# Sum of Integrals on Adjacent Intervals for Integrable Functions

## Theorem

Let $f$ be a real function which is Riemann integrable on any closed interval $\mathbb I$.

Let $a, b, c \in \mathbb I$.

Then:

$\displaystyle \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t = \int_a^b f \left({t}\right) \rd t$

### Corollary

Let $a_0, a_1, \ldots, a_n$ be real numbers, where $n \in \N$ and $n \ge 2$.

Then:

$\displaystyle \int_{a_0}^{a_n} f \left({t}\right) \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } f \left({t}\right) \rd t$

## Proof

Without loss of generality, assume $a < b$.

First let $a < c < b$.

Let $P_1$ and $P_2$ be any subdivisions of $\left[{a \,.\,.\, c}\right]$ and $\left[{c \,.\,.\, b}\right]$ respectively.

Then $P = P_1 \cup P_2$ is a subdivision of $\left[{a \,.\,.\, b}\right]$.

From the definitions of upper sum and lower sum:

$L \left({P_1}\right) + L \left({P_2}\right) = L \left({P}\right)$
$U \left({P_1}\right) + U \left({P_2}\right) = U \left({P}\right)$

We consider the lower sum.

The same conclusion can be obtained by investigating the upper sum.

By definition of definite integral:

$\displaystyle L \left({P}\right) \le \int_a^b f \left({t}\right) \rd t$

Thus, given the subdivisions $P_1$ and $P_2$, we have:

$\displaystyle L \left({P_1}\right) + L \left({P_2}\right) \le \int_a^b f \left({t}\right) \rd t$

and so:

$\displaystyle L \left({P_1}\right) \le \int_a^b f \left({t}\right) \rd t - L \left({P_2}\right)$

So, for any subdivision $P_2$ of $\left[{c \,.\,.\, b}\right]$, $\displaystyle \int_a^b f \left({t}\right) \rd t - L \left({P_2}\right)$ is an upper bound of $L \left({P_1}\right)$.

Thus:

$\displaystyle \sup_{P_1} \left({L \left({P_1}\right)}\right) \le \int_a^b f \left({t}\right) \rd t - L \left({P_2}\right)$

where $\sup_{P_1} \left({L \left({P_1}\right)}\right)$ ranges over all subdivisions of $P_1$.

Thus by definition of definite integral:

$\displaystyle \int_a^c f \left({t}\right) \rd t \le \int_a^b f \left({t}\right) \rd t - L \left({P_2}\right)$

and so:

$\displaystyle L \left({P_2}\right) \le \int_a^b f \left({t}\right) \rd t - \int_a^c f \left({t}\right) \rd t$

Similarly, we find that:

$\displaystyle \int_c^b f \left({t}\right) \rd t \le \int_a^b f \left({t}\right) \rd t - \int_a^c f \left({t}\right) \rd t$

Therefore:

$\displaystyle \int_a^b f \left({t}\right) \rd t \ge \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$

Having established the above, we now need to put it into context.

Let $P$ be any subdivision of $\left[{a \,.\,.\, b}\right]$, which may or may not include the point $c$.

Let $Q = P \cup \left\{ {c}\right\}$ be the subdivision of $\left[{a \,.\,.\, b}\right]$ obtained from $P$ by including with it, if necessary, the point $c$.

It is easy to show that $L \left({P}\right) \le L \left({Q}\right)$.

Let $Q_1$ be the subdivision of $\left[{a \,.\,.\, b}\right]$ which includes only the points of $Q$ that lie in $\left[{a \,.\,.\, c}\right]$.

Let $Q_2$ be the subdivision of $\left[{a \,.\,.\, b}\right]$ which includes only the points of $Q$ that lie in $\left[{c \,.\,.\, b}\right]$.

From the definition of lower sum:

$L \left({Q}\right) = L \left({Q_1}\right) + L \left({Q_2}\right)$

We have:

 $\displaystyle L \left({P}\right)$ $\le$ $\displaystyle L \left({Q}\right)$ $\displaystyle$ $=$ $\displaystyle L \left({Q_1}\right) + L \left({Q_2}\right)$ $\displaystyle$ $\le$ $\displaystyle \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$

So $\displaystyle \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$ is an upper bound for $L \left({P}\right)$, where $P$ is any subdivision of $\left[{a \,.\,.\, b}\right]$.

Thus:

$\displaystyle \sup_P \left({L \left({P}\right)}\right) \le \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$

Thus, by definition:

$\displaystyle \int_a^b f \left({t}\right) \rd t \le \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$

Combining this with the result:

$\displaystyle \int_a^b f \left({t}\right) \rd t \ge \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$

the result follows.

$\Box$

Now suppose $a < b < c$.

Then from the definition of definite integral:

$\displaystyle \int_c^b f \left({x}\right) \rd x := -\int_b^c f \left({x}\right) \rd x$

and it follows that:

 $\displaystyle \int_a^b f \left({t}\right) \rd t$ $=$ $\displaystyle \int_a^c f \left({t}\right) \rd t - \int_b^c f \left({t}\right) \rd t$ main result $\displaystyle$ $=$ $\displaystyle \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$

The case of $c < a < b$ is proved similarly.

Finally, suppose $a = c < b$.

Then:

 $\displaystyle \int_a^b f \left({t}\right) \rd t$ $=$ $\displaystyle 0 + \int_c^b f \left({t}\right) \rd t + \int_a^c f \left({t}\right) \rd t$ as $a = c$ $\displaystyle$ $=$ $\displaystyle \int_a^c f \left({t}\right) \rd t + \int_c^b f \left({t}\right) \rd t$ Integral on Zero Interval, as $a = c$

The case of $a < c = b$ is proved similarly.

Hence the result, from Proof by Cases.

$\blacksquare$