Positive Part of Composition of Functions

Theorem

Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.

Let $T : X \to X'$ be a $\Sigma/\Sigma'$-measurable mapping.

Let $f : X' \to \overline \R$ be a function.

Then:

$\paren {f \circ T}^+ = f^+ \circ T$

where $\paren {f \circ T}^+$ denotes the positive part of $f \circ T$.

Proof

Let $x \in X$ be such that $\map {\paren {f \circ T} } x = \map f {\map T x} \ge 0$.

Then $\map f {\map T x} = \map {f^+} {\map T x}$ by the definition of the positive part.

So:

\(\ds \map {\paren {f \circ T}^+} x\)

\(=\)

\(\ds \map {\paren {f \circ T} } x\)

Definition of Positive Part

\(\ds \)

\(=\)

\(\ds \map f {\map T x}\)

\(\ds \)

\(=\)

\(\ds \map {f^+} {\map T x}\)

Now let $x \in X$ be such that $\map {\paren {f \circ T} } x = \map f {\map T x} < 0$.

Then $\map {f^+} {\map T x} = 0$ and $\map {\paren {f \circ T}^+} x = 0$ by the definition of the positive part.

So:

$\map {\paren {f \circ T}^+} x = \map {\paren {f^+ \circ T} } x$

for all $x \in X$.

$\blacksquare$