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.
On
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 $[-1,1]$.
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 $$ \int f\,d\mu = \int \lim_{n \to \infty } f_n \,d\mu = \lim_{n \to \infty} \int f_n \, d\mu = \lim_{n \to \infty} \int_{-n}^n f(x)\,dx = \int_{-\infty}^\infty f(x)\,dx $$