Sum of Positive and Negative Parts
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Theorem
Let $X$ be a set, and let $f: X \to \overline \R$ be an extended real-valued function.
Let $f^+, f^-: X \to \overline \R$ be the positive and negative parts of $f$, respectively.
Then $\size {f} = f^+ + f^-$, where $\size {f}$ is the absolute value of $f$.
Proof
Let $x \in X$.
Suppose that $\map f x \ge 0$, where $\ge$ signifies the extended real ordering.
Then $\size {\map f x} = \map f x$, and:
- $\map {f^+} x = \map \max {\map f x, 0} = \map f x$
- $\map {f^-} x = - \map \min {\map f x, 0} = 0$
Hence $\map {f^+} x + \map {f^-} x = \map f x = \size {\map f x}$.
Next, suppose that $\map f x < 0$, again in the extended real ordering.
Then $\size {\map f x} = - \map f x$, and:
- $\map {f^+} x = \map \max {\map f x, 0} = 0$
- $\map {f^-} x = - \map \min {\map f x, 0} = - \map f x$
Hence $\map {f^+} x + \map {f^-} x = - \map f x = \size {\map f x}$.
Thus, for all $x \in X$:
- $\map {f^+} x + \map {f^-} x = \size {\map f x}$
That is, $f^+ + f^- = \size {f}$.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.7 \ \text{(v)}$, $\S 8$: Problem $6$