What is the Radon-Nikodym derivative of the pushforward of a function?

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Here is this week's POTW:

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Let $f : \Bbb R\to \Bbb R$ be the function $f(x) = x^2$. Let $\mu$ be the pushforward of the Lebesgue measure $m$ with respect to $f$. Evaluate the Radon-Nikodym derivative of $\mu$ with respect to $m$.
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No one answered this week's problem. You can read my solution below.

Spoiler

The Radon-Nikodym derivative at point $x$, $D_m\mu(x)$, shall be found by cases.

Fix $x\in \Bbb R\setminus\{0\}$. Suppose $x > 0$. For $0 < r < x$,

$$\frac{\mu(x-r,x+r)}{m(x-r,x+r)} = \frac{m(\sqrt{x-r},\sqrt{x+r})}{2r} = \frac{\sqrt{x+r} - \sqrt{x-r}}{2r} = \frac{1}{\sqrt{x+r} + \sqrt{x-r}}.$$

Since the last expression tends to $\frac{1}{2\sqrt{x}}$ as $r\to 0$, then $D_m\mu(x) = \frac{1}{2\sqrt{x}}$.

Let $x \le 0$. For all $r > 0$,

$$\frac{\mu(x-r,x+r)}{m(x-r,x+r)} = \frac{m(0,\sqrt{x+r})}{2r} = \frac{\sqrt{x+r}}{2r}.$$

The last expression tends to infinity as $r\to 0$, so $D_m\mu(x) = \infty$.

In summary,

$$D_m\mu(x) = \begin{cases}\frac{1}{2\sqrt{x}} & \text{if $x > 0$.}\\0& \text{otherwise}\end{cases}$$