Lebesgue differentiation theorem

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

Let $a_n=2^{-2^n}$ and, in angular coordinates near zero, take a bounded indicator alternating between one and zero on the rings $a_{n+1}<|t|\le a_n$. Since $a_{n+1}/a_n\to 0$, the mean over $[-a_n,a_n]$ approaches the value of the outermost ring. Its even and odd subsequences approach one and zero. Thus an integrable function can fail to have a limiting centered average at a specified point, although Lebesgue differentiation theorem ensures convergence almost everywhere.

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.