# Existence of Integral on Union of Adjacent Intervals

## 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 $\left[{c \,.\,.\, b}\right]$
$f$ 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 $\left[{c \,.\,.\, b}\right]$.

$\blacksquare$