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 $\closedint a b$.

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 and lower sums, 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:

Because $\epsilon$ was arbitrary, it follows from Condition for Darboux Integrability that $f^+$ is integrable on $\closedint a b$