Exercise 1.6.24 (Lebesgue density theorem)

Question

If $E \subset {\bf R}^d$ is Lebesgue measurable, show that almost every point in $E$ is a point of density for $E$, and almost every point in the complement of $E$ is not a point of density for $E$.

Elu Thingol · Accepted Answer

\begin{proof}
As the hint suggests, we use the Lebesgue differentiation theorem (Theorem 1.6.19). Since the indicator function $1_{E}$ is an absolutely integrable function, for almost every $x \in \mathbf{R}^d$, one has
$$\lim_{r 	o 0} \dfrac{m\left(E \cap B(x,r)ight)}{m\left(B(x,r)ight)}
= \lim_{r 	o 0} \dfrac{1}{m\left(B(x,r)ight)} \int_{B(x,r)} 1_E
= 1_{E}(x).$$
From this, we conclude that for almost every point $x \in E$ the above limit will be $1$ whereas for almost every point $x 
ot\in E$ the above limit will be $0$.

Exercise 1.6.24 (Lebesgue density theorem)

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Exercise 1.6.24 (Lebesgue density theorem)

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