= Solution
For $r>0$, define the forward and backward <one-sided interval averaging operators>
$$
A_r^+f(x)=\frac1r\int_x^{x+r}f(t)\,dt,\qquad
A_r^-f(x)=\frac1r\int_{x-r}^{x}f(t)\,dt,
$$
and set $A_0^\pm f=f$. Each positive-radius average is bounded in <absolute value> by $m_u(f)(x)$. Although $x$ is an endpoint of its averaging interval, enlarge that interval slightly so that $x$ is interior, apply the maximal bound and let the enlargement shrink to zero.
To dominate the limiting identity too without assuming the <differentiation> result we are proving, use the <sublinear operator>
$$
S(f)=m_u(f)+|f|.
$$
The maximal estimate and <Chebyshev inequality> imply
$$
\lambda\{S(f)>\alpha\}
\leq\lambda\{m_u(f)>\alpha/2\}
+\lambda\{|f|>\alpha/2\}
\leq\frac8\alpha\|f\|_1.
$$
Thus $S$ has <weak type (1,1)> and dominates every $A_r^\pm$, including $r=0$.
Smooth <compactly supported> functions are dense in $L^1$, obtained by truncating and then smoothing with a <compactly supported> <approximate identity>. For each such <continuous function> both averages tend to its value at every point. Applying part (ii) to the two families, and intersecting their full-measure convergence sets, gives $A_r^\pm f(x)\to f(x)$ <almost everywhere>.
The forward <difference quotient> of the primitive is $A_r^+f(x)$, and the backward quotient is $A_r^-f(x)$:
$$
\frac{F(x+r)-F(x)}r=A_r^+f(x),\qquad
\frac{F(x-r)-F(x)}{-r}=A_r^-f(x).
$$
They have the same limit, proving \b[<differentiability> <almost everywhere>], with the stronger conclusion
$$
\boxed{F'(x)=f(x)\quad\text{almost everywhere}.}
$$
This proves <differentiation of an indefinite Lebesgue integral> directly from the maximal estimate and density, without assuming $m_u(f)\geq|f|$ as an unproved prerequisite.
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