So I want to check if my solution is correct I must show that the function $f: [-\infty,\infty] \rightarrow \mathbb{R}$ , $x\rightarrow e^{-x^{4}}$ is Lebesgue integrable.
For showing it I have written: $\int_{[-\infty,\infty]}f(x)d\lambda =\int_{[0,\infty]}2f(x)d\lambda$
$f$ is measurable since continous, so if we find a function $g$ that is an upper bound to $2f$ in $[0,\infty]$ and that is Riemann integrable in $[0,\infty]$ we have our proof.
So from what I know $g$ can be a noncontinous function. So I can choose:
$g(x):=\left\{\begin{matrix} 2 & 0 \leq x\leq 1\\ 2/e^x & x> 1 \end{matrix}\right.$
And we have that $\int_{[0,\infty]}2f(x)dx\leq \int_{[0,\infty]}g(x)dx=\int_{[0,1]}2dx+\int_{[1,\infty]}2/e^x dx< \infty$
From Riemann Integrable follows Lebesgue inetgrable and we have our proof.
Everything right or have I done some errors?
To show that $x \mapsto e^{-x^4}$ is Lebesgue integrable on $[1,\infty)$ and complete your argument, note that
$$\int_{[1,\infty)}e^{-x^4}\chi_{[1,b]} =\int_{[1,b]}e^{-x^4} \leqslant \int_{[1,b]}e^{-x}$$ If a function is Riemann integrable on a finite interval then it is Lebesgue integrable and the integrals coincide. Hence,
$$\int_{[1,\infty)}e^{-x^4}\chi_{[1,b]} \leqslant \underbrace{\int_{[1,b]}e^{-x}}_{Lebesgue} = \underbrace{\int_1^b e^{-x}dx}_{Riemann} = e^{-1}-e^{-b}$$
By the monotone convergence theorem, it follows that
$$\int_{[1,\infty)}e^{-x^4}=\int_{[1,\infty)}\lim_{b \to \infty}e^{-x^4}\chi_{[1,b]}= \lim_{b \to \infty}\int_{[1,\infty)}e^{-x^4}\chi_{[1,b]}\\\leqslant\lim_{b \to \infty} \int_1^b e^{-x}dx=\lim_{b \to \infty}(e^{-1} - e^{-b}) = e^{-1} < \infty$$
Thus, $x \mapsto e^{-x^4}$ is Lebesgue integrable on $[1,\infty)$.