= Solution
Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the <triangle inequality> yields
$$
Df(x)\leq Mh(x)+|h(x)|.
$$
Therefore the maximal inequality and <Chebyshev's inequality> give
$$
\lambda(\{Df>2\varepsilon\})\leq\lambda(\{Mh>\varepsilon\})+\lambda(\{|h|>\varepsilon\})\leq\frac4\varepsilon\|f-g\|_1.
$$
Density makes the right side arbitrarily small. For each positive rational $\varepsilon$ the exceptional set is null, proving \b[$Df=0$ <almost everywhere>], the <Lebesgue differentiation theorem>. Values where the chosen representative is not finite already form a <null set>.
For the fundamental theorem define $F(x)=\int_a^xf(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its <measure> is small by <absolute continuity> of the integral. At a <Lebesgue point> of $f$, for $h\ne0$,
$$
\left|\frac{F(x+h)-F(x)}h-f(x)\right|\leq\frac1{|h|}\int_{x-|h|}^{x+|h|}|f(t)-f(x)|dt\to0.
$$
Thus \b[$F'=f$ <almost everywhere>] and $F(b)-F(a)=\int_a^bf$. Differentiability at every point is not claimed for an arbitrary integrable function.
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