I am stuck on the following problem:

Let $f(x,y) = \max \{ x^2 + y^2 , 1 \}$ and define a Borel measure $\mu$ on $\mathbb{R}$ by $\mu(E) : = (m \times m)(f^{-1}(E))$, where $m$ is Lebesgue measure. Find the Radon-Nikodym derivative $d \rho/d m$, where $\rho$ is the absolutely continuous part of $\mu$.

It is easy to see that $\mu$ is a positive measure (but not finite), $\mu(- \infty, 1)) = 0$, and $\mu$ is not absolutely continuous with respect to $m$ (for instance $m(\{ 1\}) = 0$ while $\mu(\{1\}) = \pi$). I feel like I have a solid understanding of the measure, specifically it is not too difficult to compute the measure of intervals, but I am unsure how to go about the actual problem.

In Folland's analysis text, his proof of the Lebesgue Radon Nikodym Theorem is not constructive and did not help me out too much. It is clear that I am looking at this problem in the wrong way, so I was hoping for a helpful hint in the right direction.

According to Lebesgue's decomposition theorem, the measure $\mu$ decomposes as $$\mu = \rho + \nu,$$ where

• $\rho$ is absolutely continuous with respect to the Lebesgue measure $m$;

• $\nu$ and $m$ are singular with respect to each other.

Furthermore, this decomposition is unique.

This uniqueness of decomposition is key. It means that a possible way of solving this problem is to guess what $\rho$ and $\nu$ might be, based on intuition, then verify our the guesses obey the above properties.

As you say, it's not hard to come up with the right intuition here. To paraphrase your description:

• The "continuous" part of the measure is supported on $(1, \infty)$. By elementary calculus, we would expect that this continuous part is given by the $\rho(E) = \int_{\mathbb R} \pi .1_{E \cap (1,\infty)} \ dm$.

• The "discrete" part of the measure is supported on $\{ 1 \}$. It is simply $\nu(E) = \pi$ if $1 \in E$, or $\nu(E) = 0$ if $1 \notin E$.

So what needs to be done to formalise this?

• We need to show that my $\rho$ really is absolutely continuous w.r.t. $m$. This should be obvious, given the definition of $\rho$. It should also be obvious what the Radon-Nikodym derivative $d\rho / dm$ is.

• We need to show that $\nu$ and $m$ are singular. This shouldn't be too difficult either.

• We need to show that $\mu = \rho + \nu$. This can be done by switching to polar coordinates using the Jacobian change of variables formula...

• wouldn't we want $\rho(E) = \int_{E} \pi \chi_{(1,\infty)} \, dr$ because we are the definition of $f$ utilizes the square of the Euclidean norm in $\mathbb{R}^2$? For instance, $\mu((4,9)) = (m \times m)(f^{-1}((4,9))) = (m \times m)(\{ (x,y) : 4 < \| (x,y) \|^2 < 9 \} ) = (m \times m)(\{ (x,y) : 2 < \| (x , y) \| < 3\}) = \pi(3^2 - 2^2) =5 \pi$? – Oiler Jun 29 '18 at 20:49
• Also, would it be enough to show that $\rho \ll m$, $\nu \perp m$, and that $\mu = \lambda + \rho$ on the open intervals of $\mathbb{R}$ since they generate the Borel $\sigma$-algebra? – Oiler Jun 29 '18 at 20:51
• @Oiler Yes, you're absolutely right on the first point! (sorry, I misread the norm) – Kenny Wong Jun 29 '18 at 22:33
• @Oiler On the second point, I did think about whether it's enough to should that $\mu = \lambda + \rho$ on open intervals. The simplest argument I could come up with is the following: If we can show $\mu = \lambda + \rho$ on open intervals, then we've also shown it on open sets (because every open set is a countable union of open intervals). Now, we use Rudin Theorem 2.18, which states that every Borel measure on $\mathbb R$ that is finite on compact subsets is a regular measure. Since this applies to both $\mu$ and $\rho + \nu$, it suffices to prove equality on open sets. – Kenny Wong Jun 29 '18 at 22:38
• @Oiler Rudin uses this kind of argument (i.e. appealing to regularity properties of Borel measures) in his proof of Theorem 2.20 (b,c,d). That's where I got the idea from. (Btw I'm referring to Rudin's Real and Complex Analysis. If you don't have the book, then I'll mention that Th 2.20 is where Rudin proves basic properties about the Lebesgue measure, by proving them on open sets.) – Kenny Wong Jun 29 '18 at 23:04