Proof

Let $[a,b]$ be a closed real interval.
Let $\mathbb R$ be the set of real numbers.
Let $f:[a,b]\mapsto \mathbb R$ a continuous function.
It is to be shown that for every $\epsilon > 0$ there exist some step functions $\phi:[a,b]\mapsto\mathbb R$ and $\psi:[a,b]\mapsto\mathbb R$ with $(i)$ $\phi(x) \le f(x)\le \psi(x)$ and $(ii)$ $|\phi(x)-\psi(x)|\le\epsilon$ for all $x\in[a,b].$ This will be shown by construction.
- Because all continuous real functions on closed intervals are uniformly continuous, so is $f$.
- This means that for every $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x)-f(y)| < \frac\epsilon2$ for all $x,y\in D$ with $|x-y| < \delta.$
- Because of the existence of arbitrarily small positive rational numbers, we can choose a natural number $n$ big enough such that $\frac 1n < \delta/(b-a)$, or, equivalently $\frac {b-a}n < \delta$.
- By setting $$t_k:=a+k\frac{b-a}n,\quad k=0,1,\ldots, n,$$ we get an equidistant partition of the closed interval $[a,b]$ with $a=t_0 < t_1 < \ldots < t_n=b.$
- Set $\phi(a):=\psi(a):=f(a)$, and for $k=1,\ldots,n$ and $t_{k-1} < x\le t_k$ set $$\phi(x):=f(t_k)-\frac\epsilon2,\quad \psi(x):=f(t_k)+\frac\epsilon2.$$
- By construction, it follows that $|\phi(x)-\psi(x)|\le\epsilon$ for all $x\in[a,b],$ thus $(2)$ is proven.
- For all $x\in ]t_{k-1},t_k]$ we have by construction $$-\frac \epsilon2 < f(x)-f(t_k) < \frac \epsilon2,$$ thus $$\phi(x) = f(t_k)-\frac \epsilon2 < f(x) < f(t_k) + \frac \epsilon2=\psi(x),$$ therefore $\phi(x) \le f(x)\le \psi(x)$ for all $x\in[a,b],$ thus $(1)$ is proven.
  ∎

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References

Bibliography

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

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Proof

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