Proof: By Induction
(related to Proposition: Integrals on Adjacent Intervals)
In the following, $[a,c]$ is a closed real interval with a real value $b$ inbetween $a < b < c.$
"$\Rightarrow$"
- By hypothesis, the function $f:[a,b]\cup[b,c]\to\mathbb R$ is Riemann-integrable on the adjacent intervals $[a,b]$ and $[b,c].$
- By the necessary and sufficient condition for integrable functions this means that for every $\epsilon > 0$ there are step functions $\phi:[a,b]\cup[b,c]\to\mathbb R$ and $\psi:[a,b]\cup[b,c]\to\mathbb R$ such that
$$\phi \le f \le \psi\quad\wedge\quad\int_a^b\psi(x)dx-\int_a^b\phi(x)dx\le\frac{\epsilon}{2}\quad\wedge\quad\int_b^c\psi(x)dx-\int_b^c\phi(x)dx\le\frac{\epsilon}{2}$$
- But the union of the intervals is $[a,b]\cup[b,c]$=$[a,c].$
- Therefore, for the given $\epsilon > 0$ there are step functions $\phi:[a,c]\to\mathbb R$ and $\psi:[a,c]\to\mathbb R$ such that
$$\int_a^c\psi(x)dx-\int_a^c\phi(x)dx=\int_a^b\psi(x)dx-\int_a^b\phi(x)dx+\int_b^c\psi(x)dx-\int_b^c\phi(x)dx=\le\epsilon.$$
- By the same necessary and sufficient condition for integrable functions this means that $f:[a,c]\to\mathbb R$ is Riemann-integrable with
$$\int_a^c f(x)dx=\int_a^b f(x)dx+\int_b^c f(x)dx.$$
"$\Leftarrow$"
The converse of the statement can be proven analogously.
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
