📂Analysis

Upper integral is greater than or equal to lower integral.

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

Theorem¹

For any partition, the Riemann(-Stieltjes) upper sum is always greater than or equal to the Riemann(-Stieltjes) lower sum.

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

Proof

Before proving, let’s assume the following:

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

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Let $P_{1}, P_{2}$ be a partition of $[a,b]$, and let $P^{\ast}$ be their common refinement. Since a refinement’s upper (lower) sum is less (more) than the partition’s, the following holds.

$$\begin{equation} \begin{aligned} &&L(P_{1},f,\alpha) \le L(P^{\ast},f,\alpha) &\le U(P^{\ast},f,\alpha) \le U(P_{2},f,\alpha) \\ \implies&& L(P_{1},f,\alpha) &\le U(P_{2},f,\alpha) \end{aligned} \label{eq1} \end{equation}$$

Here, fix $P_{2}$ and for all $P_{1}$, take $\sup$. Then, by the definition of the lower sum, the following is obtained.

$$\sup\limits_{P_{1}} L(P_{1},f,\alpha) = \underline {\int _{a} ^b} f d\alpha \le U(P_{2}, f , \alpha)$$

Similarly, by taking $\inf$ for $P_{2}$ in the above equation, the upper sum’s definition yields the following.

$$\underline {\int _{a} ^b} f d\alpha \le \inf \limits_{P_{2}} U(P_{2}, f , \alpha) = \overline {\int _{a} ^b} f d\alpha$$

Therefore, the following result is obtained.

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

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Corollary

For any two partitions, the Riemann-Stieltjes upper sum is always greater than or equal to the lower sum.

$$L(P_{1},f,\alpha) \le U(P_{2}, f, \alpha) \quad \forall P_{1},\ P_{2}$$

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It holds by $\eqref{eq1}$.