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:
\(\ds \map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }\) | \(=\) | \(\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} \cup \set {x \in N : \size {\map f x} \ge \alpha} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } + \map \mu {\set {x \in N : \size {\map f x} \ge \alpha} }\) | from the countable additivity of $\mu$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} } + \map \mu N\) | Measure is Monotone | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {\set {x \in X \setminus N : \size {\map f x} \ge \alpha} }\) |
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.
$\blacksquare$
Sources
- 2014: Loukas Grafakos: Classical Fourier Analysis (3rd ed.) ... (previous) ... (next): $1.1.1$: The Distribution Function