# Showing a product of two Lebesgue integrals is $\geq 1$ if the product of the integrands is $\geq 1$

Let $$\mu$$ be a probability measure on a set $$X$$, i.e. $$\mu(X)=1$$, and let $$f$$ and $$g$$ be positive measurable functions on $$X$$. Show that if $$fg\geq1$$, then the integral of $$f$$ times the integral of $$g$$ is greater than or equal to $$1$$.

I'm not sure how to approach this. Since $$f$$ and $$g$$ are positive, the integral of $$f$$ and the integral of $$g$$ are equal to the $$L_1$$ norm of $$f$$ and the $$L_1$$ norm of $$g$$, but I'm not sure why $$fg\geq 1$$ implies that the product of their $$L_1$$ norms is greater than or equal to $$1$$. Considering that $$fg$$ is involved, I was thinking of using Holder's inequality, but I'm not sure what $$p$$ and $$q$$ to use.

This relies more on an arithmetic trick than on analysis. Since $$fg \ge 1$$ you have also that $$\sqrt{fg} \ge 1$$ so that $$1 \le \int_X \sqrt{fg} \le \left(\int_X f \right)^{1/2} \left( \int_X g \right)^{1/2}$$ by Holder's inequality. Now square both sides.
• I don't understand your application of Holder's inequality. What are your $p$ and $q$? – Keshav Srinivasan Nov 8 '18 at 20:31
• $2$ and $2$. $\mbox{}$ – Umberto P. Nov 8 '18 at 20:32
• And how do we know that $\sqrt f$ and $\sqrt g$ are in $L_2$? – Keshav Srinivasan Nov 8 '18 at 20:35
• You don't. Since $f$ and $g$ are positive the integrals are all defined, even if possibly infinite. – Umberto P. Nov 8 '18 at 20:39