Fix $M > 0$. For $x \in B_M(0)$ and $r \in (0, 1)$, we have $B_r(x) \subseteq B_{M+1}(0)$, so $E \cap B_r(x) = (E \cap B_{M+1}(0)) \cap B_r(x)$. The [function](/page/Function) $f := \mathbb{1}_{E \cap B_{M+1}(0)}$ satisfies $f \in L^1(\mathbb{R}^n)$ since $\mathcal{L}^n(B_{M+1}(0)) < \infty$. By the [Lebesgue Differentiation Theorem](/theorems/74), for a.e. $x \in B_M(0)$: \begin{align*} \lim_{r \downarrow 0} \frac{1}{\mathcal{L}^n(B_r(x))} \int_{B_r(x)} |f(y) - f(x)| \, d\mathcal{L}^n(y) = 0. \end{align*} Since $f = \mathbb{1}_{E \cap B_{M+1}(0)}$ and $x \in B_{M+1}(0)$, we have $f(x) = \mathbb{1}_E(x)$, and the [integral](/page/Integral) on the left equals $\frac{\mathcal{L}^n(E \cap B_r(x))}{\mathcal{L}^n(B_r(x))} - \mathbb{1}_E(x)$ in absolute value (splitting into the cases $x \in E$ and $x \notin E$). Hence the density ratio converges to $\mathbb{1}_E(x)$ for a.e. $x \in B_M(0)$. The exceptional set $A_M$ has measure zero. Since $\{x \in \mathbb{R}^n : \text{claim fails}\} = \bigcup_{M=1}^\infty A_M$ is a countable union of measure-zero [sets](/page/Set), it has measure zero.