Negative 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 $\map {\paren {f \circ T} } x = \map f {\map T x} \le 0$.

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

So:

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

Then $\map {f^-} {\map T x} = 0$ 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 $x \in X$.