Existence of Integral on Union of Adjacent Intervals

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Theorem

Let $f$ be a real function defined on a closed interval $\closedint a b$ where $a < b$.

Let $c$ be a point in $\openint a b$.


Then:

$f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$

if and only if:

$f$ is Darboux integrable on $\closedint a b$.


Proof

Necessary Condition

We need to prove that $f$ is Darboux integrable on $\closedint a b$.

To do this it suffices to show that for all $\epsilon > 0$, there exists a subdivision $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$.

Here, $\map U S$ and $\map L S$ are, respectively, the upper and lower sums of $f$ on $\closedint a b$ with respect to the subdivision $S$.


Let a strictly positive $\epsilon$ be given.

Since $f$ is Darboux integrable on $\closedint a c$, we know that a subdivision $S_1$ of $\closedint a c$ exists such that $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$.

Since $f$ is Darboux integrable on $\closedint c b$, we know that a subdivision $S_2$ of $\closedint c b$ exists such that $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$.


Define the subdivision $S = S_1 \cup S_2$.

We observe that $S$ is a subdivision of $\closedint a b$.


We get by the definition of upper sum:

\(\displaystyle \map U S\) \(=\) \(\displaystyle \map U {S_1} + \map U {S_2}\) since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent

Also, by the definition of lower sum:

\(\displaystyle \map L S\) \(=\) \(\displaystyle \map L {S_1} + \map L {S_2}\) since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent

These two equations give:

\(\displaystyle \map U S – \map L S\) \(=\) \(\displaystyle \map U {S_1} + \map U {S_2} – \paren {\map L {S_1} + \map L {S_2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map U {S_1} – \map L {S_1} + \map U {S_2} – \map L {S_2}\)
\(\displaystyle \) \(<\) \(\displaystyle \dfrac \epsilon 2 + \dfrac \epsilon 2\) since $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$ and $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

This shows that $S$ satisfies $\map U S – \map L S < \epsilon$.

We conclude from this that $f$ is Darboux integrable on $\closedint a b$ since $S$ is a subdivision of $\closedint a b$.

$\Box$


Sufficient Condition

We need to prove that $f$ is Darboux integrable on $\closedint a c$ and $\closedint c b$.


Let a strictly positive $\epsilon$ be given.

The Darboux integrability of $f$ on $\closedint a b$ implies that a subdivision $P$ of $\closedint a b$ exists such that $\map U P – \map L P < \epsilon$.

Here, $\map U P$ and $\map L P$ are, respectively, the upper and lower sums of $f$ on $\closedint a b$ with respect to the subdivision $P$.


Define the subdivision $S = P \cup \set c$.

We observe that $S$ equals $P$ if $c$ is a point in $P$, otherwise $S$ is a finer subdivision than $P$.


We have

\(\displaystyle \epsilon\) \(>\) \(\displaystyle \map U P – \map L P\)
\(\displaystyle \) \(\ge\) \(\displaystyle \map U S – \map L P\) as $\map U S \le \map U P $ by the definition of upper sum and the fact that $S$ refines $P$
\(\displaystyle \) \(\ge\) \(\displaystyle \map U S – \map L S\) as $\map L S \ge \map L P $ by the definition of lower sum and the fact that $S$ refines $P$

This shows that $S$ satisfies $\map U S – \map L S < \epsilon$.


Define:

$S_1 = S \cap \set {x: x \le c}$
$S_2 = S \cap \set {x: x \ge c}$

We observe:

$S_1$ is a subdivision of $\closedint a c$.
$S_2$ is a subdivision of $\closedint c b$.
$S_1$ and $S_2$ are adjacent.
The union of $S_1$ and $S_2$ equals $S$.


We get by the definition of upper sum:

\(\displaystyle \map U S\) \(=\) \(\displaystyle \map U {S_1} + \map U {S_2}\) since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent

Also, by the definition of lower sum:

\(\displaystyle \map L S\) \(=\) \(\displaystyle \map L {S_1} + \map L {S_2}\) since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent


We have

\(\displaystyle \epsilon\) \(>\) \(\displaystyle \map U S – \map L S\)
\(\displaystyle \) \(=\) \(\displaystyle \map U {S_1} + \map U {S_2} – \paren {\map L {S_1} + \map L {S_2} }\) as $\map U S = \map U {S_1} + \map U {S_2}$ and $\map L S = \map L {S_1} + \map L {S_2}$
\(\displaystyle \) \(=\) \(\displaystyle \map U {S_1} – \map L {S_1} + \map U {S_2} – \map L {S_2}\)
\(\displaystyle \) \(\ge\) \(\displaystyle \map U {S_1} – \map L {S_1}\) as $\map U {S_2} – \map L {S_2} \ge 0$ by the definitions of upper and lower sums

This shows that $S_1$ satisfies $\map U {S_1} – \map L {S_1} < \epsilon$.

A similar deduction focusing on $S_2$ instead of $S_1$ shows that $S_2$ satisfies $\map U {S_2} – \map L {S_2} < \epsilon$.


$\map U {S_1} – \map L {S_1} < \epsilon$ gives that $f$ is Darboux integrable on $\closedint a c$ since $S_1$ is a subdivision of $\closedint a c$.

$\map U {S_2} – \map L {S_2} < \epsilon$ gives that $f$ is Darboux integrable on $\closedint c b$ since $S_2$ is a subdivision of $\closedint c b$.

$\blacksquare$