📂Analysis

Necessary and Sufficient Conditions for Riemann(-Stieltjes) Integrability

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

Theorem¹

A necessary and sufficient condition for a function $f$ to be Riemann(-Stieltjes) integrable on $[a,b]$ is that for every $\epsilon >0$, there exists a partition $P$ of $[a,b]$ that satisfies $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.

$$\begin{equation} f \in \mathscr{R} (\alpha) \text{ on } [a,b] \\ \iff \forall\epsilon >0, \exists P\text{ s.t. } U(P,f,\alpha) - L(P,f,\alpha) < \epsilon \end{equation}$$

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This condition is practically used when proving integrability.

Proof

Given the assumptions below:

• $f : [a,b] \to \mathbb{R}$ is bounded.
• $\alpha : [a,b] \to \mathbb{R}$ is a monotonically increasing function.
• $P$ is referred to as a partition of $[a,b]$.

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• $(\implies)$

  Suppose $f$ is a Riemann(-Stieltjes) integrable function and $\epsilon > 0$ is given. By the definition of lower and upper integration, for every partition $P$, the following holds.

  $$L(P,f,\alpha) \le \underline{\int _{a}^{b}} f d\alpha = \int _{a} ^{b} f d\alpha$$

  Therefore, there exists a partition $P_{1}$ that satisfies:

  $$\begin{equation} \int _{a} ^{b} f d\alpha - L(P_{1},f,\alpha) < \frac{\epsilon}{2} \end{equation}$$

  Similarly, the following is true.

  $$\int _{a} ^{b} f d\alpha = \overline {\int _{a} ^{b}} f d\alpha \le U(P,f,\alpha)$$

  Thus, there exists a partition $P_{2}$ that satisfies:

  $$\begin{equation} U(P_2,f,\alpha) - \int _{a}^{b} f d\alpha < \frac{\epsilon}{2} \end{equation}$$

  Now, let’s call $P^{\ast}$ the common refinement of $P_{1}$ and $P_{2}$. Then, since the refinement of the upper(sum) or lower(sum) is smaller(or larger) than the partition, by $(2)$, $(3)$, the following holds.

  $$\begin{align*} U(P^{\ast},f,\alpha) &\le U(P_2,f,\alpha) \\ &\lt {\color{blue}\int _{a} ^{b} f d\alpha} + \frac{\epsilon}{2} \\ &\lt {\color{blue} L(P_{1},f,\alpha) + \frac{\epsilon}{2} } + \frac{\epsilon}{2} \\ &= L(P_{1},f,\alpha) + \epsilon \\ &\le L(P^{\ast},f,\alpha) + \epsilon \end{align*}$$

  Therefore, there exists a partition $P^{\ast}$ that satisfies $U(P^{\ast},f,\alpha)-L(P^{\ast},f,\alpha) < \epsilon$.

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• $(\impliedby)$

  Assume for all $\epsilon >0$, there exists a partition $P$ of $[a,b]$ that satisfies $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$. By the definition of upper and lower integration, the following equation holds.

  $$L(P,f,\alpha) \le \underline {\int _{a} ^{b}} f d\alpha \le \overline{ \int _{a} ^{b}}f d\alpha \le U(P,f,\alpha)$$

  If $A<B<C<D$, then $C-B<D-A$. Thus, using the assumption and the above equation, we obtain:

  $$0 \le \overline {\int _{a}^{b}} f d\alpha -\underline{\int _{a} ^{b}} f d\alpha < \epsilon$$

  For this to be true for all positive numbers $\epsilon$, the following must hold:

  $$\overline {\int _{a}^{b}} f d\alpha -\underline{\int _{a} ^{b}} f d\alpha=0$$

  Therefore, the following is true, and it is the definition of integrability for $f$, meaning $f$ is integrable.

  $$\overline {\int _{a}^{b}} f d\alpha =\underline{\int _{a} ^{b}} f d\alpha$$

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Corollaries²

• (a) If for some partition $P$ and $\varepsilon >0$, $(1)$ holds, then $(1)$ also holds for all refinements of $P$.

• (b) If for a partition $P=\left\{ x_{0},\cdots,x_{n} \right\}$, $(1)$ holds and we denote it by $s_{i},t_{i}\in [x_{i-1},x_{i}]$, then the following inequality is true. $$\sum \limits _{i=1} ^{n} \left| f(s_{i}) -f(t_{i}) \right| \Delta \alpha_{i} <\varepsilon$$

• (c) If $f$ is integrable and the assumption of (b) holds, then the following formula is true. $$\left| \sum \limits _{i=1} ^{n} f(t_{i})\Delta \alpha_{i} - \int _{a} ^{b}f (x)d\alpha (x) \right| < \varepsilon$$

Proof

(a)

Let’s call $P^{\ast}$ a refinement of $P$. Then, by the properties of refinements, the following holds.

$$U(P^{\ast},f,\alpha) -L(P^{\ast},f,\alpha)<U(P,f,\alpha) -L(P,f,\alpha) <\varepsilon$$

Thus, (a) is true.

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(b)

Let’s denote the following for $x\in[x_{i-1},x_{i}]$:

$$M_{i}=\sup f(x) \quad \text{and} \quad m_{i}=\inf f(x)$$

Then for all $s_{i},t_{i}\in [x_{i-1},x_{i}]$, the following is true.

$$\left| f(s_{i})-f(t_{i}) \right| < M_{i}-m_{i},\quad i=1,\cdots,n$$

Therefore, by the definition of upper and lower sum, the following holds.

$$\begin{align*} \sum \limits _{i=1} ^{n} \left| f(s_{i})-f(t_{i}) \right| \Delta \alpha_{i} &< \sum \limits _{i=1} ^{n}(M_{i}-m_{i})\Delta \alpha_{i} \\ &=U(P,f,\alpha)-L(P,f,\alpha) \\ &< \varepsilon \end{align*}$$

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(c)

Continuing with the notation used in the previous proofs, it is self-evident by the definition of upper and lower sums that the following is true.

$$L(P,f,\alpha) \le \sum \limits _{i=1} ^{n} f(t_{i})\Delta \alpha_{i} \le U(P,f,\alpha)$$

Also, by the definition of integration, the following is obviously true.

$$L(P,f,\alpha) \le \int _{a} ^{b} f(x)d\alpha (x) \le U(P,f,\alpha)$$

Therefore, by the above two formulas, the following is true.

$$\left| \sum \limits _{i=1} ^{n} f(t_{i})\Delta \alpha_{i} - \int _{a} ^{b}f (x)d\alpha (x) \right| < \varepsilon$$

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