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$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\ep}{\epsilon} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

The nice answer by user michael still needs some brushing up, which is what I have done here. I think this could help readers appreciate michael's answer, the central point of which is using the Vitali covering lemma:

Let $A\subseteq\R^d$ be such that $\mu(\R^d\setminus A) = 0$ and $\lambda(A) = 0$. For any real $\epsilon>0$, there is an open set $O_\ep$ such that $ A \subset O_\ep$ and $\lambda(O_\ep) < \epsilon$. Take any real $c>0$ and let \begin{equation} A_c:=\{x\in A\colon \liminf_{r\downarrow0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}< c\}. \end{equation} For each $x\in A_c$ pick $r_x>0$ such that \begin{equation} B(x, r_x) \subset O_\ep\quad\text{and}\quad \frac {\mu(B(x,5r_x))}{\lambda(B(x,5r_x))}<c. \end{equation} By the Vitali covering lemma, there is a countable set $J\subseteq A_c$ such that the balls $B(x,r_x)$ are disjoint for distinct $x\in J$ and \begin{equation} A_c \subseteq \bigcup_{x\in J} B(x,5r_x). \end{equation} But then \begin{multline} \mu(A_c) \le \sum_{x\in J}\mu(B(x,5r_x))\le c \sum_{x\in J} \lambda(B(x,5r_x)) \\ \le c\,5^d \sum_{x\in J} \lambda(B(x,r_x)) \le c\,5^d \la(O_\ep)\le c\,5^d\ep. \end{multline} Letting $\ep\downarrow0$, we see that $\mu(A_c)=0$, for all real $c>0$. This implies that $\mu$-a.e. $$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\to0.$$

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\ep}{\epsilon} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

The nice answer by user michael still needs some brushing up, which is what I have done here. I think this could help readers appreciate michael's answer, the central point of which is using the Vitali covering lemma:

Let $A\subseteq\R^d$ be such that $\mu(\R^d\setminus A) = 0$ and $\lambda(A) = 0$. For any real $\epsilon>0$, there is an open set $O_\ep$ such that $ A \subset O_\ep$ and $\lambda(O_\ep) < \epsilon$. Take any real $c>0$ and let \begin{equation} A_c:=\{x\in A\colon \liminf_{r\downarrow0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}< c\}. \end{equation} For each $x\in A_c$ pick $r_x>0$ such that \begin{equation} B(x, r_x) \subset O_\ep\quad\text{and}\quad \frac {\mu(B(x,5r_x))}{\lambda(B(x,5r_x))}<c. \end{equation} By the Vitali covering lemma, there is a countable set $J\subseteq A_c$ such that the balls $B(x,r_x)$ are disjoint for distinct $x\in J$ and \begin{equation} A_c \subseteq \bigcup_{x\in J} B(x,5r_x). \end{equation} But then \begin{multline} \mu(A_c) \le \sum_{x\in J}\mu(B(x,5r_x))\le c \sum_{x\in J} \lambda(B(x,5r_x)) \\ \le c\,5^d \sum_{x\in J} \lambda(B(x,r_x)) \le c\,5^d \la(O_\ep)\le c\,5^d\ep. \end{multline} Letting $\ep\downarrow0$, we see that $\mu(A_c)=0$, for all real $c>0$. This implies that $\mu$-a.e. $$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\downarrow0.$$

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

Yes, the desired result holds for any mutually singular Radon measures $\mu$ and $\nu$, say on $\R^d$. Indeed, let $f$ and $g$ be, respectively, the Radon--Nikodym densities of $\mu$ and $\nu$ with respect to the measure $\rho:=\mu+\nu$. Then $fg=0$ $\rho$-a.e. The key observation is that, for each real $r>0$, the function $$x\mapsto f_r(x):=\frac{\mu(B(x,r))}{\la(B(x,r))}$$ is the convolution $f*k_r$ of $f$ with the kernel function $k_r$ given by the formula \begin{equation} k_r(x):=\frac1{r^d}\,k(x/r), \quad k:=\frac1{\la(B(0,1))}\,I_{B(0,1)}, \end{equation} where $I_{B(0,1)}$ is the indicator function of $B(0,1)$ and $\la$ is the Lebesgue measure. So, by a well-known fact (see e.g. theorem 8.15 in G.B. Folland Real analysis, Pure and Applied Mathematics, John Wiley & Sons Inc, New York (1984)), $f_r\to f$ and, similarly, $g_r\to g$ $\rho$-a.e., for the similarly defined $g_r$; the convergence everywhere here is as $r\downarrow0$. So, $f_r/g_r\to f/g$ $\rho$-a.e. and hence $\mu$-a.e.; also, $f/g=\infty$ $\mu$-a.e. Thus, $f_r/g_r\to\infty$ $\mu$-a.e., and so, \begin{equation} \frac{\mu(B(x,r))}{\nu(B(x,r))}=\frac{f_r(x)}{g_r(x)}\to\infty \end{equation} for $\mu$-almost all $x$, as desired.

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\ep}{\epsilon} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

The nice answer by user michael still needs some brushing up, which is what I have done here. I think this could help readers appreciate michael's answer, the central point of which is using the Vitali covering lemma:

Let $A\subseteq\R^d$ be such that $\mu(\R^d\setminus A) = 0$ and $\lambda(A) = 0$. For any real $\epsilon>0$, there is an open set $O_\ep$ such that $ A \subset O_\ep$ and $\lambda(O_\ep) < \epsilon$. Take any real $c>0$ and let \begin{equation} A_c:=\{x\in A\colon \liminf_{r\downarrow0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}< c\}. \end{equation} For each $x\in A_c$ pick $r_x>0$ such that \begin{equation} B(x, r_x) \subset O_\ep\quad\text{and}\quad \frac {\mu(B(x,5r_x))}{\lambda(B(x,5r_x))}<c. \end{equation} By the Vitali covering lemma, there is a countable set $J\subseteq A_c$ such that the balls $B(x,r_x)$ are disjoint for distinct $x\in J$ and \begin{equation} A_c \subseteq \bigcup_{x\in J} B(x,5r_x). \end{equation} But then \begin{multline} \mu(A_c) \le \sum_{x\in J}\mu(B(x,5r_x))\le c \sum_{x\in J} \lambda(B(x,5r_x)) \\ \le c\,5^d \sum_{x\in J} \lambda(B(x,r_x)) \le c\,5^d \la(O_\ep)\le c\,5^d\ep. \end{multline} Letting $\ep\downarrow0$, we see that $\mu(A_c)=0$, for all real $c>0$. This implies that $\mu$-a.e. $$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\to0.$$

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

Yes, the desired result holds for any mutually singular Radon measures $\mu$ and $\nu$, say on $\R^d$. Indeed, let $f$ and $g$ be, respectively, the Radon--Nikodym densities of $\mu$ and $\nu$ with respect to the measure $\rho:=\mu+\nu$. Then $fg=0$ $\rho$-a.e. The key observation is that, for each real $r>0$, the function $$x\mapsto f_r(x):=\frac{\mu(B(x,r))}{\la(B(x,r))}$$ is the convolution $f*k_r$ of $f$ with the kernel function $k_r$ given by the formula \begin{equation} k_r(x):=\frac1{r^d}\,k(x/r), \quad k:=\frac1{\la(B(0,1))}\,I_{B(0,1)}, \end{equation} where $I_{B(0,1)}$ is the indicator function of $B(0,1)$ and $\la$ is the Lebesgue measure. So, $f_r\to f$ and, similarly, $g_r\to g$ $\rho$-a.e., for the similarly defined $g_r$; the convergence everywhere here is as $r\downarrow0$. So, $f_r/g_r\to f/g$ $\rho$-a.e. and hence $\mu$-a.e.; also, $f/g=\infty$ $\mu$-a.e. Thus, $f_r/g_r\to\infty$ $\mu$-a.e., and so, \begin{equation} \frac{\mu(B(x,r))}{\nu(B(x,r))}=\frac{f_r(x)}{g_r(x)}\to\infty \end{equation} for $\mu$-almost all $x$, as desired.

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

Yes, the desired result holds for any mutually singular Radon measures $\mu$ and $\nu$, say on $\R^d$. Indeed, let $f$ and $g$ be, respectively, the Radon--Nikodym densities of $\mu$ and $\nu$ with respect to the measure $\rho:=\mu+\nu$. Then $fg=0$ $\rho$-a.e. The key observation is that, for each real $r>0$, the function $$x\mapsto f_r(x):=\frac{\mu(B(x,r))}{\la(B(x,r))}$$ is the convolution $f*k_r$ of $f$ with the kernel function $k_r$ given by the formula \begin{equation} k_r(x):=\frac1{r^d}\,k(x/r), \quad k:=\frac1{\la(B(0,1))}\,I_{B(0,1)}, \end{equation} where $I_{B(0,1)}$ is the indicator function of $B(0,1)$ and $\la$ is the Lebesgue measure. So, by a well-known fact (see e.g. theorem 8.15 in G.B. Folland Real analysis, Pure and Applied Mathematics, John Wiley & Sons Inc, New York (1984)), $f_r\to f$ and, similarly, $g_r\to g$ $\rho$-a.e., for the similarly defined $g_r$; the convergence everywhere here is as $r\downarrow0$. So, $f_r/g_r\to f/g$ $\rho$-a.e. and hence $\mu$-a.e.; also, $f/g=\infty$ $\mu$-a.e. Thus, $f_r/g_r\to\infty$ $\mu$-a.e., and so, \begin{equation} \frac{\mu(B(x,r))}{\nu(B(x,r))}=\frac{f_r(x)}{g_r(x)}\to\infty \end{equation} for $\mu$-almost all $x$, as desired.