Measure density (${\rho }_{\mu ,A}(x)$)

For a measurable set $A\subseteq {\mathbb{R}}^n$, its density at $x$ with respect to a locally finite measure $\mu$ iswhen this limit exists.

The spherical derivative compares the mass of shrinking centered open balls with their Lebesgue measure, when the limit exists. This is a density of one measure relative to Euclidean volume, rather than the relative density of a set within one fixed measure. A singular finite Borel measure has spherical derivative zero Lebesgue almost everywhere.

For a finite singular Borel measure, choose compact subsets of a Lebesgue-null carrier whose omitted mass is arbitrarily small. Outside any such compact subset, sufficiently small balls see only the omitted measure. The uncentered maximal weak-type inequality bounds the volume of points where the omitted measure has large density. Letting its total mass tend to zero shows that the upper spherical density is zero almost everywhere. This argument does not require the null carrier itself to be closed or separated from every point outside it.

If $f$ is locally Lebesgue integrable on ${\mathbb{R}}^n$, then for Lebesgue almost every $x$,

A point $x\in {\mathbb{R}}^n$ is a Lebesgue point of a locally Lebesgue integrable function $f$ whenThe Lebesgue differentiation theorem says that almost every point is a Lebesgue point.

If $g$ is locally Lebesgue integrable andthen $G^{\prime }(x)=g(x)$ at every Lebesgue point of $g$, and hence almost everywhere.

For every Lebesgue measurable set $A\subseteq {\mathbb{R}}^n$, its Lebesgue density is $1$ at almost every point of $A$ and $0$ at almost every point of its complement.

If a measurable set $E\subseteq \mathbb{R}$ has positive measure, then for every $\epsilon >0$ there is a bounded interval $I$ such thatChoose a density-one point of a finite positive-measure portion of $E$ using the Lebesgue density theorem, and take a sufficiently small interval centred there.