Positive Part of Darboux Integrable Function is Integrable
Theorem
Let $f$ be a real function that is Darboux integrable over $\closedint a b$.
Let $f^+$ be the positive part of $f$.
Then $f^+$ is Darboux integrable over $\closedint a b$.
Corollary
Let $f$ be a real function that is Darboux integrable over $\closedint a b$.
Let $f^-$ be the negative part of $f$.
Then $f^-$ is Darboux integrable over $\closedint a b$.
Proof
Let $\epsilon > 0$ be a strictly positive real number.
By Condition for Darboux Integrability, there is a finite subdivision $P = \sequence {x_i}_{0 \mathop \le i \mathop \le n}$ such that:
- $\map {U_f} P - \map {L_f} P < \epsilon$
where $\map {U_f} P$ and $\map {L_f} P$ are the upper Darboux sum and lower Darboux sum, respectively.
Consider the terms of $\map {U_f} P$ and $\map {L_f} P$ for an arbitrary index $i$.
By Supremum does not Precede Infimum:
- $m_i^{\paren f} \le M_i^{\paren f}$
where $m_i^{\paren f}$ and $M_i^{\paren f}$ are respectively the infimum and supremum of $f$ on $\closedint {x_{i - 1} } {x_i}$.
Therefore, there are three cases to consider:
If $0 \le m_i^{\paren f} \le M_i^{\paren f}$:
- In this case, it follows from the definition of infimum that:
- $\forall x \in \closedint {x_{i - 1} } {x_i}: \map f x \ge 0$
- Therefore, on that interval, $\map {f^+} x = \map f x$.
- So:
- $M_i^{\paren {f^+} } - m_i^{\paren f} = M_i^{\paren f} - m_i^{\paren f}$
If $m_i^{\paren f} < 0 \le M_i^{\paren f}$:
- Because $\map {f^+} x \ge 0$, it follows that $m_i^{\paren {f^+} } \ge 0 > m_i^{\paren f}$.
- As $0 \le M_i^{\paren f}$, it follows that $\sup \paren {f \closedint a b \cup \set 0} = M_i^{\paren f}$.
- But $f^+ \closedint {x_{i - 1} } {x_i} \subseteq f \closedint {x_{i - 1} } {x_i} \cup \set 0$, so by Supremum of Subset:
- $M_i^{\paren {f^+} } \leq M_i^{\paren f}$
- Therefore:
- $M_i^{\paren {f^+} } - m_i^{\paren {f^+} } < M_i^{\paren f} - m_i^{\paren f}$
If $m_i^{\paren f} \le M_i^{\paren f} < 0$:
- Because $M_i^{\paren f} < 0$, it follows that $f$ is strictly negative on $\closedint {x_{i - 1} } {x_i}$.
- Thus, $\forall x \in \closedint {x_{i - 1} } {x_i}: \map {f^+} x = 0$.
- Therefore:
- $m_i^{\paren {f^+} } = M_i^{\paren {f^+} } = 0$.
- And it follows that:
- $M_i^{\paren {f^+} } - m_i^{\paren {f^+} } = 0 \le M_i^{\paren f} - m_i^{\paren f}$
Then, in every case:
- $M_i^{\paren {f^+} } - m_i^{\paren {f^+} } \le M_i^{\paren f} - m_i^{\paren f}$
But:
\(\ds \map {U_{f^+} } P - \map {L_{f^+} } P\) | \(=\) | \(\ds \sum_i M_i^{\paren {f^+} } \paren {x_i - x_{i - 1} } - \sum_i m_i^{\paren {f^+} } \paren {x_i - x_{i - 1} }\) | Definition of Upper Darboux Sum and Definition of Lower Darboux Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_i \paren {M_i^{\paren {f^+} } - m_i^{\paren {f^+} } } \paren {x_i - x_{i - 1} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_i \paren {M_i^{\paren f } - m_i^{\paren f } } \paren {x_i - x_{i - 1} }\) | Inequality holds termwise | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {U_f} P - \map {L_f} P\) | Definition of Upper Darboux Sum and Definition of Lower Darboux Sum | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
Because $\epsilon$ was arbitrary, it follows from Condition for Darboux Integrability that $f^+$ is integrable on $\closedint a b$
$\blacksquare$