[proofplan] We reduce to the case of finite total measure by localising to compact subsets. The proof has three parts. First, we establish a key covering lemma: if $D_\mu \nu \le \alpha$ on a set $A$, then $\nu(A) \le \alpha \mu(A)$ (and symmetrically for the reverse inequality), proved via a Vitali covering argument and outer regularity. Second, we use this lemma to show $D_\mu \nu$ is finite $\mu$-a.e. (the set where the upper derivative is infinite has $\mu$-measure zero by sending $\alpha \to \infty$) and that the limit exists $\mu$-a.e. (the set where liminf $<$ limsup has $\mu$-measure zero by a rational sandwich argument). Third, we prove $\mu$-measurability by expressing $D_\mu \nu$ as a pointwise limit of ratios $\nu(B(x,r))/\mu(B(x,r))$, each of which is $\mu$-measurable by upper semicontinuity of the ball measure map. [/proofplan] [step:Reduce to the case of finite total measure by localising to compact balls] Since $\mu$ and $\nu$ are [Radon measures](/page/Derivative%20of%20a%20radon%20measure) on $\mathbb{R}^n$, they are inner regular with respect to compact sets. Write $\mathbb{R}^n = \bigcup_{m=1}^{\infty} \overline{B}(0, m)$. The derivative $D_\mu \nu(x) = \lim_{r \to 0} \nu(B(x,r)) / \mu(B(x,r))$ depends only on the behaviour of $\mu$ and $\nu$ in arbitrarily small neighbourhoods of $x$, so it suffices to prove the result on each compact set $\overline{B}(0, m)$. The restrictions $\mu|_{\overline{B}(0,m)}$ and $\nu|_{\overline{B}(0,m)}$ are finite Radon measures. We therefore assume $\mu(\mathbb{R}^n) < \infty$ and $\nu(\mathbb{R}^n) < \infty$ for the remainder of the proof. [guided] Why can we reduce to finite total measure? The derivative $D_\mu \nu(x) = \lim_{r \to 0} \nu(B(x,r)) / \mu(B(x,r))$ is a local quantity: it depends only on the values of $\mu$ and $\nu$ on balls centred at $x$ of arbitrarily small radius. The behaviour of $\mu$ and $\nu$ far from $x$ is irrelevant to the limit. Since $\mu$ and $\nu$ are [Radon measures](/page/Derivative%20of%20a%20radon%20measure) on $\mathbb{R}^n$, they assign finite measure to every compact set. In particular, for each $m \ge 1$, the restrictions $\mu|_{\overline{B}(0,m)}$ and $\nu|_{\overline{B}(0,m)}$ are finite Radon measures. Write $\mathbb{R}^n = \bigcup_{m=1}^{\infty} \overline{B}(0, m)$. For each $m$, the derivative $D_\mu \nu(x)$ for $x \in \overline{B}(0, m)$ depends only on $\mu$ and $\nu$ restricted to a neighbourhood of $x$ (contained in $\overline{B}(0, m+1)$ for small enough radii). So if we prove the result — existence, finiteness, and measurability of $D_\mu \nu$ — on each $\overline{B}(0, m)$ using the finite measures $\mu|_{\overline{B}(0,m)}$ and $\nu|_{\overline{B}(0,m)}$, we obtain the result on all of $\mathbb{R}^n$. Why does this union argument work? The set of exceptional points where $D_\mu \nu$ fails to exist or is infinite is $\mu$-null on each $\overline{B}(0, m)$. Since $\mathbb{R}^n = \bigcup_m \overline{B}(0, m)$, the full exceptional set is a countable union of $\mu$-null sets, hence $\mu$-null. We may therefore assume without loss of generality that $\mu(\mathbb{R}^n) < \infty$ and $\nu(\mathbb{R}^n) < \infty$ for the remainder of the proof. This finiteness is used in two essential places: the covering lemma (next step) needs $\nu(\mathbb{R}^n) < \infty$ to bound $\mu(I) \le \nu(\mathbb{R}^n)/\alpha \to 0$, and the Vitali covering argument requires finite ambient measure to guarantee convergent series. [/guided] [/step] [step:Establish a covering lemma translating pointwise derivative bounds into measure bounds] [claim:Covering lemma] Fix $0 < \alpha < \infty$. - (i) If $A \subset \{x \in \mathbb{R}^n \mid D_\mu \nu(x) \le \alpha\}$, then $\nu(A) \le \alpha \mu(A)$. - (ii) If $A \subset \{x \in \mathbb{R}^n \mid D_\mu \nu(x) \ge \alpha\}$, then $\nu(A) \ge \alpha \mu(A)$. [/claim] [proof] **Part (i).** Fix $\varepsilon > 0$ and let $U \supset A$ be an open set. For each $a \in A$, the condition $D_\mu \nu(a) \le \alpha$ means that for every $\varepsilon > 0$, there exist arbitrarily small radii $r > 0$ with $\nu(B(a, r)) \le (\alpha + \varepsilon) \mu(B(a, r))$. Define the family \begin{align*} \mathcal{F} := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \le (\alpha + \varepsilon) \mu(B(a, r)) \}. \end{align*} Since $D_\mu \nu(a) \le \alpha$ for each $a \in A$, the family $\mathcal{F}$ contains balls of arbitrarily small radius centred at each point of $A$, so $\mathcal{F}$ is a fine cover of $A$. Since $\nu$ is a Radon measure, the [Vitali Covering Theorem](/theorems/15) applies to $\nu$ and the fine cover $\mathcal{F}$. It produces a countable pairwise disjoint subfamily $\mathcal{G} \subset \mathcal{F}$ such that \begin{align*} \nu\!\Bigl(A \setminus \bigcup_{B \in \mathcal{G}} B\Bigr) = 0. \end{align*} Using subadditivity of $\nu$, the defining property of $\mathcal{F}$, and disjointness of $\mathcal{G}$ with $\bigcup \mathcal{G} \subset U$: \begin{align*} \nu(A) \le \sum_{B \in \mathcal{G}} \nu(B) \le (\alpha + \varepsilon) \sum_{B \in \mathcal{G}} \mu(B) \le (\alpha + \varepsilon)\, \mu(U). \end{align*} Since $\mu$ is a [Radon measure](/page/Derivative%20of%20a%20radon%20measure), it is outer regular: $\mu(A) = \inf\{\mu(U) \mid U \supset A,\; U \text{ open}\}$. Taking the infimum over all open $U \supset A$ gives $\nu(A) \le (\alpha + \varepsilon)\, \mu(A)$. Sending $\varepsilon \to 0$ yields $\nu(A) \le \alpha\, \mu(A)$. **Part (ii).** Fix $\varepsilon > 0$ with $0 < \varepsilon < \alpha$, and let $U \supset A$ be an open set. For each $a \in A$, the condition $D_\mu \nu(a) \ge \alpha$ means there exist arbitrarily small radii $r > 0$ with $\nu(B(a, r)) \ge (\alpha - \varepsilon)\, \mu(B(a, r))$. Define \begin{align*} \mathcal{F}' := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \ge (\alpha - \varepsilon)\, \mu(B(a, r)) \}. \end{align*} This family is again a Vitali cover of $A$: it contains balls of arbitrarily small radius centred at each point of $A$. The [Vitali Covering Theorem](/theorems/15) applies because $\mu$ is a Radon measure and $\mathcal{F}'$ is a fine cover of $A$. It produces a countable pairwise disjoint subfamily $\mathcal{G}' \subset \mathcal{F}'$ with $\mu(A \setminus \bigcup_{B \in \mathcal{G}'} B) = 0$. Then \begin{align*} \nu(A) \ge \nu\!