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 (unknown function "\mathscr"): {\displaystyle P_{\epsilon_1} \in \mathscr{P}[a, c]} such that for all partitions 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' \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:

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

• Hence Failed to parse (syntax error): {\displaystyle \int_a^b f(x) \: d \alpha (x)} exists and:

<math>\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = \int_a^c f(x) \: d \alpha (x) + \int_c^b f(x) \: d \alpha (x) \quad \blacksquare \end{al