# Applying the definition of Lebesgue Integral to specific functions

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.

• 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. – Mark Viola Aug 31 '15 at 21:52
• Indeed, you are right. :) – Kolmin Aug 31 '15 at 21:58

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.