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 $\int _{a}^{c}f(x)\;d\alpha (x)=A$ and $\int _{c}^{b}f(x)\;d\alpha (x)=B$ for some $A,B\in \mathbb {R}$. Let $\epsilon >0$ be given.

• Since $\int _{a}^{c}f(x)\;d\alpha (x)=A$ we have that for $\epsilon _{1}={\frac {\epsilon }{2}}>0$ there exists a partition $P_{\epsilon _{1}}\in {\mathcal {P}}[a,c]$ such that for all partitions $P'\in {\mathcal {P}}[a,c]$ 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 $\int _{c}^{b}f(x)\;d\alpha (x)=B$ we have that for $\epsilon _{2}={\frac {\epsilon }{2}}>0$ there exists a partition $P_{\epsilon _{2}}\in {\mathcal {P}}[c,b]$ such that for all partitions $P''\in {\mathcal {P}}[c,b]$ 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 $P\in {\mathcal {P}}[a,b]$ 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 $\int _{a}^{b}f(x)\;d\alpha (x)$ exists and:

${\begin{aligned}\quad \int _{a}^{b}f(x)\;d\alpha (x)=\int _{a}^{c}f(x)\;d\alpha (x)+\int _{c}^{b}f(x)\;d\alpha (x)\quad \blacksquare \end{aligned}}$