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

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

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

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

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

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

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

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

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

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

Therefore, $$\int_a^b f \left({t}\right) dt \ge \int_a^c f \left({t}\right) dt + \int_c^b f \left({t}\right) dt$$.


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

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

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

Combining this with the result $$\int_a^b f \left({t}\right) dt \ge \int_a^c f \left({t}\right) dt + \int_c^b f \left({t}\right) dt$$, the result follows.