Past exam of the real and harmonic analysis course of the University of Oxford / 2008 / 4 / a / iii / Solution

For a set $M$ of finite measure apply the preceding result to $f=1_M$. Since its local average minus its point value is bounded in absolute value by its mean absolute oscillation, it givesFor a general measurable $M$, apply the same argument to $1_{M\cap [-j,j]}\in L^1(\mathbb{R})$. At every interior point of $(-j,j)$, sufficiently short averages agree with those of $1_M$. The countable union of these interiors covers the line, and a countable union of null exceptional sets remains null. This proves the Lebesgue density theorem without a finite-measure assumption on $M$. The source's $\chi$ is the indicator of $M$.