Sum of Integrals on Adjacent Intervals for Integrable Functions

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Theorem

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

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


Then:

$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$


Corollary

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


Then:

$\ds \int_{a_0}^{a_n} \map f t \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$


Proof

Lemma

Let $\closedint a b$ be a closed real interval.

Let $c$ be a real number.

Let $a < c < b$.


Let $f$ be a real function defined on $\closedint a b$.

Let $\map L S$ be the lower Darboux sum of $f$ on $\closedint a b$ where $S$ is a subdivision of $\closedint a b$.


Let $P$ and $Q$ be finite subdivisions of $\closedint a b$.

Let:

$Q = P \cup \set c$.


Then:

$\map L Q \ge \map L P$

$\Box$


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

First let $a < c < b$.

Let $P_1$ and $P_2$ be any subdivisions of $\closedint a c$ and $\closedint c b$ respectively.

Then $P = P_1 \cup P_2$ is a subdivision of $\closedint a b$.


From the definitions of upper Darboux sum and lower Darboux sum:

$\map L {P_1} + \map L {P_2} = \map L P$
$\map U {P_1} + \map U {P_2} = \map U P$


We consider the lower Darboux sum.

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

By definition of Darboux integral:

$\ds \map L P \le \int_a^b \map f t \rd t$

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

$\ds \map L {P_1} + \map L {P_2} \le \int_a^b \map f t \rd t$

and so:

$\ds \map L {P_1} \le \int_a^b \map f t \rd t - \map L {P_2}$


So, for any subdivision $P_2$ of $\closedint c b$, $\ds \int_a^b \map f t \rd t - \map L {P_2}$ is an upper bound of $\map L {P_1}$.

Thus:

$\ds \map {\sup_{P_1} } {\map L {P_1} } \le \int_a^b \map f t \rd t - \map L {P_2}$

where $\map {\sup_{P_1} } {\map L {P_1} }$ ranges over all subdivisions of $P_1$.

Thus by definition of definite integral:

$\ds \int_a^c \map f t \rd t \le \int_a^b \map f t \rd t - \map L {P_2}$

and so:

$\ds \map L {P_2} \le \int_a^b \map f t \rd t - \int_a^c \map f t \rd t$


Similarly, we find that:

$\ds \int_c^b \map f t \rd t \le \int_a^b \map f t \rd t - \int_a^c \map f t \rd t$

Therefore:

$\ds \int_a^b \map f t \rd t \ge \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$


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

Let $P$ be any subdivision of $\closedint a b$, which may or may not include the point $c$.

Let $Q = P \cup \set c$ be the subdivision of $\closedint a b$ obtained from $P$ by including with it, if necessary, the point $c$.


We have $\map L P \le \map L Q$ by the lemma.


Let $Q_1$ be the subdivision of $\closedint a b$ which includes only the points of $Q$ that lie in $\closedint a c$.

Let $Q_2$ be the subdivision of $\closedint a b$ which includes only the points of $Q$ that lie in $\closedint c b$.


From the definition of lower Darboux sum:

$\map L Q = \map L {Q_1} + \map L {Q_2}$

We have:

\(\ds \map L P\) \(\le\) \(\ds \map L Q\)
\(\ds \) \(=\) \(\ds \map L {Q_1} + \map L {Q_2}\)
\(\ds \) \(\le\) \(\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t\)


So $\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$ is an upper bound for $\map L P$, where $P$ is any subdivision of $\closedint a b$.


Thus:

$\ds \map {\sup_P} {\map L P} \le \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$

Thus, by definition:

$\ds \int_a^b \map f t \rd t \le \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$

Combining this with the result:

$\ds \int_a^b \map f t \rd t \ge \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$

the result follows.

$\Box$

Now suppose $a < b < c$.

Then from the definition of Darboux integral:

$\ds \int_c^b \map f t \rd t := -\int_b^c \map f t \rd t$

and it follows that:

\(\ds \int_a^b \map f t \rd t\) \(=\) \(\ds \int_a^c \map f t \rd t - \int_b^c \map f t \rd t\) main result
\(\ds \) \(=\) \(\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t\)

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


Finally, suppose $a = c < b$.

Then:

\(\ds \int_a^b \map f t \rd t\) \(=\) \(\ds 0 + \int_c^b \map f t \rd t + \int_a^c \map f t \rd t\) as $a = c$
\(\ds \) \(=\) \(\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t\) Integral on Zero Interval, as $a = c$

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

Hence the result, from Proof by Cases.

$\blacksquare$


Sources

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