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If $g$ is locally Lebesgue integrable andthen $G^{\prime }(x)=g(x)$ at every Lebesgue point of $g$, and hence almost everywhere.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

Work first with essential bounds. On a Euclidean ball $B_R(x_0)⋐B_1$, put $M={\mathrm{esssup}}_{B_R}u$, $m={\mathrm{essinf}}_{B_R}u$ and $\omega =M-m$. The functions $M-u$ and $u-m$ are nonnegative weak solutions, hence weak supersolutions. In $B_{R/2}$, at least one of the sets $\{u\le (M+m)/2\}$ and $\{u\ge (M+m)/2\}$ has at least half the measure.

In the first case, the Weak Harnack inequality applied to $M-u$ givesIn the second case, applying it to $u-m$ gives the same lower bound for the improvement of the minimum. Thus the oscillation decay estimate isThis measure argument works even when $q<1$, avoiding a false use of the triangle inequality in that range.

Iteration yields ${\mathrm{osc}}_{B_r(x_0)}u\le C(r/R{)}^{\mu }{\mathrm{osc}}_{B_R(x_0)}u$, with, for example, $\mu =\min \{1/2,-\log \tau /\log 2\}\in (0,1)$. At Lebesgue points the shrinking oscillations define a unique continuous representative, and the estimate extends to every point by continuity. For $x,y\in B_{1/4}$ at distance less than $1/4$, apply the estimate to a Euclidean ball centred at $x$ with initial radius $1/2$ and final radius comparable to $|x-y|$. This Euclidean ball stays in $B_1$. Larger distances use the trivial bound $2\Vert u{\Vert }_{\infty }$. We obtainThe constants depend only on $n,\lambda ,\Lambda$. This derives the relevant conclusion of the De Giorgi-Nash-Moser theorem directly from part (a).

For an integrable function $f:{\mathbb{R}}^n\to \mathbb{C}$, the Lebesgue differentiation theorem states thatfor almost everywhere $x\in {\mathbb{R}}^n$. In particular,at every Lebesgue point of $f$.

Almost every $x\in \mathbb{R}$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,For $t<0$, the same estimate over $[x+t,x]$ gives the identical limit. Thereforeat every Lebesgue point of $g$, so $G$ is differentiable $\lambda$-almost everywhere. This is the differentiation of an indefinite Lebesgue integral.