\Bigl(\bigcup_{B \in \mathcal{G}'} B \cap A\Bigr) = \sum_{B \in \mathcal{G}'} \nu(B \cap A) \le \sum_{B \in \mathcal{G}'} \nu(B) \ge (\alpha - \varepsilon) \sum_{B \in \mathcal{G}'} \mu(B) \ge (\alpha - \varepsilon)\, \mu(A), \end{align*} The first inequality uses monotonicity of $\nu$. The first equality uses countable additivity of $\nu$ on the pairwise disjoint family $\mathcal{G}'$. The second inequality uses the defining property of $\mathcal{F}'$. The final inequality uses $\mu(A) = \mu(A \cap \bigcup \mathcal{G}') \le \sum_{B \in \mathcal{G}'} \mu(B)$, which holds since $\mu(A \setminus \bigcup \mathcal{G}') = 0$ and the balls are disjoint. Sending $\varepsilon \to 0$ gives $\nu(A) \ge \alpha\, \mu(A)$. [/proof] [guided] This lemma is the core technical tool. It translates pointwise derivative bounds into global measure comparisons using a Vitali covering argument. The idea: if $D_\mu \nu(x) \le \alpha$ at every point of $A$, then near each $x \in A$ the ratio $\nu(B)/\mu(B)$ is at most $\alpha + \varepsilon$ for small balls. By selecting a disjoint subcover of $A$ from such balls via Vitali, we sum the local bounds to obtain a global bound. **Part (i) in detail.** Fix $\varepsilon > 0$ and an open set $U \supset A$. For each $a \in A$, since $D_\mu \nu(a) \le \alpha$, we can find balls $B(a, r) \subset U$ of arbitrarily small radius satisfying $\nu(B(a, r)) \le (\alpha + \varepsilon)\, \mu(B(a, r))$. The collection \begin{align*} \mathcal{F} := \{ B(a, r) \subset U \mid a \in A,\; \nu(B(a, r)) \le (\alpha + \varepsilon)\, \mu(B(a, r)) \} \end{align*} is a Vitali cover of $A$: it contains balls of arbitrarily small radius centred at every point of $A$. The [Vitali Covering Theorem](/theorems/15) for Radon measures provides a countable disjoint sub-collection $\mathcal{G} \subset \mathcal{F}$ covering $A$ up to $\nu$-measure zero: $\nu(A \setminus \bigcup_{B \in \mathcal{G}} B) = 0$. Summing the local estimates over $\mathcal{G}$: \begin{align*} \nu(A) \le \sum_{B \in \mathcal{G}} \nu(B) \le (\alpha + \varepsilon) \sum_{B \in \mathcal{G}} \mu(B) \le (\alpha + \varepsilon)\, \mu(U). \end{align*} The first inequality uses $\nu(A \setminus \bigcup \mathcal{G}) = 0$, the second uses the defining property of $\mathcal{F}$, and the third uses disjointness of $\mathcal{G}$ with $\bigcup \mathcal{G} \subset U$. Since $\mu$ is Radon, it is outer regular: $\mu(A) = \inf\{\mu(U) \mid U \supset A,\; U \text{ open}\}$. Taking the infimum over $U$: $\nu(A) \le (\alpha + \varepsilon)\, \mu(A)$. Sending $\varepsilon \to 0$: $\nu(A) \le \alpha\, \mu(A)$. **Part (ii).** The argument is parallel but with the inequality reversed. For each $a \in A$ with $D_\mu \nu(a) \ge \alpha$, we find balls satisfying $\nu(B(a,r)) \ge (\alpha - \varepsilon)\, \mu(B(a,r))$, form a Vitali cover, extract a disjoint subcover $\mathcal{G}'$ with $\mu(A \setminus \bigcup \mathcal{G}') = 0$, and obtain \begin{align*} \nu(A) \ge \sum_{B \in \mathcal{G}'} \nu(B) \ge (\alpha - \varepsilon) \sum_{B \in \mathcal{G}'} \mu(B) \ge (\alpha - \varepsilon)\, \mu(A). \end{align*} Sending $\varepsilon \to 0$ gives $\nu(A) \ge \alpha\, \mu(A)$. Why does Vitali cover $A$ up to $\nu$-null in part (i) but up to $\mu$-null in part (ii)? In part (i) we need the uncovered remainder to have zero $\nu$-measure for the first inequality in the chain. In part (ii) we need $\mu(A) \le \sum \mu(B)$, so the remainder must be $\mu$-null. Both applications are valid since $\mu$ and $\nu$ are both Radon measures, and the [Vitali Covering Theorem](/theorems/15) applies to any Radon measure. [/guided] [/step] [step:Show $D_\mu \nu$ is finite $\mu$-a.e. by sending the covering lemma bound to infinity] Define $I := \{x \in \mathbb{R}^n \mid D_\mu \nu(x) = +\infty\}$, where $D_\mu \nu(x) = +\infty$ means $\limsup_{r \to 0} \nu(B(x,r))/\mu(B(x,r)) = +\infty$. For each $\alpha > 0$, we have $I \subset \{x \mid D_\mu \nu(x) \ge \alpha\}$. By part (ii) of the covering lemma: \begin{align*} \nu(I) \ge \alpha\, \mu(I), \end{align*} hence $\mu(I) \le \nu(I)/\alpha \le \nu(\mathbb{R}^n)/\alpha$. Since $\nu(\mathbb{R}^n) < \infty$ (by the reduction to finite measures), sending $\alpha \to \infty$ gives $\mu(I) = 0$. [guided] We need to show that $D_\mu \nu$ is finite $\mu$-a.e. Let $I = \{x \in \mathbb{R}^n \mid D_\mu \nu(x) = +\infty\}$. For any $\alpha > 0$, the set $I$ is contained in $\{x \mid D_\mu \nu(x) \ge \alpha\}$, so the covering lemma (part ii) gives \begin{align*} \nu(I) \ge \alpha\, \mu(I). \end{align*} Rearranging: $\mu(I) \le \nu(I)/\alpha \le \nu(\mathbb{R}^n)/\alpha$. Since $\nu(\mathbb{R}^n) < \infty$ (we reduced to finite total measure in the first step), the right-hand side tends to zero as $\alpha \to \infty$. Therefore $\mu(I) = 0$, and $D_\mu \nu$ is finite $\mu$-a.e. The key idea is that having an infinite derivative at $x$ means $\nu$ assigns disproportionately large mass near $x$ relative to $\mu$. But $\nu$ has finite total mass, so this can only happen on a $\mu$-negligible set. [/guided] [/step] [step:Show the limit exists $\mu$-a.e. via a rational sandwich argument] For rational $0 < a < b$, define \begin{align*} R(a, b) := \Bigl\{ x \in \mathbb{R}^n \;\Big|\; \liminf_{r \to 0} \frac{\nu(B(x,r))}{\mu(B(x,r))} < a < b < \limsup_{r \to 0} \frac{\nu(B(x,r))}{\mu(B(x,r))} < \infty \Bigr\}. \end{align*} On $R(a,b)$, the upper derivative exceeds $b$, so $R(a,b) \subset \{x \mid D_\mu \nu(x) \ge b\}$ (where $D_\mu \nu$ denotes the upper derivative). The covering lemma (part ii) gives $\nu(R(a,b)) \ge b\, \mu(R(a,b))$. On $R(a,b)$, the lower derivative is below $a$, so $R(a,b) \subset \{x \mid D_\mu \nu(x) \le a\}$ (where $D_\mu \nu$ denotes the lower derivative). The covering lemma (part i) gives $\nu(R(a,b)) \le a\, \mu(R(a,b))$. Combining: \begin{align*} b\, \mu(R(a,b)) \le \nu(R(a,b)) \le a\, \mu(R(a,b)). \end{align*} Since $b > a$, this forces $\mu(R(a,b)) = 0$. The set where the limit does not exist finitely is \begin{align*} I \cup \bigcup_{\substack{a, b \in \mathbb{Q} \\ 0 < a < b}} R(a, b), \end{align*} which is a countable union of $\mu$-null sets (we showed $\mu(I) = 0$ in the previous step and $\mu(R(a,b)) = 0$ for each pair of rationals), hence $\mu$-null. Therefore $D_\mu \nu(x)$ exists and is finite for $\mu$-a.e. $x \in \mathbb{R}^n$. [guided] We now show that the limit $\lim_{r \to 0} \nu(B(x,r))/\mu(B(x,r))$ exists $\mu$-a.e. The limit fails to exist finitely precisely when either $\limsup = \infty$ (already handled: the set $I$ has $\mu(I) = 0$) or $\liminf < \limsup$ with both finite. For the second case, we use a rational sandwich. If $\liminf_{r \to 0} \nu(B(x,r))/\mu(B(x,r)) < \limsup_{r \to 0} \nu(B(x,r))/\mu(B(x,r))$ with both finite, then by density of the rationals there exist $a, b \in \mathbb{Q}$ with $\liminf < a < b < \limsup$. So the "bad" set is contained in \begin{align*} \bigcup_{\substack{a, b \in \mathbb{Q} \\ 0 < a < b}} R(a, b). \end{align*} On $R(a,b)$, the upper derivative exceeds $b$. We apply the covering lemma (part ii) with $\alpha = b$ to the set $R(a,b)$: this gives $\nu(R(a,b)) \ge