3.13. Integrators of Bounded Variation

Recall in the previous section we always assume that the integrator \(\alpha\) is increasing. Now, we want to extend the theorems about existence of integrals to the case when the integrator is of bounded variation.

For example, we want to generalize Theorem 3.16 to that assuming \(\alpha\) is of bounded variation on \([a, b]\), if \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), then \(f^2 \in \mathfrak{R}(\alpha)\) on \([a, b]\).

The key of achieving this is that a function \(\alpha\) of bounded variation can be written as a difference of two increasing functions \(\alpha_1\) and \(\alpha_2\), (Theorem 2.5).

\[\alpha = \alpha_1 - \alpha_2 \]

Then, to exploit the condition that \(f \in \mathfrak{R}(\alpha)\), we would want that \(f \in \mathfrak{R}(\alpha_1)\) and \(f \in \mathfrak{R}(\alpha_2)\). But this is not true in general due to the nonuniqueness of the decomposition of \(\alpha\) into two increasing functions. For example, consider \(\alpha(x) = 0\) on \([a, b]\). We can write \(\alpha\) as a difference of two identity functions \(\alpha_1(x) = \alpha(x) = x\). Then the Dirichlet function \(\ind_{\Q}\) is not integrable w.r.t. \(\alpha_1\) nor \(\alpha_2\). But it is integrable w.r.t. \(\alpha\) for the integrator is constant.

However, if we decompose \(\alpha\)(in the canonical way) as

\[\alpha(x) = V_a^x(\alpha) - [ V_a^x(\alpha) - \alpha(x) ]\]

then we will see in Theorem 3.18 that \(f\) is also integrable w.r.t. \(V_a^x(\alpha)\) and \(V_a^x(\alpha) - \alpha(x)\).

Lemma 3.1

Assume \(\alpha\) is of bounded variation on \([a, b]\). If \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), then for any given \(\varepsilon > 0\) there exists a partition \(P_\varepsilon\) of \([a, b]\) such that for all refinement \(P \supseteq P_\varepsilon\), \(P = \{x_0, \ldots, x_n\}\), we have

(3.33)\[\sum_{k=1}^n \omega_f(I_k) \abs{\Delta \alpha_k} < \varepsilon\]

Note

If \(\alpha\) is increasing, then what this lemma states is exactly that \(f\) satisfies the Riemann’s condition.

The idea of the proof is as follows. The appearance of the oscillation \(\omega_f(I_k)\) in (3.33) suggests us to consider the difference \(f(t_k) - f(t^\prime_k)\). We have for all \(t_k, t^\prime_k \in [x_{k-1}, x_k]\) that

\[\abs{\sum_{k=1}^n [ f(t_k) - f(t^\prime_k) ] \Delta \alpha_k} < c \varepsilon\]

The difficulty is that \(\Delta \alpha_k\) is not nonnegative anymore as in the previous section. To solve this, we simply consider the set \(A\) of indices of which \(\Delta \alpha_k \geq 0\) and the set \(B\) of indices of which \(\Delta \alpha_k < 0\) separately. Then we have

\[\sum_{k=1}^n [ f(t_k) - f(t^\prime_k) ]\Delta\alpha_k = \sum_{k \in A}[ f(t_k) - f(t^\prime_k) ]\abs{\Delta \alpha_k} + \sum_{k \in B}[ f(t_k^\prime) - f(t_k) ]\abs{\Delta \alpha_k}\]

The rest of the proof is done by choosing the \(t_k\) and \(t^\prime_k\) and then comparing \(\abs{f(t_k) - f(t^\prime_k)}\) with \(\omega_f(I_k)\).

Proof. If \(V_a^b(\alpha) = 0\), then \(\alpha\) is constant and hence the conclusion is trivial. In the following context, we assume \(V_a^b(\alpha) > 0\).

