📂Analysis

Linearity of Riemann(-Stieltjes) Iintegral

Theorem¹

• This article is based on the Riemann-Stieltjes integral. If set to $\alpha=\alpha (x)=x$, it is the same as the Riemann integral.

1. Let’s say $f$ is integrable by Riemann(-Stieltjes) from $[a,b]$. Then, for a constant $c\in \mathbb{R}$, $cf$ is also integrable from $[a,b]$, and its value is as follows.

$$ \int_{a}^{b}cf d\alpha = c\int_{a}^{b}f d\alpha $$

2. Let two functions $f_{1}$, $f_{2}$ be integrable by Riemann(-Stieltjes) from $[a,b]$. Then, $f_{1}+f_{2}$ is also integrable, and its value is as follows.

$$ \int _{a} ^{b}(f_{1}+f_{2})d\alpha = \int _{a} ^{b} f_{1}d\alpha + \int_{a}^{b} f_{2} d\alpha $$

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It means that integration is linear.

$$ \int _{a} ^{b}(f_{1}+cf_{2})d\alpha = \int _{a} ^{b} f_{1}d\alpha + c\int_{a}^{b} f_{2} d\alpha $$

The reason for specifically mentioning sums and constant multiples separately is because they are proven separately.

Auxiliary Theorem

For a function $f$ that is integrable by Riemann(-Stieltjes) from $[a,b]$ and any positive number $\varepsilon> 0$, there exists a partition $P$ of $[a,b]$ that satisfies the following equation.

$$ \begin{align} U(P,f,\alpha) \lt \int_{a}^{b}f d\alpha +\varepsilon \tag{L1} \\ \int_{a}^{b}f d\alpha -\varepsilon \lt L(P,f,\alpha) \tag{L2} \end{align} $$

$U$, $L$ are respectively the Riemann(-Stieltjes) upper sum and lower sum.

Proof

$\eqref{L1}$

Let’s say an arbitrary positive number $\varepsilon \gt 0$ is given. Then, by the necessary and sufficient condition for integrability, there exists a partition $P$ that satisfies the following equation.

$$ U(P,f,\alpha)-L(P,f,\alpha) \lt \varepsilon $$

Since $L(P,f,\alpha) \le \displaystyle \int_{a}^{b}fd\alpha$, the following holds.

$$ U(P,f,\alpha)-\int_{a}^{b}f d\alpha\le U(P,f,\alpha)-L(P,f,\alpha) \lt \varepsilon $$

Therefore, the summary is as follows.

$$ U(P,f,\alpha ) \lt \int_{a}^{b}f d\alpha +\varepsilon $$

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$\eqref{L2}$

As in the proof of $\eqref{L1}$, there exists a partition $P$ that satisfies the following.

$$ U(P,f,\alpha)-L(P,f,\alpha) \lt \varepsilon $$

Since $\displaystyle \int_{a}^{b}fd\alpha \le U(P,f,\alpha)$, the following holds.

$$ \int_{a}^{b}f d\alpha-L(P,f,\alpha)\le U(P,f,\alpha)-L(P,f,\alpha) \lt \varepsilon $$

Therefore, the summary is as follows.

$$ \int_{a}^{b}f d\alpha -\varepsilon \lt L(P,f,\alpha) $$

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Proof

When $f_{1}, f_{2}, f$ is integrable, it will be shown that $f_{1}+f_{2}, cf$ is also integrable and that its actual value is equal to $\displaystyle \int f_{1} + \int f_{2}, c\int f$.

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1.

• Case 1. $c=0$

  It is obvious that $cf=0$ is integrable. It is also obvious that the following equation holds.

  $$ \int_{a}^{b}0fd\alpha=0=0\int_{a}^{b}fd\alpha $$

• Case 2. $c>0$

  Let’s say an arbitrary positive number $\varepsilon >0$ is given. Then, by the necessary and sufficient condition for integrability, there exists a partition $P=\left\{ a=x_{0} \lt \cdots \lt x_{i} \lt \cdots \lt x_{n}=b\right\}$ that satisfies the following.

  $$ \begin{equation} U(P,f,\alpha) - L(P,f,\alpha)<\frac{\varepsilon}{c} \end{equation} $$

  Let’s set it as follows.

  $$ \begin{align*} M_{i} &= \sup _{[x_{i-1}, x_{i}]} f(x) \\ m_{i} &= \inf _{[x_{i-1}, x_{i}]} f(x) \\ M_{i}^{c} &= \sup _{[x_{i-1}, x_{i}]} cf(x) \\ m_{i}^{c} &= \inf _{[x_{i-1}, x_{i}]} cf(x) \end{align*} $$