b\, \mu(R(a,b))$. Simultaneously, the lower derivative on $R(a,b)$ is below $a$. We apply the covering lemma (part i) with $\alpha = a$: this gives $\nu(R(a,b)) \le a\, \mu(R(a,b))$. Combining: \begin{align*} b\, \mu(R(a,b)) \le \nu(R(a,b)) \le a\, \mu(R(a,b)). \end{align*} Since $b > a$, this forces $\mu(R(a,b)) = 0$. The set $\mathbb{Q} \times \mathbb{Q}$ is countable, so $\bigcup_{a < b} R(a,b)$ is a countable union of $\mu$-null sets, hence $\mu$-null. Together with $\mu(I) = 0$, we conclude that $D_\mu \nu(x)$ exists and is finite for $\mu$-a.e. $x \in \mathbb{R}^n$. This establishes the first assertion of the theorem. [/guided] [/step] [step:Establish upper semicontinuity of the ball measure map $x \mapsto \mu(B(x,r))$] [claim:Upper semicontinuity of ball measure] For each fixed $r > 0$, the map $x \mapsto \mu(B(x, r))$ is upper semicontinuous on $\mathbb{R}^n$, and the same holds for $\nu$. [/claim] [proof] Let $(y_k)_{k \ge 1}$ be a sequence in $\mathbb{R}^n$ with $y_k \to x$. For any $z \in \mathbb{R}^n$, if $\limsup_{k \to \infty} \mathbb{1}_{B(y_k, r)}(z) = 1$, then $z \in B(y_{k_l}, r)$ for some subsequence, so $|z - y_{k_l}| < r$, and taking $l \to \infty$ gives $|z - x| \le r$, hence $z \in \overline{B}(x, r)$. This shows $\limsup_{k \to \infty} \mathbb{1}_{B(y_k, r)} \le \mathbb{1}_{\overline{B}(x, r)}$ pointwise. For $k$ large enough that $B(y_k, r) \subset B(x, 2r)$, define the non-negative functions $g_k := \mathbb{1}_{B(x, 2r)} - \mathbb{1}_{B(y_k, r)} \ge 0$. Then $\liminf_{k \to \infty} g_k \ge \mathbb{1}_{B(x, 2r)} - \mathbb{1}_{\overline{B}(x, r)}$ pointwise. The sequence $(g_k)$ is non-negative and bounded by $1$, so Fatou's Lemma applies with respect to the finite measure $\mu|_{B(x,2r)}$: \begin{align*} \int_{\mathbb{R}^n} \liminf_{k \to \infty} g_k \, d\mu \le \liminf_{k \to \infty} \int_{\mathbb{R}^n} g_k \, d\mu. \end{align*} The left-hand side is at least $\mu(B(x, 2r)) - \mu(\overline{B}(x, r))$. The right-hand side equals $\mu(B(x, 2r)) - \limsup_{k \to \infty} \mu(B(y_k, r))$. Therefore \begin{align*} \limsup_{k \to \infty} \mu(B(y_k, r)) \le \mu(\overline{B}(x, r)). \end{align*} For the derivative $D_\mu \nu$, we take the limit as $r \to 0$ over a continuous parameter. The set of radii $r > 0$ for which $\mu(\partial B(x, r)) > 0$ is at most countable: the sets $\{\partial B(x, r)\}_{r > 0}$ are pairwise disjoint for distinct $r$, and $\mu$ has finite total mass, so at most countably many can carry positive $\mu$-measure. For all other $r$, we have $\mu(\overline{B}(x,r)) = \mu(B(x,r))$, so $\limsup_{k \to \infty} \mu(B(y_k, r)) \le \mu(B(x, r))$, confirming upper semicontinuity at generic radii. This suffices for establishing measurability of the derivative in the next step. [/proof] [guided] Why is upper semicontinuity the right property here? We need to show $f_r(x) = \nu(B(x,r))/\mu(B(x,r))$ is measurable, which requires $x \mapsto \mu(B(x,r))$ and $x \mapsto \nu(B(x,r))$ to be measurable. Upper semicontinuous functions are Borel measurable (the sublevel sets $\{f \le t\}$ are closed), so establishing upper semicontinuity is sufficient. The key difficulty is that the open ball $B(x,r)$ changes as $x$ varies: adding a small perturbation to $x$ shifts the ball, and points near the boundary $\partial B(x,r)$ may enter or leave. The set can only grow