Let \(\varepsilon > 0\) be arbitrary. Because \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), there exists a partition \(P_\varepsilon\) of \([a, b]\) such that for any refinement \(P=\{x_0, \ldots, x_n\}\) and any list of representatives \(T\) of \(P\), we have

\[\abs{S(P, T, f, \alpha) - \int_a^b f \dif \alpha} < \varepsilon / 4 \]

It then follows that

\[\abs{\sum_{k=1}^n [ f(t_k) - f(t^\prime_k) ] \Delta \alpha_k} < \varepsilon / 2 \quad\forall t_k, t^\prime_k \in[x_{k-1}, x_k]\]

Let subsets \(A\) and \(B\) of indices \(\{1, \ldots, n\}\) be defined by

\[\begin{split}A & = \set{\Delta \alpha_k \geq 0}{k \in \{1, \ldots, n\}}\\ B & = \set{\Delta \alpha_k < 0}{k \in \{1, \ldots, n\}}\end{split}\]

Clearly, \(A\) and \(B\) forms a partition of \(\{1, \ldots, n\}\), i.e., \(A \cup B = \{1, \ldots, n\}\) and \(A \cap B = \emptyset\). It then follows that

\[\sum_{k=1}^n [ f(t_k) - f(t^\prime_k) ]\Delta\alpha_k = \sum_{k \in A}[ f(t_k) - f(t^\prime_k) ]\abs{\Delta \alpha_k} + \sum_{k \in B}[ f(t_k^\prime) - f(t_k) ]\abs{\Delta \alpha_k}\]

Therefore,

(3.34)\[\abs{ \sum_{k \in A} [ f(t_k) - f(t^\prime_k) ] \abs{\Delta \alpha_k} + \sum_{k \in B} [ f(t_k^\prime) - f(t_k) ] \abs{\Delta \alpha_k} } < \varepsilon / 2 \quad\forall t_k, t^\prime_k \in[x_{k-1}, x_k]\]

Now, we will strategically select \(t_k\) and \(t^\prime_k\) to accomplish our goal.

1. For \(k \in A\), we may select \(t_k\) and \(t^\prime_k\) such that \(f(t_k) - f(t^\prime_k) > \max \left\{ 0, \omega_f(I_k) - \frac{\varepsilon}{2 V_a^b(\alpha)} \right\}\), and

2. For \(k \in B\), we may select \(t_k\) and \(t^\prime_k\) such that \(f(t^\prime_k) - f(t_k) > \max \left\{ 0, \omega_f(I_k) - \frac{\varepsilon}{2 V_a^b(\alpha)} \right\}\).

Plugging the particular choices of \(t_k\)’s and \(t^\prime_k\)’s into (3.34), we obtain

\[\begin{split}\varepsilon / 2 & > \abs{ \sum_{k \in A} [ f(t_k) - f(t^\prime_k) ] \abs{\Delta \alpha_k} + \sum_{k \in B} [ f(t_k^\prime) - f(t_k) ] \abs{\Delta \alpha_k} }\\& = \sum_{k \in A}[ f(t_k) - f(t^\prime_k) ]\abs{\Delta \alpha_k} + \sum_{k \in B}[ f(t_k^\prime) - f(t_k) ]\abs{\Delta \alpha_k}\\& > \sum_{k=1}^n \left(\omega_f(I_k) - \frac{\varepsilon}{2 V_a^b(\alpha)}\right)\abs{\Delta \alpha_k}\\& = \sum_{k=1}^n \omega_f(I_k) \abs{\Delta \alpha_k} - \frac{\varepsilon}{2 V_a^b(\alpha)}\sum_{k=1}^n \abs{\Delta \alpha_k}\\& \geq\sum_{k=1}^n \omega_f(I_k) \abs{\Delta \alpha_k} - \varepsilon / 2 \end{split}\]

where the last inequality follows from the property of total variations \(\sum_{k=1}^n \abs{\Delta \alpha_k} \leq V_a^b(\alpha)\). Moving the term \(-\varepsilon / 2\) to the left in the above inequality yields

\[\sum_{k=1}^n \omega_f(I_k) \abs{\Delta \alpha_k} < \varepsilon\]

Theorem 3.18

Assume \(\alpha\) is of bounded variation on \([a, b]\). If \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\) then \(f \in \mathfrak{R}(V_a^x(\alpha))\) on \([a, b]\).

And of course, applying the linearity, we also have \(f \in \mathfrak{R}(V_a^x(\alpha) - \alpha)\) on \([a, b]\).

Proof. Suppose \(f\) is bounded by \(M > 0\).

