10
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Shouldn't you also have shown that the measure $\lambda_X$ is $\sigma$-finite? $\endgroup$ Commented Sep 29, 2021 at 11:57

1 Answer 1

Reset to default
4
$\begingroup$

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, $$ \dfrac{\mathrm{d}\lambda_{X}}{\mathrm{d}\lambda}{}={}\dfrac{1}{\sqrt{x}}{\bf{1}}_{x>0}\,. $$
$\endgroup$
2
  • $\begingroup$ @user182551 Hi, do you have any questions concerning this? Feel free to ask if you do. $\endgroup$ Commented Apr 3, 2015 at 19:10
  • 4
    $\begingroup$ Hi @ki3i, sorry to revive this old question. I was wondering if you could explain why it is sufficient to consider sets of the form $(a,b)$ when demonstrating absolute continuity? In light of math.stackexchange.com/questions/1581455/…, it appears as though in general it is insufficient to consider absolute continuity restricted to a generating set of the sigma algebra. Thanks in advance! $\endgroup$ Commented Dec 7, 2020 at 16:54

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.