Calculation of Radon–Nikodym derivative Suppose the function $X \colon \mathbb{R} \longrightarrow \mathbb{R} \colon x \longmapsto X(x) : = x^2$. 
I want to calculate the Radon–Nikodym derivative $\frac{\text{d}\lambda_X}{\text{d}\lambda}$, where $\lambda$ denotes the Lebesgue measure and $\lambda_X$ is the Pushforward measure of $X$ with respect to $\lambda$.
To calculate $\frac{\text{d}\lambda_X}{\text{d}\lambda}$ I first need to show that $\lambda_X \ll \lambda$. Now we have
$$ \lambda(X^{-1}(a,b))=\lambda(\sqrt{a},\sqrt{b})=\sqrt{b}-\sqrt{a}$$
and
$$ \lambda(a,b)=0 \Longrightarrow \lambda_X(a,b)=\lambda(\sqrt{a},\sqrt{b})=0$$
and hence $\lambda_X \ll \lambda$.
Now my questions are: 


*

*Is there a way to write the measure $\lambda_X$ explicitly down?

*How can I finally calculate $\frac{\text{d}\lambda_X}{\text{d}\lambda}$?


Calculating $\frac{\text{d}\lambda_X}{\text{d}\lambda}$ means, finding a measurable function $\varphi \geq 0$ with
$$\lambda_X(A) = \int_{A} \varphi \, \text{d} \lambda. $$
EDIT: By trying I found out that $\varphi(x)=2 \frac{1}{\sqrt{x}}$. Is this correct?
 A: If $X \colon \mathbb{R} \longrightarrow \mathbb{R^{+}}\bigcup\left\{0\right\}$ is defined as $ X(x) = x^2$, then observe that, for $b\ge a\ge 0$,
$$
\begin{eqnarray*}
\lambda_{X}\left(a\,,b\right){}:={}\lambda\left(X^{-1}\left(a\,,b\right)\right)&{}={}&\lambda\left(\left\{\left(-\sqrt{b}\,,-\sqrt{a}\right)\bigcup\left(\sqrt{a}\,,\sqrt{b}\right)\right\}\right)\newline
&&\newline
&{}={}&\lambda\left(-\sqrt{b}\,,-\sqrt{a}\right)+\lambda\left(\sqrt{a}\,,\sqrt{b}\right)\newline
&&\newline
&{}={}&2\left(\sqrt{b}-\sqrt{a}\right)\,.
\end{eqnarray*}
$$
Clearly, $\lambda_{X}$ is a non-negative, sigma-additive, sigma-finite set function, since $\lambda$ is and $X$ is Borel-measurable. And, $\lambda_X\ll\lambda$, since
\begin{eqnarray*}
\lambda\left(a\,,b\right){}={}0&{}\implies{}&b-a{}={}0\newline
&{}\implies{}& b{}={}a\newline
&{}\implies{}&\sqrt{b}{}={}\sqrt{a}\newline
&{}\implies{}&2\left(\sqrt{b}-\sqrt{a}\right){}={}0\newline
&{}\implies{}&\lambda\left(X^{-1}\left(a\,,b\right)\right)=0\,.
\end{eqnarray*}
Furthermore, by the fundamental theorem of calculus and up to sets of measure zero, 
$$
\begin{eqnarray*}
\lambda_{X}\left(a\,,b\right)&{}={}&2\left(\sqrt{b}-\sqrt{a}\right){}\overbrace{=}^{\scriptstyle Riemann\ Integral}{}\int\limits^{\infty}_{0}{\bf{1}}_{\left\{(a, b)\right\}}(x)\,\dfrac{1}{\sqrt{x}}\ \mathrm{d}x\newline
&&\newline
&{}\overbrace{=}^{\scriptstyle Lebesgue\ Integral}{}&\int\limits_{\mathbb{R}_{\ge0}}{\bf{1}}_{\left\{(a, b)\right\}}(x)\,\dfrac{1}{\sqrt{x}}\ \mathrm{d}\lambda(x)\newline
&&\newline
&{}={}&\int\limits_{(a, b)}\,\dfrac{1}{\sqrt{x}}\ \mathrm{d}\lambda(x)\newline
&&\newline
&{}={}&\int\limits_{(a,b)}\dfrac{\mathrm{d}\lambda_{X}}{\mathrm{d}\lambda}\ \mathrm{d}\lambda\,.
\end{eqnarray*}
$$

Therefore, up to sets of measure zero,
$$
\dfrac{\mathrm{d}\lambda_{X}}{\mathrm{d}\lambda}{}={}\dfrac{1}{\sqrt{x}}{\bf{1}}_{x>0}\,.
$$

