The Additivity Theorem

The Additivity Theorem for Riemann Integrable Functions: Let $f$ be a real-valued function on the interval $[a,b]$ , and let $c\in (a,b)$ . Then, $f$ is Riemann integrable on $[a,b]$ if and only if it is also Riemann integrable on $[a,c]$ and $[c,b]$ . In this case,

$\int _{a}^{b}f=\int _{a}^{c}f+\int _{c}^{b}f$

Proof: Suppose that

Failed to parse (syntax error): {\displaystyle \int_a^c f(x) \: d \alpha (x) = A}

and

Failed to parse (syntax error): {\displaystyle \int_c^b f(x) \: d \alpha (x) = B}

for some

A,B\in \mathbb {R}

. Let

\epsilon >0

be given.

Since $Failed to parse (syntax error): {\displaystyle \int_a^c f(x) \: d \alpha (x) = A}$ we have that for $\epsilon _{1}={\frac {\epsilon }{2}}>0$ there exists a partition $Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{\epsilon_1} \in \mathscr{P}[a, c]}$ such that for all partitions $Failed to parse (unknown function "\mathscr"): {\displaystyle P' \in \mathscr{P}[a, c]}$ finer than $P_{\epsilon _{1}}$ , ( $P_{\epsilon _{1}}\subseteq P'$ ) and for any choice of $t_{k}$ 's in each $k^{\mathrm {th} }$ subinterval we have that:

{\begin{aligned}\quad \mid S(P',f,\alpha )-A\mid <\epsilon _{1}={\frac {\epsilon }{2}}\quad (*)\end{aligned}}

Similarly, since $Failed to parse (syntax error): {\displaystyle \int_c^b f(x) \: d \alpha (x) = B}$ we have that for $\epsilon _{2}={\frac {\epsilon }{2}}>0$ there exists a partition $Failed to parse (unknown function "\mathscr"): {\displaystyle P_{\epsilon_2} \in \mathscr{P}[c, b]}$ such that for all partitions $Failed to parse (unknown function "\mathscr"): {\displaystyle P'' \in \mathscr{P}[c, b]}$ finer than $P_{\epsilon _{2}}$ , $(P_{\epsilon _{2}}\subseteq P''$ ) and for any choice of $u_{k}$ 's in each $k^{\mathrm {th} }$ subinterval we have that:

{\begin{aligned}\quad \mid S(P'',f,\alpha )-B\mid <\epsilon _{2}={\frac {\epsilon }{2}}\quad (**)\end{aligned}}

Let $P_{\epsilon }=P_{\epsilon _{1}}\cup P_{\epsilon _{2}}$ . Then $P_{\epsilon }$ is a partition of $[a,b]$ and for all partitions $Failed to parse (unknown function "\mathscr"): {\displaystyle P \in \mathscr{P}[a, b]}$ finer than $P_{\epsilon }$ , ( $P_{\epsilon }\subseteq P$ ) we must have that $(*)$ and $(**)$ hold. Then for any choice of $v_{k}$ 's in each $k^{\mathrm {th} }$ subinterval we have that: