Survival Function preserves Inequality

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

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


 * $\size f \le \size g$ $\mu$-almost everywhere.

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

Then:


 * $F_f \le F_g$

Proof
We aim to show that:


 * $\map {F_f} \alpha \le \map {F_g} \alpha$ for all $\alpha \in \hointr 0 \infty$.

That is:


 * $\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} }$ for all $\alpha \in \hointr 0 \infty$.

Since $f \le g$ $\mu$-almost everywhere, there exists a $\mu$-null set such that:


 * if $x \in X$ has $\size {\map f x} > \size {\map g x}$ then $x \in N$.

From Measurable Functions Determine Measurable Sets, in particular we have:


 * $\set {x \in X : \size {\map f x} > \size {\map g x} }$ is $\mu$-null

from Null Sets Closed under Subset.

Let $\alpha \in \hointr 0 \infty$.

We have:

Swapping $f$ for $g$ in this computation also gives:


 * $\ds \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} } = \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge \alpha} }$

So we aim to show that:


 * $\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X \setminus N : \size {\map g x} \ge \alpha} }$

From Measure is Monotone, it suffices to show:


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

Let $x \in X \setminus N$ have:


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

Then, we have:


 * $\alpha \le \size {\map f x} \le \size {\map g x}$

So we have:


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

So, from the definition of set inclusion, we have:


 * $\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \subseteq \set {x \in X \setminus N : \size {\map g x} \ge \alpha}$ for all $\alpha \in \hointr 0 \infty$

and hence the demand.