# Proof

We prove both implications:

### "$$\Leftarrow$$"

• Now, assume that $$f$$ is bounded and that for every $$\epsilon > 0$$ there exist some step functions $$\phi:[a,b]\mapsto\mathbb R$$ and $$\psi:[a,b]\mapsto\mathbb R$$ with $\phi\le f\le \psi\quad\quad( * )$ and $\int_a^b\psi(x)dx-\int_a^b\phi(x)dx\le \epsilon.$
• This means that, as $$\epsilon$$ tends to $$0$$, the Riemann lower integral of the step function $$\phi$$ and the Riemann upper integral of the step function $$\psi$$ tend to each other: $\lim_{\epsilon\to 0}\int_{a~*}^{b}\phi(x)dx=\lim_{\epsilon\to 0}\int_{a}^{b~*}\psi(x)dx.\quad\quad ( * * )$
• On the other hand, from $$( * )$$, from the monotony of Riemann integrals for step functions, and from the fact that convergence preserves lower and upper bounds, it follows. $\lim_{\epsilon\to 0}\int_{a~*}^{b}\phi(x)dx\le \int_{a~*}^{b}f(x)dx \le \int_{a}^{b}f(x)dx\le \int_{a}^{b~*}f(x)dx \le \lim_{\epsilon\to 0}\int_{a}^{b~*}\psi(x)dx.$
• Together with $$( * * )$$, this means that $$f$$ is Riemann integrable, because in fact, $\int_{a~*}^{b}f(x)dx = \int_{a}^{b}f(x)dx= \int_{a}^{b~*}f(x)dx.$

Github: ### References

#### Bibliography

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