# Proof

Let$$\phi,\psi\in T[a,b]$$ be step functions. According to the definition of the Riemann integral for step functions, both integrals $\int_a^b\phi(x)dx=\sum_{i=1}^n\phi(x_i)(x_i-x_{i-1})\quad\quad\text{and}\quad\quad\int_a^b\psi(x)dx=\sum_{i=1}^n\psi(x_i)(x_i-x_{i-1})$ can be calculated without loss of generality using the same partition $$a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b$$. Therefore, the stated

linearity rules:

$\int_a^b(\phi+\psi)(x)dx=\int_a^b\phi(x)dx+\int_a^b\psi(x)dx$ $\int_a^b(\lambda\cdot \phi)(x)dx=\lambda\cdot\int_a^b\phi(x)dx\quad\quad(\text{for all }\lambda\in\mathbb R)$

and the monotony rule:

$\phi\le \psi\Rightarrow \int_a^b\phi(x)dx\le \int_a^b\psi(x)dx$

are trivial. In the monotony rule, "$$\phi\le \psi$$" means the order relation for step functions.

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983