Difference 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 $f = f^+ - f^-$.
Proof
Let $x \in X$.
Consider the following four cases for the value of $\map f x$ in $\overline{\R}$:
$(1): \quad \map f x = -\infty$
- By ordering on extended reals:
- $\map {f^+} x = \map \max {0, \map f x} = \map \max {0, -\infty} = 0$
- $\map {f^-} x = - \map \min {0, \map f x} = - \map \min {0, -\infty} = +\infty$
- By extended real subtraction, it thus follows that:
- $\map {f^+} x - \map {f^-} x = 0 - \paren {+\infty} = - \infty = \map f x$
$(2): \quad \map f x \in \openint {-\infty} 0$
- Since $\map f x < 0$, it follows that:
- $\map {f^+} x = \map \max {0, \map f x} = 0$
- $\map {f^-} x = - \map \min {0, \map f x} = - \map f x$
- Thence it immediately follows that:
- $\map {f^+} x - \map {f^-} x = 0 - \paren {- \map f x} = \map f x$
$(3): \quad \map f x \in \hointr 0 {+\infty}$
- Since $\map f x \ge 0$, we obtain:
- $\map {f^+} x = \map \max {0, \map f x} = \map f x$
- $\map {f^-} x = - \min {0, \map f x} = 0$
- Then, these immediately imply:
- $\map {f^+} x - \map {f^-} x = \map f x - 0 = \map f x$
$(4): \quad \map f x = +\infty$
- By ordering on extended reals:
- $\map {f^+} x = \map \max {0, \map f x} = \map \max {0, +\infty} = +\infty$
- $\map {f^-} x = - \map \min {0, \map f x} = - \map \min {0, +\infty} = 0$
- By extended real subtraction, it now follows that:
- $\map {f^+} x - \map {f^-} x = +\infty - 0 = +\infty = \map f x$
In all cases, $\map {f^+} x - \map {f^-} x = \map f x$.
As $x \in X$ was arbitrary, we may conclude that:
- $\forall x \in X: \map {f^+} x - \map {f^-} x = \map f x$
That is, we have shown:
- $f = f^+ - f^-$
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.7 \ \text{(v)}$, $\S 8$: Problem $6$