  Since $c>0$, $cM_{i} = M_{i}^{c}$ holds, and $cm_{i} = m_{i}^{c}$. Then, by the definition of Riemann-(Stieltjes) sum and $(1)$, the following holds.

  $$ \begin{align} U(P,cf,\alpha)- L(P,cf,\alpha) &= \sum \limits_{i=1}^{n}M_{i}^{c}\Delta \alpha_{i}-\sum \limits_{i=1}^{n}m_{i}^{c}\Delta \alpha_{i} \nonumber\\ &= \sum \limits_{i=1}^{n}cM_{i}\Delta \alpha_{i}-\sum \limits_{i=1}^{n}cm_{i}\Delta \alpha_{i} \nonumber\\ &= c\left( \sum \limits_{i=1}^{n}M_{i}\Delta \alpha_{i}-\sum \limits_{i=1}^{n}m_{i}\Delta \alpha_{i} \right) \nonumber\\ &= c\Big[ U(P,f,\alpha)-L(P,f,\alpha)\Big] \nonumber\\ &\lt \varepsilon \end{align} $$

  Therefore, by the necessary and sufficient condition for integrability, $cf$ is integrable. Since the integral is less than the upper sum, the following holds.

  $$ c \int_{a}^{b}fd \alpha \le cU(P,f,\alpha) = U(P,cf,\alpha) $$

  Then, by $(2)$ and the Auxiliary Theorem, the following holds.

  $$ c\int _{a}^{b}f d\alpha \le U(P,cf,\alpha) lt \int _{a}^{b} cf d\alpha +\varepsilon $$

  Since $\varepsilon$ is assumed to be any positive number, the following holds.

  $$ \begin{equation} c\int_{a}^{b}fd\alpha \le \int_{a}^{b}cfd\alpha \end{equation} $$

  The process of proving the opposite inequality is similar. By $(1)$ and the Auxiliary Theorem, the following holds.

  $$ cU(P,f,\alpha) \le c\int_{a}^{b}fd\alpha +\varepsilon $$

  Also, the following equation holds.

  $$ \int_{a}^{b} cfd\alpha \le U(P,cf,\alpha)=cU(P,f,\alpha) $$

  From the above two equations, the following equation is obtained.

  $$ \int_{a}^{b} cfd \alpha \le cU(P,f,\alpha)< c\int_{a}^{b}fd\alpha +\varepsilon $$

  Since $\varepsilon$ is any positive number, the following holds.

  $$ \begin{equation} \int_{a}^{b} cf d\alpha \le c\int_{a}^{b}fd\alpha \end{equation} $$

  By $(3)$ and $(4)$, the following holds.

  $$ \int_{a}^{b}cfd\alpha = c\int_{a}^{b}fd\alpha $$

• Case 3. $c=-1$

  The proof process is similar to Case 2. First, let’s say an arbitrary positive number $\varepsilon$ is given. Since $f$ is integrable, by the necessary and sufficient condition for integrability, there exists a partition $P$ for the given $\varepsilon$ that satisfies the following.

  $$ U(P,f,\alpha) - L(P,f,\alpha) <\varepsilon $$

  Now, let’s set it as follows.

  $$ \begin{align*} M_{i} &= \sup _{[x_{i-1},x_{i}]}f \\ m_{i} &= \inf_{[x_{i-1},x_{i}]}f \\ M_{i}^{\ast} &= \sup _{[x_{i-1},x_{i}]}(-f) \\ m_{i}^{\ast} &= \inf_{[x_{i-1},x_{i}]}(-f) \end{align*} $$

  Since $M_{i}=-m_{i}^{\ast}$ and $m_{i}=-M_{i}^{\ast}$, $M_{i}-m_{i}=M_{i}^{\ast}-m_{i}^{\ast}$ holds. Therefore, the following holds.

  $$ \begin{align*} U(P,-f,\alpha)-L(P,-f,\alpha) &= \sum\limits_{i=1}^{n}M_{i}^{\ast}\Delta \alpha_{i}-\sum\limits_{i=1}^{n}m_{i}^{\ast}\Delta \alpha_{i} \\ &= \sum\limits_{i=1}^{n}M_{i}\Delta \alpha_{i} - \sum\limits_{i=1}^{n}m_{i}\Delta\alpha_{i} \\ &= U(P,f,\alpha) -L(P,f,\alpha) \\ &\lt \varepsilon \end{align*} $$

  Therefore, $-f$ is integrable.

