Bound on Survival Function of Pointwise Product

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions.

Let $F_{f g}$ be the survival function of the pointwise product $f g$.

Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively.

Then:


 * $\ds \map {F_{f g} } {\alpha \beta} \le \map {F_f} \alpha + \map {F_g} \beta$ for all $\alpha, \beta \in \hointr 0 \infty$.

Proof
Let $\alpha, \beta \in \hointr 0 \infty$.

We show that:


 * $\set {x \in X : \size {\map f x \map g x} \ge \alpha \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$

Let $x \in X$ be such that $\size {\map f x \map g x} \ge \alpha \beta$.

If $\size {\map f x \map g x} = \infty$, we have $\size {\map f x} = \infty$ or $\size {\map g x}$.

Then we either have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$, so that:


 * $x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$.

Now suppose that $\size {\map f x \map g x} < \infty$.

Suppose that $\size {\map f x} < \alpha$ and $\size {\map g x} < \beta$, then $\size {\map f x \map g x} = \size {\map f x} \size {\map g x} < \alpha \beta$

contradicting that:


 * $\size {\map f x \map g x} \ge \alpha \beta$

So we must have $\size {\map f x} \ge \alpha$ or $\size {\map g x} \ge \beta$.

So we have:


 * $x \in \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$.

again in this case.

So:


 * $\set {x \in X : \size {\map f x \map g x} \ge \alpha \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$

by the definition of set inclusion.

Then we have: