Sum of Integrals on Adjacent Intervals for Continuous Functions

Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a .. b}\right]$.

Let $a < c < b$.

Then:
 * $\displaystyle \int_a^c f \left({t}\right) dt + \int_c^b f \left({t}\right) dt = \int_a^b f \left({t}\right) dt$

Proof
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 that:
 * $L \left({P_1}\right) + L \left({P_2}\right) = L \left({P}\right)$
 * $U \left({P_1}\right) + U \left({P_2}\right) = U \left({P}\right)$

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

We have $\displaystyle L \left({P}\right) \le \int_a^b f \left({t}\right) dt$ by definition.

Thus, given the subdivisions $P_1$ and $P_2$, we have:
 * $\displaystyle L \left({P_1}\right) + L \left({P_2}\right) \le \int_a^b f \left({t}\right) dt$

and so:
 * $\displaystyle L \left({P_1}\right) \le \int_a^b f \left({t}\right) dt - L \left({P_2}\right)$

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

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

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

Thus by definition:
 * $\displaystyle \int_a^c f \left({t}\right) dt \le \int_a^b f \left({t}\right) dt - L \left({P_2}\right)$

and so:
 * $\displaystyle L \left({P_2}\right) \le \int_a^b f \left({t}\right) dt - \int_a^c f \left({t}\right) dt$

Similarly, we find that:
 * $\displaystyle \int_c^b f \left({t}\right) dt \le \int_a^b f \left({t}\right) dt - \int_a^c f \left({t}\right) dt$

Therefore:
 * $\displaystyle \int_a^b f \left({t}\right) dt \ge \int_a^c f \left({t}\right) dt + \int_c^b f \left({t}\right) dt$


 * 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 $P_1$ be the subdivisions of $\left[{a .. b}\right]$ which includes the points of $Q$ that lie in $\left[{a .. c}\right]$.

Let $P_2$ be the subdivisions of $\left[{a .. b}\right]$ which includes the points of $Q$ that lie in $\left[{c .. b}\right]$.

We have $L \left({P}\right) \le L \left({Q}\right) = L \left({P_1}\right) + L \left({P_2}\right)$.

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

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

Thus, by definition:
 * $\displaystyle \int_a^b f \left({t}\right) dt \le \int_a^c f \left({t}\right) dt + \int_c^b f \left({t}\right) dt$

Combining this with the result:
 * $\displaystyle \int_a^b f \left({t}\right) dt \ge \int_a^c f \left({t}\right) dt + \int_c^b f \left({t}\right) dt$

the result follows.