How to go about showing that f(x) = $e^{-|x|}$ is Lebesgue integrable over $\mathbb R$?

Question

How would I go about showing that f(x) = $e^{-|x|}$ is Lebesgue integrable over $\mathbb R$?

Would the Monotone Convergence Theorem be useful?

I've also thought about using the following fact: Let f be a nonegative Lebesgue measurable function. Then $\lim_{n \to \infty} \int_{[-n.n]} f\,d\lambda$ = $\int_\mathbb R f\,d\lambda$.

Answer 1

The restriction to $I_n=[-n,n]$ is continuous, so Lebesgue integrable. If you do the integral, it's bounded by $2$. So take $f_n$ to be $f$ multiplied by the characteristic function of $I_n$, and apply monotone convergence.

Answer 2

We know that for some $M>0$, $e^{u}>u^{2}$ for all $u\geq M$, so $e^{-|x|}<|x|^{-2}$ for all $x$ with $|x|\geq M$. Now \begin{align*} \int_{\bf{R}}e^{-|x|}dx&=\int_{-M}^{M}e^{-|x|}dx+\int_{|x|\geq M}e^{-|x|}dx\\ &\leq 2M+\int_{|x|\geq M}\dfrac{1}{|x|^{2}}dx\\ &=2M+2\int_{M}^{\infty}\dfrac{1}{x^{2}}dx\\ &=2M+\dfrac{2}{M}\\ &<\infty. \end{align*}