in the limit (since a point inside $B(y_k, r)$ for infinitely many $k$ lands in the closed ball $\overline{B}(x,r)$), which is exactly why we get an upper bound $\limsup \mu(B(y_k,r)) \le \mu(\overline{B}(x,r))$ rather than equality. The gap between $\mu(\overline{B}(x,r))$ and $\mu(B(x,r))$ is $\mu(\partial B(x,r))$, which is positive for at most countably many radii $r$ (since the boundaries $\partial B(x,r)$ are disjoint for distinct $r$ and $\mu$ has finite total mass). Since the derivative takes $r \to 0$ over a continuous parameter, we can always find a sequence of "good" radii $r_k \to 0$ with $\mu(\partial B(x, r_k)) = 0$, at which upper semicontinuity gives the full conclusion $\limsup \mu(B(y_k, r_k)) \le \mu(B(x, r_k))$. [/guided] [/step] [step:Conclude $\mu$-measurability of $D_\mu \nu$ as a pointwise limit of measurable ratios] For each $r > 0$, the map $x \mapsto \mu(B(x, r))$ is upper semicontinuous by the previous step, hence Borel [measurable](/page/Measurable%20Functions). The same holds for $x \mapsto \nu(B(x, r))$. Define the ratio function \begin{align*} f_r: \mathbb{R}^n &\to [0, +\infty] \\ x &\mapsto \begin{cases} \dfrac{\nu(B(x, r))}{\mu(B(x, r))} & \text{if } \mu(B(x, r)) > 0, \\ +\infty & \text{otherwise}. \end{cases} \end{align*} Since the numerator and denominator are Borel measurable, and the set $\{x \mid \mu(B(x,r)) > 0\}$ is Borel (as the preimage of $(0, \infty)$ under an upper semicontinuous function), the ratio $f_r$ is $\mu$-measurable for each $r > 0$. The derivative is the pointwise limit along the sequence $r = 1/k$: \begin{align*} D_\mu \nu(x) = \lim_{k \to \infty} f_{1/k}(x) \quad \text{for } \mu\text{-a.e. } x. \end{align*} Since a pointwise $\mu$-a.e. limit of $\mu$-measurable functions is $\mu$-measurable, $D_\mu \nu$ is $\mu$-measurable. This completes the proof of both assertions: $D_\mu \nu$ exists and is finite $\mu$-a.e. (established in the previous steps), and $D_\mu \nu$ is $\mu$-measurable. [guided] We now tie the proof together. The previous steps established that $D_\mu \nu(x) = \lim_{r \to 0} \nu(B(x,r))/\mu(B(x,r))$ exists and is finite for $\mu$-a.e. $x$. It remains to show this limit is a $\mu$-measurable function. The strategy is standard: express $D_\mu \nu$ as a pointwise limit of measurable functions, then invoke closure of measurability under pointwise limits. For each $r > 0$, the map $x \mapsto \mu(B(x,r))$ is upper semicontinuous (Step 5), hence Borel measurable. The same holds for $x \mapsto \nu(B(x,r))$. The ratio $f_r(x) = \nu(B(x,r))/\mu(B(x,r))$ (with the convention $+\infty$ when the denominator vanishes) is then Borel measurable, because it is a ratio of Borel measurable functions on the Borel set $\{x \mid \mu(B(x,r)) > 0\}$. Why do we restrict to the sequence $r = 1/k$ rather than taking $r \to 0$ over all positive reals? Because a pointwise limit of a countable sequence of measurable functions is measurable, while uncountable operations do not preserve measurability in general. Since the limit exists $\mu$-a.e. along any sequence $r_k \to 0$ (including $r_k = 1/k$), restricting to this countable sequence suffices. The function $D_\mu \nu(x) = \lim_{k \to \infty} f_{1/k}(x)$ is therefore a pointwise $\mu$-a.e. limit of $\mu$-measurable functions, hence $\mu$-measurable. This completes the proof. [/guided] [/step]