> [!definition] > > Let $\bracs{E_r}_{r > 0} \subset \cb(\real^n)$ be a family of Borel sets. They **shrink nicely** to $x \in \real^n$ if > 1. $E_r \subset B(x, r)$ for each $r$. > 2. There exists $\alpha > 0$, such that for all $r > 0$, $m(E_r) > \alpha m(B(x, r))$. > [!theorem] Lebesgue Differentiation Theorem > > Let $f \in \loci$ be a [[Locally Integrable|locally integrable]] function, $x \in L_f$ be a point in its Lebesgue set, and $\bracs{E_r}_{r > 0}$ be a family of Borel sets that shrink nicely to $x$. Then > $ > \lim_{r \to 0}\frac{1}{m(E_r)}\int_{E_r}\abs{f(y) - f(x)}dy = 0 > $ > and > $ > \lim_{r \to 0}\int_{E_r} f(y)dy = f(x) > $ > *Proof*. Let $\alpha$ such that $m(E_r) > \alpha m(B(x, r))$ for all $r > 0$. Then > $ > \begin{align*} > \frac{1}{m(E_r)}\int_{E_r}\abs{f(y) - f(x)}dy &\le \frac{1}{m(E_r)}\int_{B(x, r)}\abs{f(y) - f(x)}dy \\ > &\le \frac{1}{m(B(x, r))}\int_{B(x, r)}\abs{f(y) - f(x)}dy > \end{align*} > $ > which goes to $0$ as $r \to 0$. > [!theorem] > > Let $\nu$ be a [[Regular Measure|regular]] [[Signed Measure|signed]]/[[Complex Measure|complex]] measure, and let > $ > d\nu = d\lambda + fdm \quad d\lambda \perp fdm > $ > be its [[Lebesgue-Radon-Nikodym Theorem|Lebesgue decomposition and Radon-Nikodym derivative]]. Then > $ > \lim_{r \to 0}\frac{\nu(E_r)}{m(E_r)} = f(x) \quad m\text{-a.e.} > $ > for every $\bracs{E_r}_{r > 0}$ that shrinks nicely to $x$. > > *Proof*. Let $\alpha > 0$ such that $m(E_r) > \alpha m(B(x, r))$ for all $r$, then > $ > \abs{\frac{\lambda(E_r)}{m(E_r)}} \le \frac{\abs{\lambda}(E_r)}{m(E_r)} \le \frac{\abs{\lambda}(B(x, r))}{m(E_r)} \le \frac{\abs{\lambda}(B(x, r))}{\alpha m(B(x, r))} > $ > and we can assume without loss of generality that $\lambda$ is positive. Let $A \in \cb(\real)$ be a support of $m$ such that $\lambda(A) = m(A^c) = 0$. Let > $ > F_k = \bracs{x \in A: \limsup_{r \to 0}\frac{\lambda(B(x, r))}{m(B(x, r))} > \frac{1}{k}} > $ > Let $\varepsilon > 0$, then by regularity, there exists $U_\varepsilon \supset A$ [[Open Set|open]] such that $\lambda(U_\varepsilon) < \varepsilon$. Let $x \in F_k$, then there exists $r_x > 0$ such that $\lambda(B(x, r_x)) > m(B(x, r_x))/k$. Take $V_\varepsilon = \bigcup_{x \in F_k}B(x, r_x)$, then there exists $\bracs{x_j}_1^J$ such that $\bracs{B(x_j, r_j)}_1^J$ are disjoint with > $ > \begin{align*} > m(V_\varepsilon) &\le 3^n\sum_{j \in [J]}m(B(x_j, r_j)) \\ > &\le 3^nk\sum_{j \in [J]}\lambda(B(x_j, r_j)) \\ > &\le 3^nk\lambda(V_\varepsilon) \le 3^nk\lambda(U_\varepsilon) \\ > m(F_k)&\le 3^nk\varepsilon > \end{align*} > $ > Since $\varepsilon$ is arbitrary, $m(F_k) = 0$ for all $k \in \nat$, so $\frac{d\nu}{dm}$ is equal to the given limit $m$-a.e.