Why $e^{-x^2}$ is Lebesgue integrable over $\mathbb R$?

Question

Why $x\longmapsto e^{-x^2}$ is Lebesgue integrable over $\mathbb R$ ? How to justify it rigorously ?

I know that on compact, if it's Riemann integrable, then it's also Lebesgue, but how does it work over $\mathbb R$ ? Because I know that Riemann integrable doesn't implies Lebesgue integrable.

@Did: Ok, but why $\frac{1}{1+x^2}$ Lebesgue integrable ? How can I show it rigorously ? — user330587, May 23 '17 at 15:57
@user330587 any non-negative Riemann integrable function is also Lebesgue integrable — Ben Grossmann, May 23 '17 at 16:44
@user330587 to prove this rigorously, it suffices to use the monotone convergence theorem — Ben Grossmann, May 23 '17 at 16:46

score 1 · Answer 1 · answered May 23 '17 at 15:27

1

Recall that one has: $$e^{-x^2}=o_{x\to\pm\infty}\left(\frac{1}{x^2}\right).$$ Therefore, since $2>1$, $\displaystyle x\mapsto\frac{1}{x^2}$ is Lebesgue-integrable over $]-\infty,-1[\cup]1,+\infty[$ and so is $x\mapsto e^{-x^2}$. Finally, notice that $x\mapsto e^{-x^2}$ is continuous over $[-1,1]$ which is compact so that $x\mapsto e^{-x^2}$ is Lebesgue-integrable over $[-1,1]$.

answered May 23 '17 at 15:27

C. Falcon

19,042

The question is for all of $\Bbb{R}.$ – Chickenmancer May 23 '17 at 15:31
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@Chickenmancer And this is precisely what I have shown since it is Lebesgue-integrable over $]\infty,-1[\cup]1,+\infty[$ and $[-1,1]$ whose union is $\mathbb{R}$. – C. Falcon May 23 '17 at 15:32
Why is $\frac{1}{x^2}$ lebesgue integrable over $]-\infty ,-1[\cup]1,+\infty [$ ? – user330587 May 23 '17 at 15:40
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@user330587 $\int_1^{\infty} \frac 1{x^2} dx$ is finite, as is $\int_{-\infty}^{-1} \frac 1{x^2} dx$ – Ben Grossmann May 23 '17 at 15:50

score 1 · Accepted Answer · answered May 23 '17 at 16:51

If $f \geq 0$ is Riemann integrable over $\Bbb R$, then it is also Lebesgue integrable.

To see this, it suffices to use the monotone convergence theorem. Define $$ f_n(x) = \begin{cases} f(x) & -n \leq x \leq n\\ 0 & \text{otherwise} \end{cases} $$ Note that $f_n \to f$ pointwise and for all $x \in \Bbb R$, $n \geq m \implies f_n(x) \geq f_m(x)$. By the monotone convergence theorem, we may conclude that $$ \int f\,d\mu = \int \lim_{n \to \infty } f_n \,d\mu = \lim_{n \to \infty} \int f_n \, d\mu = \lim_{n \to \infty} \int_{-n}^n f(x)\,dx = \int_{-\infty}^\infty f(x)\,dx $$

Why $e^{-x^2}$ is Lebesgue integrable over $\mathbb R$?

2 Answers2