Lebesgue integral of a ratio of Lebesgue densities

Question

I need a hint to solve the following problem:

$P$ is a probability mass on $\mathcal B(\mathbb R)$ with a Lebesgue density $h$, $f$ is another Lebesgue density. I need to show that $\int \frac{h(x)}{f(x)}dP \ge 1$.

Answer 1

Note that $$ \int\frac{h}{f}\,{\rm d}P=\int\frac{h^{2}}{f}\,{\rm d}\lambda, $$ where $\lambda$ is the Lebesgue measure on $\mathbb{R}$.

But by Cauchy Schwarz, we have $$ 1=\int h\,{\rm d}\lambda=\int\frac{h}{\sqrt{f}}\cdot\sqrt{f}\,{\rm d}\lambda\leq\sqrt{\int\frac{h^{2}}{f}\,{\rm d}\lambda}\cdot\sqrt{\int f\,{\rm d}\lambda}=\sqrt{\int\frac{h^{2}}{f}\,{\rm d}\lambda}=\sqrt{\int\frac{h}{f}\,{\rm d}P}. $$ The only slight problem (which I leave to you) is discussion of why there is no problem with dividing by zero, i.e. $\frac{h\left(x\right)}{\sqrt{f\left(x\right)}}$ does not make sense if $f\left(x\right)=0$.