Let \(\varepsilon > 0\) be arbitrary. By Lemma 3.1, there exists a partition \(P^\prime_\varepsilon\) of \([a, b]\) such that for all \(P \supseteq P^\prime_\varepsilon\), we have

(3.35)\[\sum_{k}\omega_f(I_k) \abs{\Delta \alpha_k} < \varepsilon / 2\]

By the definition of total variations and Proposition 2.4, there exists a partition \(P^{\prime\prime}_\varepsilon\) such that for all its refinement we have

(3.36)\[V_a^b(\alpha) < \sum_{k}\abs{\Delta \alpha_k} + \frac{\varepsilon}{4M}\]

Let \(P_\varepsilon = P^\prime_\varepsilon \cup P^{\prime\prime}_\varepsilon\). For any its refinement \(P\), both (3.35) and (3.36) hold. And we have

\[\begin{split}\sum_{k}\omega_f(I_k) (V_{x_{k-1}}^{x_k}(\alpha) - \abs{\Delta \alpha_k}) & \leq 2M \sum_{k}(V_{x_{k-1}}^{x_k}(\alpha) - \abs{\Delta \alpha_k}) \\& = 2M \left( V_a^b(\alpha) - \sum_{k}\abs{\Delta \alpha_k}\right)\end{split}\]

Then applying (3.36) yields

(3.37)\[\sum_{k}\omega_f(I_k) (V_{x_{k-1}}^{x_k}(\alpha) - \abs{\Delta \alpha_k}) < \varepsilon / 2\]

Adding (3.35) and (3.37), we obtain

\[\sum_{k}\omega_f(I_k) V_{x_{k-1}}^{x_k}(\alpha) < \varepsilon\]

This implies that \(f\) satisfies the Riemann’s condition w.r.t. \(V_a^x(\alpha)\) on \([a, b]\) and hence the proof is complete.

Theorem 3.19

Assume \(\alpha\) is of bounded variation on \([a, b]\). If \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\) then \(f\) is also integrable on any subinterval. That is, if \([c, d] \subseteq[a, b]\), then \(f \in \mathfrak{R}(\alpha)\) on \([a, d]\).

Proof. Thanks to Theorem 3.18 and Theorem 3.2, we only need to prove this theorem for increasing integrators. In what follows, we assume \(\alpha\) is increasing.

We are going to show that \(f\) satisfies the Riemann’s condition w.r.t. \(\alpha\) on \([c, d]\). Let \(\varepsilon > 0\) be arbitrary. Because \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), there exists a partition \(P_\varepsilon\) of \([a, b]\) such that

\[U(P_\varepsilon, f, \alpha) - L(P_\varepsilon, f, \alpha) < \varepsilon\]

Let \(P^\prime = P_\varepsilon \cup \{c, d\}\). Since \(P^\prime\) is a refinement of \(P_\varepsilon\). It also holds that

\[U(P^\prime, f, \alpha) - L(P^\prime, f, \alpha) < \varepsilon\]

Let \(P = P^\prime \cap [c, d]\). We note that \(P\) is a partition of \([c, d]\). And if we write

\[U(P^\prime, f, \alpha) - L(P^\prime, f, \alpha) = \sum_{k \in J}\omega_f(I_k) \Delta\alpha_k \]

then we will find that \(U(P, f, \alpha) - L(P, f, \alpha)\) is the sum of a subcollection of terms from the above sum. That is, we can write

\[U(P, f, \alpha) - L(P, f, \alpha) = \sum_{k \in K}\omega_f(I_k) \Delta\alpha_k \]

where \(K \subseteq J\). Since each term \(\omega_f(I_k) \Delta \alpha_k\) is nonnegative, we have

\[U(P, f, \alpha) - L(P, f, \alpha) \leq U(P^\prime, f, \alpha) - L(P^\prime, f, \alpha) < \varepsilon\]

This implies that \(f\) satisfies the Riemann’s condition w.r.t. \(\alpha\) on \([c, d]\).

Theorem 3.20

Assume \(f, g \in \mathfrak{R}(\alpha)\) on \([a, b]\) where the integrator \(\alpha\) is increasing. Define functions \(F\) and \(G\) on \([a, b]\) by

\[F(x) = \int_a^x f \dif\alpha\]

and

\[G(x) = \int_a^x g \dif\alpha\]

Then \(f \in \mathfrak{R}(G)\), \(g \in \mathfrak{R}(F)\) and \(fg \in \mathfrak{R}(\alpha)\) on \([a, b]\). And we have

\[\int_a^b f g \dif\alpha = \int_a^b f \dif G = \int_a^b g \dif F \]

Proof. We first show that \(F\) and \(G\) are well defined. When \(x = a\), the lower and upper limits are equal, hence the integrals are zeros by the definition. And when \(a < x \leq b\), by Theorem 3.19, \(f\) and \(g\) are integrable w.r.t. \(\alpha\) on \([a, x]\).

