= Solution
Write $\lambda_n$ for <Lebesgue measure> on $\mathbb R^n$. The <Prékopa–Leindler inequality> says that if $0<\theta<1$ and nonnegative <measurable functions> $f,g,h$ satisfy
$$
h((1-\theta)x+\theta y)\geq f(x)^{1-\theta}g(y)^\theta\qquad(x,y\in\mathbb R^n),
$$
then their <Lebesgue integrals> satisfy
$$
\boxed{\int h\geq\left(\int f\right)^{1-\theta}\left(\int g\right)^\theta.}
$$
It 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> gives
$$
a(u):=x'(u)=\frac{F}{f(x(u))}>0,\qquad b(u):=y'(u)=\frac{G}{g(y(u))}>0.
$$
For 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>:
$$
\int_{\mathbb R}h(s)\,ds\geq\int_0^1 h(z(u))z'(u)\,du.
$$
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 $\int_{z^{-1}(I)}z'\leq\lambda_1(I)$, 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 obtain
$$
\begin{aligned}
h(z(u))z'(u)&\geq F^{1-\theta}G^\theta\frac{(1-\theta)a(u)+\theta b(u)}{a(u)^{1-\theta}b(u)^\theta}\\
&\geq F^{1-\theta}G^\theta.
\end{aligned}
$$
Integrating over $0<u<1$ proves the one-dimensional <Prékopa–Leindler inequality>.
For higher dimensions, use <mathematical induction> and <Tonelli theorem>. Write $x=(x',s)$ and set
$$
F_1(x')=\int_{\mathbb R}f(x',s)\,ds,\qquad G_1(y')=\int_{\mathbb R}g(y',t)\,dt,\qquad H_1(z')=\int_{\mathbb R}h(z',r)\,dr.
$$
Apply the one-dimensional <Prékopa–Leindler inequality> to the last-coordinate slices, with $z'=(1-\theta)x'+\theta y'$. It gives
$$
H_1((1-\theta)x'+\theta y')\geq F_1(x')^{1-\theta}G_1(y')^\theta.
$$
Apply 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. \b[The result therefore holds for general nonnegative measurable functions.]
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