Computing Lesbegue Integral Using the definition of super level set Use the definition of the Lebesgue integral in terms of level sets to compute: $$\int_{\mathbb{R^2}} e^{-\mid x\mid^2} dx$$
Also, I am required to verify my answer using polar coordinates. 
My Approach
Defining a super level set for $f(x)$ $$s_{f}(t):=\{x \in \mathbb{R^2}:f(x) \gt t\}$$
After solving I got these cases for $s_{f}(t)$
$$s_{f}(t)=\begin{cases}
\mathbb{R^2}, t \leq 0\\
0 \lt \mid x \mid \lt \sqrt{ln\mid 1/t \mid}, 0 \lt t \leq 1\\
\phi,t \gt 1
\end{cases}$$
My question: How to proceed forward and evaluate the Lebesgue Integral?
I know that the Lebesgue Integral formula that is : $$\int_{\mathbb{R^2}} f d\mu = \int_{0}^{\infty} \mu(s_f(t)) dt$$
$\mu$ is the measure.
 A: For $t> 0$ we have that
$$
e^{-|x|^2}>t\iff -|x|^2>\ln t\iff |x|<\sqrt{-\ln t}\tag1
$$
Also we knows that (as is shown in the book)
$$
\int_{\Bbb R ^2}e^{-|x|^2}\mu_2(dx)=\int_{[0,\infty)}\mu_2(S_f(t))\; \mu_1(dt)=\int_0^\infty\mu_2(S_f(t))\; dt\tag2
$$
where $\mu_n$ is the Lebesgue measure in $\Bbb R^n$ and the RHS above is an improper integral of Riemann. By last note that
$$
\mu_2(S_f(t))=\mu_2\left(\left\{x\in \Bbb R ^2:e^{-|x|^2}>t\right\}\right)
=\begin{cases}
\infty ,& t=0\\
-\pi\ln t,& t\in(0,1)\\
0, &\text{ otherwise }
\end{cases}\tag3
$$
because the Lebesgue measure in $\Bbb R^2$ is the notion of area, in this case the area of circles of infinite radius (if $t=0$), finite radius (if $t\in(0,1)$), or the area of the empty set when $t\geqslant  1$, that in measure theory is set to zero. And so
$$
\begin{align*}
\int_{\Bbb R ^2}e^{-|x|^2}\mu_2(dx)&=\int_{(0,1)}-\pi\ln t\,\mu_1(dt)\\
&=-\pi\lim_{s\to 0^+}\int_s^1\ln t\; dt\\
&=-\pi\lim_{s\to 0^+}[t\ln t-t]^{1}_{s}\\
&=\pi
\end{align*}\tag4
$$
Of course using polar coordinates we get the same answer, that is
$$
\int_{\Bbb R^2}e^{-|x|^2}\mu_2(dx)=\int_0^{2\pi}\int_0^\infty re^{-r^2}\;dr\; d\alpha=-\pi\int_0^\infty d(e^{-r^2})=\pi\tag5
$$
because $[e^{-r^2}]'=-2re^{-r^2}$. Also we can use Fubini's theorem to evaluate this integral, that is, if $x:=(x_1,x_2)$ then
$$
\begin{align*}
\int_{\Bbb R^2}e^{-|x|^2}\mu_2(dx)&=\int_{\Bbb R^2}e^{-x_1^2-x_2^2}\mu_2(d(x_1,x_2))\\
&=\left(\int_{\Bbb R}e^{-x_1^2}\mu_1(dx_1)\right)\left(\int_{\Bbb R}e^{-x_2^2}\mu_1(dx_2)\right)\\
&=\left(\int_{\Bbb R}e^{-s^2}\mu_1(ds)\right)^2\\
&=(\sqrt\pi)^2\\
&=\pi
\end{align*}\tag6
$$
provided that $\int_{\Bbb R}e^{-s^2}\, \mu_1(ds)=\sqrt\pi$, as can be seen here.