  As in the proof of Case 2., by the Auxiliary Theorem, the following holds.

  $$ U(P,-f,\alpha) \lt \int_{a}^{b}(-f)d\alpha +\varepsilon $$

  Also, the following equation holds.

  $$ -\int_{a}^{b}fd\alpha\le -L(P,f,\alpha)=U(P,-f,\alpha) \lt \int_{a}^{b}(-f)d\alpha + \varepsilon $$

  Since $\varepsilon$ is any positive number, the following holds. $$ -\int_{a}^{b}fd\alpha \le \int_{a}^{b}(-f)d\alpha $$

  Then, by the Auxiliary Theorem, the following equation holds.

  $$ \int_{a}^{b}(-f)d\alpha -\varepsilon \lt L(P,-f,\alpha)=-U(P,f,\alpha)\le-\int_{a}^{b}fd\alpha $$

  Since $\varepsilon$ is any positive number, the following holds.

  $$ \int_{a}^{b}(-f)d\alpha \le -\int_{a}^{b}fd\alpha $$

  Therefore, the following is obtained.

  $$ \int_{a}^{b}(-f)d\alpha =-\int_{a}^{b}fd\alpha $$

• Case 4. $c \lt 0 \quad \text{and} \quad c\ne -1$

  It holds by Case 2. and Case 3.

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2.

Let’s say $f=f_{1}+f_{2}$. Let $P$ be any partition of $[a,b]$. Then, by the definition of Riemann(-Stieltjes) upper and lower sums, the following holds.

$$ \begin{equation} \begin{aligned} L(P,f_{1},\alpha) + L(P,f_{2},\alpha)& \le L(P,f,\alpha) \\ &\le U(P,f,\alpha) \\ &\le U(P,f_{1},\alpha) +U(P,f_{2},\alpha) \end{aligned} \end{equation} $$

Let’s say an arbitrary positive number $\varepsilon > 0$ is given. Then, by the necessary and sufficient condition for integrability, there exists a partition $P_{j}$ that satisfies the following.

$$ U(P_{j},f_{j},\alpha)-L(P_{j},f_{j},\alpha)<\varepsilon,\quad (j=1,2) $$

Now, let’s consider $P$ again as a common refinement of $P_{1}$ and $P_{2}$. Then, by $(5)$, the following holds.

$$ \begin{align*} U(P,f,\alpha)-L(P,f,\alpha) &\le \left[ U(P,f_{1},\alpha)-L(P,f_{1},\alpha) \right] + \left[ U(P,f_{2},\alpha)-L(P,f_{2},\alpha) \right] \\ &< \varepsilon \end{align*} $$

Therefore, by the necessary and sufficient condition for integrability, $f$ is integrable. Then, by the Auxiliary Theorem, the following equation holds.

$$ U(P,f_{j},\alpha)<\int _{a}^{b}f_{j}d\alpha+\varepsilon,\quad (j=1,2) $$

Furthermore, by definition, since the upper sum is greater than the integral, the following holds.

$$ \int_{a}^{b}fd\alpha \le U(P,f,\alpha) $$

From the above equation and the third inequality of $(5)$, the following holds.

$$ \begin{align*} \int_{a}^{b}fd\alpha &\le U(P,f,\alpha) \\ &\le U(P,f_{1},\alpha)+U(P,f_{2},\alpha) \\ &< \int_{a}^{b}f_{1}d\alpha +\int_{a}^{b}f_{2}d\alpha + 2\varepsilon \end{align*} $$

Since $\varepsilon$ is any positive number, the following holds.

$$ \begin{equation} \int_{a}^{b} fd\alpha \le \int_{a}^{b}f_{1}d\alpha + \int_{a}^{b} f_{2}d\alpha \label{6} \end{equation} $$

Proving the opposite direction of the inequality completes the proof. Since it’s already shown above that a constant multiple of an integrable function is also integrable, $-f_{1}, -f_{2}$ is known to be integrable. Therefore, repeating the above process for these two functions yields the following equation.

$$ \int_{a}^{b}(-f)d\alpha \le \int_{a}^{b}(-f_{1})d\alpha + \int_{a}^{b} (-f_{2})d\alpha $$

Furthermore, since $\displaystyle \int (-f)d\alpha=-\int fd\alpha$, by multiplying both sides by $-1$, the following is obtained.

$$ \begin{equation} \int_{a}^{b}fd\alpha \ge \int_{a}^{b}f_{1}d\alpha + \int_{a}^{b} f_{2}d\alpha \label{7} \end{equation} $$

Therefore, by $(6)$ and $(7)$, the following is obtained.

$$ \int_{a}^{b}fd\alpha = \int_{a}^{b}f_{1}d\alpha + \int_{a}^{b} f_{2}d\alpha $$

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