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Assume in addition that the probability space is a metric space, continuous integrable functions are dense in $L^1$, and the diameters of its partition cells tend to zero at each point. For a continuous $g$, the oscillation average tends to zero because every point in $M_k(x)$ is within its diameter of $x$. For $h=f-g$, the upper limit of the oscillation of $f$ is bounded by the partition maximal function of $h$ plus $|h(x)|$. The first-crossing maximal inequality for refining partitions and Chebyshev's inequality bound the measure where this exceeds $2\epsilon$ by $2\Vert h{\Vert }_1/\epsilon$. Approximation proves the displayed almost everywhere assertion. Refinement supplies the maximal estimate; shrinking diameter supplies the approximation argument. Without shrinking diameter, a constant partition generally converges only to its cell averages.

The first-crossing maximal inequality for refining partitions follows from countable disjointness and the strict average bound:Thus this partition maximal operator has weak-type constant one.

The conclusion of differentiation along refining metric partitions isThe crucial geometric property is shrinking diameter, not merely refinement: since $x\in M_k(x)$ and $\mathrm{diam}{M}_k(x)<1/k$, continuity makes the corresponding oscillation tend to zero for every continuous test function. Continuous $L^1$ approximation and the maximal inequality then transfer this to $f$, exactly as before: the exceptional-set bound is $2\Vert f-g{\Vert }_1/\epsilon$ for an oscillation threshold $2\epsilon$. Countability makes the maximal function measurable, and positive cell measure makes every average defined. Refinement supplied the disjoint first-crossing argument; shrinking diameter supplies differentiation.