3.2. Linear Properties

The following theorem shows the linearity of integrals in the fashion of the integrands.

Theorem 3.1

If \(f, g \in \mathfrak{R}(\alpha)\) on \([a, b]\), then \(c_1 f + c_2 g \in \mathfrak{R}(\alpha)\) on \([a, b]\). And

(3.2)\[\int_a^b c_1 f + c_2 g \dif\alpha = c_1 \int_a^b f \dif\alpha + c_2 \int_a^b g \dif\alpha\]

Proof. Let \(\varepsilon > 0\) be arbitrary. Because \(f\) and \(g\) are both Riemann integrable on \([a, b]\), there exists a partition \(P_\varepsilon\) of \([a, b]\) such that for any \(P \supseteq P_\varepsilon\) and set of representatives \(T\) of \(P\) satisfying

(3.3)\[\abs{S(P,T,f,\alpha) - \int_a^b f \dif \alpha} < \frac{\varepsilon}{\abs{c_1} + \abs{c_2} + 1}\quad\text{and}\quad\abs{S(P,T,g,\alpha) - \int_a^b g \dif \alpha} < \frac{\varepsilon}{\abs{c_1} + \abs{c_2} + 1}\]

Note

The reason of the choice of the small number \(\frac{\varepsilon}{\abs{c_1} + \abs{c_2} + 1}\) will be clear later. And the \(+1\) in the denominator is designed for the case that both \(c_1\) and \(c_2\) are zeros.

Consider the Riemann-Stieltjes sum \(S(P,T,c_1 f + c_2 g, \alpha)\). We have

\[\begin{split}& \quad& & \abs{S(P,T,c_1 f + c_2 g, \alpha) - c_1 \int_a^b f \dif \alpha - c_2 \int_a^b g \dif \alpha}\\& = & & \abs{ \sum_k (c_1 \Delta f_k + c_2 \Delta g_k) - c_1 \int_a^b f \dif \alpha - c_2 \int_a^b g \dif \alpha}\\& = & & \abs{ c_1 \sum_k \Delta f_k + c_2 \sum_k \Delta g_k - c_1 \int_a^b f \dif \alpha - c_2 \int_a^b g \dif \alpha}\\& = & & \abs{ c_1 S(P,T,f,\alpha) + c_2 S(P,T,g,\alpha) - c_1 \int_a^b f \dif \alpha - c_2 \int_a^b g \dif \alpha}\\& \leq& & \abs{c_1}\abs{ S(P,T,f,\alpha) - \int_a^b f \dif \alpha } + \abs{c_2}\abs{ S(P,T,g,\alpha) - \int_a^b g \dif \alpha }\end{split}\]

Applying (3.3), we obtain

\[\begin{split}\abs{S(P,T,c_1 f + c_2 g, \alpha) - c_1 \int_a^b f \dif \alpha - c_2 \int_a^b g \dif \alpha}& < \abs{c_1}\frac{\varepsilon}{\abs{c_1} + \abs{c_2} + 1} + \abs{c_2}\frac{\varepsilon}{\abs{c_1} + \abs{c_2} + 1}\\& = \frac{\varepsilon (\abs{c_1} + \abs{c_2})}{\abs{c_1} + \abs{c_2} + 1}\\& < \varepsilon\end{split}\]

This shows that \(c_1 f + c_2 g\) is also Riemann integrable on \([a, b]\), and (3.2) is satisfied.

Analogously, we can prove that the integral is linear in the integrators.

Theorem 3.2

If \(f \in \mathfrak{R}(\alpha)\) and \(f \in \mathfrak{R}(\beta)\) on \([a, b]\), then \(f \in \mathfrak{R}(c_1 \alpha + c_2 \beta)\) on \([a, b]\). And

\[\int_a^b f \dif(c_1 \alpha + c_2 \beta) = c_1 \int_a^b f \dif\alpha + c_2 \int_a^b f \dif\beta\]

Theorem 3.3

Assume \(c \in (a, b)\). If two of the three integrals in (3.4) exist, then the other one also exists, and (3.4) holds.

(3.4)\[\int_a^c f \dif\alpha + \int_c^b f \dif\alpha = \int_a^b f \dif\alpha\]