The conclusion that \(fg \in \mathfrak{R}(\alpha)\) on \([a, b]\) is exactly Theorem 3.17.

In the following, we will only prove that \(f \in \mathfrak{R}(G)\) and

\[\int_a^b f g \dif\alpha = \int_a^b f \dif G \]

Suppose \(f\) is bounded by \(M > 0\). Let \(\varepsilon > 0\) be arbitrary. Because \(g \in \mathfrak{R}(\alpha)\) on \([a, b]\), equivalently, \(g\) satisfies the Riemann’s condition. There exists a partition \(P^\prime_\varepsilon\) such that

(3.38)\[\sum_{k=1}^n \omega_g(I_k) \Delta\alpha_k < \frac{\varepsilon}{2M}\quad\forall P \supseteq P^\prime_\varepsilon\]

Because the product \(fg\) is integrable w.r.t. \(\alpha\), there exists a partition \(P^{\prime\prime}_\varepsilon\) such that

(3.39)\[\abs{S(P, T, fg, \alpha) - \int_a^b f g \dif \alpha} < \varepsilon / 2 \quad\forall P \supseteq P^{\prime\prime}_\varepsilon\quad\forall T \text{ of } P\]

Let \(P_\varepsilon = P^\prime_\varepsilon \cup P^{\prime\prime}_\varepsilon\). Let \(P \supseteq P_\varepsilon\) and \(T\) be any list of representatives of \(P\). We will compare \(S(P, T, f, G)\) and \(S(P, T, fg, \alpha)\). We have

(3.40)\[\begin{split}\abs{S(P, T, f, G) - S(P, T, fg, \alpha)}& = \abs{\sum_{k=1}^n f(t_k) [\Delta G_k - g(t_k) \Delta \alpha_k]}\nonumber\\& \leq\sum_{k=1}^n \abs{f(t_k)}\abs{\Delta G_k - g(t_k) \Delta \alpha_k}\nonumber\\& \leq M \sum_{k=1}^n \abs{\Delta G_k - g(t_k) \Delta \alpha_k}\end{split}\]

where

\[\Delta G_k = \int_{x_{k-1}}^{x_k} g \dif\alpha\]

Note

The well-definedness of \(\Delta G_k\) also follows from Theorem 3.19.

Let \(m_k = \inf_{x \in [x_{k-1}, x_k]} g(x)\) and \(M_k = \sup_{x \in [x_{k-1}, x_k]} g(x)\). Because \(m_k \leq g(x) \leq M_k \; \forall x \in [x_{k-1}, x_k]\), by the comparison theorem of integrals (Corollary 3.1), we have

\[m_k \Delta\alpha_k = \int_a^n m_k \dif\alpha\leq\int_{x_{k-1}}^{x_k} g \dif\alpha\leq\int_a^n M_k \dif\alpha = M_k \Delta\alpha_k \]

That is,

\[m_k \Delta\alpha_k \leq\Delta G_k \leq M_k \Delta\alpha_k \]

It then follows that

(3.41)\[\abs{\Delta G_k - g(t_k) \Delta \alpha_k}\leq\omega_g(I_k) \Delta\alpha_k\]

Combining (3.40), (3.41) and (3.38) yields

(3.42)\[\abs{S(P, T, f, G) - S(P, T, fg, \alpha)}\leq M \sum_{k=1}^n \omega_g(I_k) \Delta\alpha_k < M \frac{\varepsilon}{2M} =\varepsilon / 2\]

Finally, compare (3.42) and (3.39) and we may conclude that \begin{multline*} \abs{S(P, T, f, G) - \int_a^b f g \dif \alpha} \ \leq \abs{S(P, T, f, G) - S(P, T, fg, \alpha)}

• \abs{S(P, T, fg, \alpha)- \int_a^b f g \dif \alpha} < \varepsilon \end{multline*} This implies that \(f \in \mathfrak{R}(G)\) on \([a, b]\) and \(\int_a^b f \dif G = \int_a^b f g \dif \alpha\).