Riemann Integral for Bounded Functions

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Now we will define the integral for bounded functions:

It can be seen from the figures above that the integral can be seen as the signed area between the function graph and the \(x\)-axis. Signed means that the area of the negative parts of the function has to be counted negative.

By the above definition, we can directly deduce that for \(\phi\in\mathcal{T}([a,b])\) holds \[\overline{\int_a^b}\phi(x)\, dx=\underline{\int_a^b}\phi(x)\, dx={\int_a^b}\phi(x)\, dx.\]

For general bounded functions \(f:[a,b]\to\mathbb{R}\) holds \[\underline{\int_a^b}f(x)\, dx\leq \overline{\int_a^b}f(x)\, dx.\] Note that in general the upper integral does not coincide with the lower integral. For instance, consider the function \(f: [0,1] \to\mathbb{R}\) with \[f(x)=\begin{cases}1&:\;x\in\mathbb{Q}\\0&:x\in\mathbb{R}\backslash\mathbb{Q}\end{cases}\] Then we have \[\overline{\int_0^1}f(x)\, dx=1>0=\underline{\int_0^1}f(x)\, dx.\]

Proof:
(i) Let \(\varepsilon>0\). Then there exist \(\phi,\psi\in\mathcal{T}([a,b])\) with \(\phi\geq f\), \(\psi\geq g\) and \[\overline{\int_a^b}f(x)\, dx+\frac\varepsilon2\geq{\int_a^b}\phi(x)\, dx,\qquad \overline{\int_a^b}g(x)\, dx+\frac\varepsilon2\geq{\int_a^b}\psi(x)\, dx.\] Then \(f+g\leq\phi+\psi\) and \[\begin{aligned} \overline{\int_a^b}f(x)+g(x)\, dx=&\inf\left\{\int_a^b\zeta(x)\, dx\;:\; \zeta\in\mathcal{T}([a,b])\text{ with }\zeta\geq f+g\right\}\\ \leq& {\int_a^b}\phi(x)+\psi(x)\, dx\\ \leq& \left(\overline{\int_a^b}f(x)\, dx+\frac\varepsilon2\right)+\left(\overline{\int_a^b}g(x)\, dx+\frac\varepsilon2\right)\\=& \overline{\int_a^b}f(x)\, dx+\overline{\int_a^b}g(x)\, dx+\varepsilon. \end{aligned}\] Since this holds true for all \(\varepsilon>0\), we conclude \[\overline{\int_a^b}f(x)+g(x)\, dx\leq\overline{\int_a^b}f(x)\, dx+\overline{\int_a^b}g(x)\, dx.\] (ii): Analogous to (i).
(iii): We only show the statement for the upper integral. The other result can be shown analogously.
Let \(\varepsilon>0\). Then there exists some \(\phi\in\mathcal{T}([a,b])\) with \(\phi\geq f\) and \[\overline{\int_a^b}f(x)\, dx+\varepsilon\geq{\int_a^b}\phi(x)\, dx.\] Then \(\lambda \phi\geq\lambda f\) and \[\overline{\int_a^b} \lambda f(x)\, dx\leq{\int_a^b}\lambda\phi(x)\, dx=\lambda{\int_a^b}\phi(x)\, dx\leq \lambda\overline{\int_a^b} f(x)\, dx+\lambda\varepsilon.\] Since this holds true for all \(\varepsilon>0\), we conclude \[\overline{\int_a^b}\lambda f(x)\, dx\leq\lambda \overline{\int_a^b}f(x)\, dx.\] The opposite inequality follows from the previous one applied to \(g(x):=\lambda f(x)\) and \(\mu:=\frac{1}{\lambda}>0\): \[\overline{\int_a^b} f(x)\, dx = \overline{\int_a^b} \mu g(x)\, dx \leq \mu \overline{\int_a^b}g(x)\, dx = \frac{1}{\lambda} \overline{\int_a^b}\lambda f(x)\, dx.\] Multiplying this inequality by \(\lambda>0\) yields \[\lambda \overline{\int_a^b} f(x)\, dx \leq \overline{\int_a^b}\lambda f(x)\, dx.\]

(iv): This result can be shown by using (iii) and the fact \[\overline{\int_a^b}-f(x)\, dx=-\underline{\int_a^b}f(x)\, dx.\] \(\Box\)

We obviously have that \(\mathcal{T}([a,b])\subset\mathcal{R}([a,b])\). In the following we state that monotonic functions as well as continuous functions are Riemann-integrable. The proof is not presented here.

The following result is straightforward: