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