Almost every point is a Lebesgue point of a locally integrable function

Let [L3, L5, L6, given, construct] $D:=\{a+ib:a,b\in \mathbb{Q}\}\mathrm{.}$ By [L3], $D$ is countable. For each $c\in D$, the function $u_c:=\mathrm{\mid }f-c\mathrm{\mid }$ is locally integrable, so [L6] and [L2] give a measurable ${\lambda }_n$-null set $N_c$ containing all exceptions such that ${\lim }_{r\to 0^{+}}{A}_r{u}_c(x)=u_c(x)=\mathrm{\mid }f(x)-c\mathrm{\mid }$ for every $x\mathrm{\notin }{N}_c$. Put $N:={\bigcup }_{c\in D}{N}_c\mathrm{.}$ By [L5], $N$ is null.

Fix $x\mathrm{\notin }N$ and let $\epsilon >0$. By density [L4], choose [step 1.1, L4, algebra] $c\in D$ with $\mathrm{\mid }f(x)-c\mathrm{\mid }<\epsilon$. Then for every $r>0$, $\frac{1}{\lambda (B(x,r))}{\int }_{B(x,r)}\mathrm{\mid }f(y)-f(x)\mathrm{\mid }d\lambda (y)\le A_r\mathrm{\mid }f-c\mathrm{\mid }(x)+\mathrm{\mid }c-f(x)\mathrm{\mid }\mathrm{.}$ Taking $r\to 0^{+}$ and using step 1.1 gives ${\mathrm{lim\hspace{0.17em}sup}}_{r\to 0^{+}}\frac{1}{\lambda (B(x,r))}{\int }_{B(x,r)}\mathrm{\mid }f(y)-f(x)\mathrm{\mid }d\lambda (y)\le 2\epsilon \mathrm{.}$ Since $\epsilon$ is arbitrary, the limit is $0$.

Step 2.1 holds for every $x\mathrm{\notin }N$, and $N$ is null. By [L1] and [L2], [L1, L2, step 1.1, step 2.1] the Lebesgue set of the class of $f$ has full measure.