Welcome guest
You're not logged in.
259 users online, thereof 0 logged in

## Proof: (related to "Addition and Scalar Multiplication of Riemann Upper and Lower Integrals")

• By hypothesis, $[a,b]$ is a closed interval and $$f,g:[a,b]\mapsto\mathbb R$$ are bounded on $$[a,b]$$.
• Let $\epsilon > 0$ be an arbitrarily small real number, and let $T[a,b]$ be the set of all step functions on $[a,b].$
• By definition of the Riemann upper integrals there are step functions $\phi,\psi\in T[a,b],$ such that $\phi(x)\ge f(x), \psi(x)\ge g(x)$ for all $x\in[a,b]$ and $$\int_a^{b}\phi(x)dx\le \int_a^{b~*}f(x)dx+\frac{\epsilon}{2},\quad\quad\int_a^{b}\psi(x)dx\le \int_a^{b~*}g(x)dx+\frac{\epsilon}{2}.$$
• Thus,
$$\int_a^{b}(\phi+\psi)(x)dx\le \int_a^{b~*}f(x)dx+\int_a^{b~*}g(x)dx+\epsilon.$$
• But, by construction, $(\phi+\psi)(x)\ge (f+g)(x)$ for all $x\in[a,b],$ it follows
$$\int_a^{b~*}(f+g)(x)dx\le \int_a^{b}(\phi+\psi)(x)dx\le \int_a^{b~*}f(x)dx+\int_a^{b~*}g(x)dx+\epsilon.$$
• Since $\epsilon$ was arbitrary, it follows
$$\int_a^{b~*}(f+g)(x)dx\le \int_a^{b~*}f(x)dx+\int_a^{b~*}g(x)dx.$$
• By definition of the Riemann upper and lower integrals, we have $$\int_{a~*}^{b}f(x)dx=-\int_{a}^{b~*}(-f)(x)dx\quad\quad\int_{a~*}^{b}g(x)dx=-\int_{a}^{b~*}(-g)(x)dx.\label{E20175}\tag{*}$$
• Therefore, by a)
$$\int_{a~*}^{b}(f+g)(x)dx\ge \int_{a~*}^{b}f(x)dx+\int_{a~*}^{b}g(x)dx.$$
• Let $\lambda \ge 0$.
• In the casse $\lambda = 0,$ the equation is trivial.
• So let $\lambda > 0.$ We have to show that
$$\lambda\int_a^{b~*}f(x)dx-\epsilon\le \int_a^{b~*}(\lambda f)(x)dx\le\lambda\int_a^{b~*}f(x)dx+\epsilon$$
for all $\epsilon > 0.$
• By definition of the Riemann upper and lower integrals, there is a step function $\phi\in T[a,b]$ with $\phi(x)\ge f(x)$ for all $x\in[a,b]$ and $$\int_{a}^{b}(\lambda \phi)(x)dx\le \lambda\int_{a}^{b~*}f(x)dx+\epsilon.$$
• Since $\lambda\phi(x)\ge \lambda f(x)$ for all $x\in[a,b],$ it follows
$$\int_{a}^{b~*}(\lambda f)(x)dx\le \int_{a}^{b}(\lambda \phi)(x)dx\le \lambda\int_{a}^{b~*}f(x)dx+\epsilon.$$
• Analogously, we can show
$$\int_{a}^{b~*}(\lambda f)(x)dx\ge \int_{a}^{b}(\lambda \phi)(x)dx\ge \lambda\int_{a}^{b~*}f(x)dx-\epsilon.$$
• Altogether, it follows
$$\int_{a}^{b~*}(\lambda f)(x)dx=\lambda\int_{a}^{b~*}f(x)dx$$
for all $\lambda\ge 0.$
• Analogously, by $(\ref{E20175})$ and c) it follows
$$\int_{a~*}^{b}(\lambda f)(x)dx=\lambda\int_{a~*}^{b}f(x)dx$$
for all $\lambda < 0.$
• From $(\ref{E20175})$ it follows
$$\int_{a~*}^{b}(\lambda\cdot f)(x)dx=\lambda\cdot\int_{a}^{b~*}f(x)dx$$
for all $\lambda < 0.$
q.e.d

| | | | created: 2020-01-09 06:07:54 | modified: 2020-01-09 06:07:54 | by: bookofproofs | references: [581]

(none)