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

Proof
WLOG 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.

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

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

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

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

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

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

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

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

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

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


 * 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 $Q_1$ 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:

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

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

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

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

the result follows.

Now suppose $a < b < c$.

Then from the definition of definite integral:
 * $\displaystyle \int_c^b f\left({x}\right) \ \mathrm dx := -\int_b^c f\left({x}\right) \ \mathrm dx$

and it follows that:

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

Finally, suppose $a = c < b$.

Then:

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

Hence the result, from Proof by Cases.