Negative Part of Composition of Functions
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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:
\(\ds \map {\paren {f \circ T}^-} x\) | \(=\) | \(\ds -\map {\paren {f \circ T} } x\) | Definition of Negative Part | |||||||||||
\(\ds \) | \(=\) | \(\ds -\map f {\map T x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {f^-} {\map T x}\) |
Now let $x \in X$ be such that $\map {\paren {f \circ T} } x = \map f {\map T x} \ge 0$.
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 $x \in X$.
$\blacksquare$