A locally integrable function can fail to differentiate on a null set

The function $f$ takes only the values $0$ and $1$, so it is measurable and [given, algebra] bounded by $1$. Hence it is locally integrable on $\mathbb{R}$.

For $r_m:=2^{-2m}$, the set on which $f=1$ inside $(-r_m,r_m)$ is [L1, given, algebra] exactly the disjoint union ${\bigcup }_{k\ge m}(2^{-2k-1},2^{-2k}),$ whose total length is ${\sum }_{k\ge m}{2}^{-2k-1}=\frac{2^{-2m-1}}{1-2^{-2}}=\frac{2^{-2m+1}}{3}\mathrm{.}$ Therefore $A_{r_m}f(0)=\frac{1}{2r_m}{\int }_{-r_m}^{r_m}f(x)dx=\frac{1}{3}\mathrm{.}$

For $s_m:=2^{-2m-1}$, the set on which $f=1$ inside $(-s_m,s_m)$ is [L1, step 1.2, algebra] exactly ${\bigcup }_{k\ge m+1}(2^{-2k-1},2^{-2k}),$ whose total length is ${\sum }_{k\ge m+1}{2}^{-2k-1}=\frac{2^{-2m-1}}{3}\mathrm{.}$ Hence $A_{s_m}f(0)=\frac{1}{2s_m}{\int }_{-s_m}^{s_m}f(x)dx=\frac{1}{6}\mathrm{.}$

Steps 1.2 and 2.1 give two sequences of radii tending to $0$ along which [L2, step 1.2, step 2.1] $A_{r}f(0)$ tends to different values. So $A_{r}f(0)$ has no limit as $r\to 0^{+}$. This does not contradict [L2], because the exceptional set here is the singleton null set $\{0\}$.