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For nonempty open sets $A,B\subset {\mathbb{R}}^n$, the Lebesgue measure of their Minkowski sum satisfiesIt also holds for compact sets and in standard measurable-set formulations with the appropriate measurability qualification. Normalize both volumes to one and apply the Prékopa–Leindler inequality to indicator functions.

First take nonempty bounded open sets $A,B\subset {\mathbb{R}}^n$. Their Minkowski sum is open and hence a Lebesgue measurable set. Apply the Prékopa–Leindler inequality to their indicator functions and the indicator function of $(1-\theta )A+\theta B$. This gives the multiplicative formTo obtain the usual additive Brunn–Minkowski inequality, put $a={\lambda }_n(A{)}^{1/n}$, $b={\lambda }_n(B{)}^{1/n}$, $A_0=A/a$ and $B_0=B/b$. Both normalized open sets have Lebesgue measure one. For $\theta =b/(a+b)$,The multiplicative bound and the scaling rule for Lebesgue measure implyReplacing $A,B$ by $A\cap B(0,R),B\cap B(0,R)$ and using continuity from below extends this form of the Brunn–Minkowski inequality to all nonempty open sets, including those of infinite Lebesgue measure. This open-set form is sufficient here and avoids any measurability qualification for sums of arbitrary measurable sets.

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.