemann integral.

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