Existence of Integral on Union of Adjacent Intervals

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Theorem

Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$, $a < b$.

Let $c$ be a point in $\left({a \,.\,.\, b}\right)$.


Then:

$f$ is Riemann integrable on $\left[{a \,.\,.\, c}\right]$ and $\left[{c \,.\,.\, b}\right]$

if and only if:

$f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.


Proof

Necessary Condition

We need to prove that $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.

To do this it suffices to show that for all $\epsilon > 0$, there exists a subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ such that $U \left({S}\right) – L \left({S}\right) < \epsilon$.

Here, $U \left({S}\right)$ and $L \left({S}\right)$ are, respectively, the upper and lower sums of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to the subdivision $S$.


Let a strictly positive $\epsilon$ be given.

Since $f$ is Riemann integrable on $\left[{a \,.\,.\, c}\right]$, we know that a subdivision $S_1$ of $\left[{a \,.\,.\, c}\right]$ exists such that $U \left({S_1}\right) – L \left({S_1}\right) < \dfrac \epsilon 2$.

Since $f$ is Riemann integrable on $\left[{c \,.\,.\, b}\right]$, we know that a subdivision $S_2$ of $\left[{c \,.\,.\, b}\right]$ exists such that $U \left({S_2}\right) – L \left({S_2}\right) < \dfrac \epsilon 2$.


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

We observe that $S$ is a subdivision of $\left[{a \,.\,.\, b}\right]$.


We get by the definition of upper sum:

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

Also, by the definition of lower sum:

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

These two equations give:

\(\displaystyle U \left({S}\right) – L \left({S}\right)\) \(=\) \(\displaystyle U \left({S_1}\right) + U \left({S_2}\right) – \left[{L \left({S_1}\right) + L \left({S_2}\right)}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle U \left({S_1}\right) – L \left({S_1}\right) + U \left({S_2}\right) – L \left({S_2}\right)\)
\(\displaystyle \) \(\lt\) \(\displaystyle \dfrac \epsilon 2 + \dfrac \epsilon 2\) since $U \left({S_1}\right) – L \left({S_1}\right) < \dfrac \epsilon 2$ and $U \left({S_2}\right) – L \left({S_2}\right) < \dfrac \epsilon 2$
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

This shows that $S$ satisfies $U \left({S}\right) – L \left({S}\right) < \epsilon$.

We conclude from this that $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$ since $S$ is a subdivision of $\left[{a \,.\,.\, b}\right]$.

$\Box$


Sufficient Condition

We need to prove that $f$ is Riemann integrable on $\left[{a \,.\,.\, c}\right]$ and $\left[{c \,.\,.\, b}\right]$.


Let a strictly positive $\epsilon$ be given.

The Riemann integrability of $f$ on $\left[{a \,.\,.\, b}\right]$ implies that a subdivision $P$ of $\left[{a \,.\,.\, b}\right]$ exists such that $U \left({P}\right) – L \left({P}\right) < \epsilon$.

Here, $U \left({P}\right)$ and $L \left({P}\right)$ are, respectively, the upper and lower sums of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to the subdivision $P$.


Define the subdivision $S = P \cup \left\{{c}\right\}$.

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


We have

\(\displaystyle \epsilon\) \(\gt\) \(\displaystyle U \left({P}\right) – L \left({P}\right)\)
\(\displaystyle \) \(\ge\) \(\displaystyle U \left({S}\right) – L \left({P}\right)\) since $U \left({S}\right) \le U \left({P}\right) $ by the definition of upper sum and the fact that $S$ refines $P$
\(\displaystyle \) \(\ge\) \(\displaystyle U \left({S}\right) – L \left({S}\right)\) since $L \left({S}\right) \ge L \left({P}\right) $ by the definition of lower sum and the fact that $S$ refines $P$

This shows that $S$ satisfies $U \left({S}\right) – L \left({S}\right) < \epsilon$.


Define:

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

We observe:

  • $S_1$ is a subdivision of $\left[{a \,.\,.\, c}\right]$.
  • $S_2$ is a subdivision of $\left[{c \,.\,.\, b}\right]$.
  • $S_1$ and $S_2$ are adjacent.
  • The union of $S_1$ and $S_2$ equals $S$.


We get by the definition of upper sum:

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

Also, by the definition of lower sum:

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


We have

\(\displaystyle \epsilon\) \(\gt\) \(\displaystyle U \left({S}\right) – L \left({S}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle U \left({S_1}\right) + U \left({S_2}\right) – \left[{L \left({S_1}\right) + L \left({S_2}\right)}\right]\) since $U \left({S}\right) = U \left({S_1}\right) + U \left({S_2}\right)$ and $L \left({S}\right) = L \left({S_1}\right) + L \left({S_2}\right)$
\(\displaystyle \) \(=\) \(\displaystyle U \left({S_1}\right) – L \left({S_1}\right) + U \left({S_2}\right) – L \left({S_2}\right)\)
\(\displaystyle \) \(\ge\) \(\displaystyle U \left({S_1}\right) – L \left({S_1}\right)\) since $U \left({S_2}\right) – L \left({S_2}\right) \ge 0$ by the definitions of upper and lower sums

This shows that $S_1$ satisfies $U \left({S_1}\right) – L \left({S_1}\right) < \epsilon$.

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


$U \left({S_1}\right) – L \left({S_1}\right) < \epsilon$ gives that $f$ is Riemann integrable on $\left[{a \,.\,.\, c}\right]$ since $S_1$ is a subdivision of $\left[{a \,.\,.\, c}\right]$.

$U \left({S_2}\right) – L \left({S_2}\right) < \epsilon$ gives that $f$ is Riemann integrable on $\left[{c \,.\,.\, b}\right]$ since $S_2$ is a subdivision of $\left[{c \,.\,.\, b}\right]$.

$\blacksquare$