Let $\nu$ and $\mu$ regular measures on $\mathbb{R}$ absolutely continuous with respect to the lebesgue measure $\lambda$.
I am asked to compute the following limit: $$ \lim_{r \to 0} \frac{(\mu \times \nu )(B_r(z))}{(\mu \times \lambda)(B_{r}(z))} $$ where $B_{r}(z)$ is the open ball of radius $r$ and center $z \in \mathbb{R}^{2}$.
In order the compute this limit, I tried to use the following corollary of the Lebesgue Differentiation theorem :
Theorem. Let $\nu$ be a signed regular measure on $\mathbb{R}^{n}$ and let $\nu=\nu_a+\nu_s$ be the Lebesgue decomposition of $\nu$ with respect to the Lebesgue measure $m$, that is, $\nu_a << m$ and $\nu_s \perp m$ . Then for $m$-almost every $x \in \mathbb{R}^{n}$: $$ \lim_{r \to 0} \frac{ \nu (E_r)}{ m(E_r)}=\frac{d \nu_a}{d m}(x) $$ For every family $(E_r )_{r>0}$ that shrinks nicely to $x$. Here, $\frac{d \nu_a}{d m}$ is the Radon-Nikodym derivative of $\nu_a$ with respect to $m$.
The condition on the family means that $E_r \subset B_r(x)$ for each $r>0$ and there is a constant $\alpha>0$ , independent of $r$, such that $m(E_r)> \alpha m(B_{r}(x))$.
Once said that, in this case since $\mu << \mu$ and $\nu << \lambda $ it follows that $\mu \times \nu << \mu \times \lambda$ and since $\mu$ and $\nu$ are regular, we have that $\mu \times \nu$ is regular. On the other hand, since $\mu \times \nu << \mu \times \lambda$ the Lebesgue decomposition of $\mu \times \nu$ with respect to $\mu \times \lambda$ is:
- $(\mu \times \nu)_a=\mu \times \nu$;
- $(\mu \times \nu)_s=0$.
Therefore, since the balls $B_{r}(z)$ are a family that shrinks nicely to $z$ we would have that: $$ \lim_{r \to 0} \frac{(\mu \times \nu )(B_r(z))}{(\mu \times \lambda)(B_{r}(z))}=\frac{d(\mu \times \nu)_a}{d(\mu \times \lambda)}(z)=\frac{d(\mu \times \nu)}{d(\mu \times \lambda)}(z) $$ for $\lambda$-almost every $z \in \mathbb{R}^2$.
However, apparently my answer is wrong and I don't know why. What did I do wrong?
In advance thank you
$\mu \times \lambda$ is not Lebesgue measure on $\mathbb{R}^2$, so you can't apply your theorem. Using some heuristic calculation (assuming all densities are smooth and compactly supported), I think the answer is $g(z_2)$, where $d\nu = g\,d\lambda$.
Actually, how about we write $$\frac{(\mu \times \nu)(B_r(z))}{(\mu \times \lambda)(B_r(z))} = \frac{(\mu \times \nu)(B_r(z))/\lambda(B_r(z))}{(\mu \times \lambda)(B_r(z))/\lambda(B_r(z))}.$$ If $d\mu(x) = f(x)\,d\lambda(x)$ and $d\nu(y) = g(y)\,d\lambda(y)$, then $d(\mu \times \nu)(x, y) = f(x)g(y) \,d\lambda(x, y)$ and $d(\mu \times \lambda)(x, y) = f(x)\,d\lambda(x, y)$, so by your theorem, the above limit as $r \to 0$ is $$\frac{f(z_1)g(z_2)}{f(z_1)} = g(z_2).$$