3.12. Comparison Theorems

Theorem 3.14

Assume \(\alpha\) is increasing and \(f \geq 0\) on \([a, b]\). If \(f \in \mathfrak{R}(\alpha)\), then

\[\int_a^b f \dif\alpha\geq 0 \]

Proof. Let \(P\) be any partition of \([a, b]\). On any subinterval \([x_{k-1}, x_k]\), we have \(f(x) \geq 0\) and hence

\[\sup_{[x_{k-1}, x_k]} f(x) \Delta\alpha_k \geq 0 \]

Summing up over \(k\), we obtain

\[U(P, f, \alpha) \geq 0 \]

Finally, taking the infimum over \(P\) yields

\[\overline{\int_a^b} f \dif\alpha\geq 0 \]

Because \(f \in \mathfrak{R}(\alpha)\), its integral equals to the upper integral, hence greater than or equal to zero.

Corollary 3.1

Assume \(\alpha\) is increasing and \(f \leq g\) on \([a, b]\). If \(f, g \in \mathfrak{R}(\alpha)\), then

\[\int_a^b f \dif\alpha\leq\int_a^b g \dif\alpha\]

Theorem 3.15

Assume \(\alpha\) is increasing on \([a, b]\). If \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), then \(\abs{f} \in \mathfrak{R}(\alpha)\) on \([a, b]\), and

(3.32)\[\abs{\int_a^b f \dif \alpha}\leq\int_a^b \abs{f}\dif\alpha\]

Proof. Let \(\varepsilon > 0\) be arbitrary. Because \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\) equivalently, satisfying the Riemann’s condition, there exists a partition \(P_\varepsilon = \{x_0, \ldots, x_n\}\) such that \(U(P, f, \alpha) - L(P, f, \alpha) < \varepsilon\). On each subinterval \(I_k = [x_{k-1}, x_k]\), we have

\[\omega_{\abs{f}}(I_k) = \sup_{x, y \in I_k}( \abs{f(x)} - \abs{f(y)}) \leq\sup_{x, y \in I_k}\abs{ f(x) - f(y) } = \omega_f(I_k) \]

Multiplying by \(\Delta \alpha_k\) on both sides and then summing over \(k\), we obtain

\[\begin{split}U(P_\varepsilon, \abs{f}, \alpha) - L(P_\varepsilon, \abs{f}, \alpha) & = \sum_{k=1}^n \omega_{\abs{f}}(I_k) \Delta\alpha_k \\& \leq\sum_{k=1}^n \omega_{f}(I_k) \Delta\alpha_k \\& = U(P_\varepsilon, f, \alpha) - L(P_\varepsilon, f, \alpha) \\& < \varepsilon\end{split}\]

This implies that \(\abs{f}\) also satisfies the Riemann’s condition and hence integrable.

Then, since \(-\abs{f} \leq f \leq \abs{f}\), (3.32) is proved by applying Corollary 3.1.

Example 3.4

If \(\abs{f}\) is integrable then \(f\) does not need to be integrable. Consider a function \(f\) on \([a, b]\) defined by

\[\begin{split}f(x) = \ind_{\Q}(x) - \ind_{\R \setminus \Q}(x) = \begin{cases} 1 & x \in \Q \\ -1 & x \notin \Q \end{cases}\end{split}\]

We have \(\abs{f(x)} = 1\), which is a constant and hence integrable. But \(f\) is not.

Theorem 3.16

Assume \(\alpha\) is increasing on \([a, b]\). If \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\) then \(f^2 \in \mathfrak{R}(\alpha)\) on \([a, b]\).

Proof. Suppose \(f\) is bounded by \(M > 0\) on \([a, b]\). 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

\[U(P_\varepsilon, f, \alpha) - L(P_\varepsilon, f, \alpha) < \frac{\varepsilon}{2M}\]

We are going to show that \(f^2\) also satisfies the Riemann’s condition.

On each subinterval \(I_k\), we have

\[\begin{split}\abs{f^2(x) - f^2(y)}& = \abs{f(x) + f(y)}\abs{f(x) - f(y)}\\& \leq 2M \sup_{x, y \in I_k}\abs{f(x) - f(y)}\\& = 2M \omega_f(I_k) \end{split}\]

It then follows that

\[\omega_{f^2}(I_k) = \sup_{x, y \in I_k}\abs{f^2(x) - f^2(y)}\leq 2M \omega_f(I_k) \]

Multiplying by \(\Delta \alpha_k\) and then summing over \(k\) yields

\[\begin{split}U(P_\varepsilon, f^2, \alpha) - L(P_\varepsilon, f^2, \alpha) & = \sum_{k}\omega_{f^2}(I_k) \Delta\alpha_k \\& \leq 2M \sum_{k}\omega_f(I_k) \Delta\alpha_k \\& = 2M [U(P_\varepsilon, f, \alpha) - L(P_\varepsilon, f, \alpha)]\\& < 2M \frac{\varepsilon}{2M}\\& = \varepsilon\end{split}\]

An immediate consequence of the previous theorem is that the product of integrable functions is also integrable.

Theorem 3.17

Assume \(\alpha\) is increasing on \([a, b]\). If \(f, g \in \mathfrak{R}(\alpha)\) on \([a, b]\) then the product is also integrable: \(fg \in \mathfrak{R}(\alpha)\) on \([a, b]\).

Proof. The product \(fg\) can be written as

\[fg = \frac{1}{4}[ (f+g)^2 - (f-g)^2 ]\]

Because the sum and difference \(f \pm g\) are integrable, so do their squares \((f \pm g)^2\) by the previous theorem. Therefore, \(fg \in \mathfrak{R}(\alpha)\) on \([a, b]\).