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 $\closedint a b$

$\blacksquare$