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Let $f:[a,b]\to \mathbb{R}$ be a monotone function (say strictly increasing).

Then, do for every $\epsilon>0$ exist two step functions $h,g$ so that $g\le f\le h$ and $0\le h-g\le \epsilon$?

Does there exist some closed form of these functions like the one below?

I encountered the problem when trying to prove the Riemann Integrability of monotone functions via the traditional definition of the Riemann Integral (not the one with Darboux sums). The book I am reading proves that a continuous function is integrable via the process below:

Let $\mathcal{P}=\left\{ a=x_0<...<x_n=b \right\}$ be a partition of $[a,b]$. Define $m_i=\min_{x\in [x_{i-1},x_i]}f(x)$ and $M_i=\max_{x\in [x_{i-1},x_i]}f(x)$. By the Extreme Value theorem $M_i,m_i$ are well defined. We approximate $f$ with step functions: \begin{gather}g=m_1\chi_{[x_0,x_1]}+\sum_{i=2}^nm_i\chi_{(x_{i-1},x_i]}\\ h=M_1\chi_{[x_0,x_1]}+\sum_{i=2}^nM_i\chi_{(x_{i-1},x_i]}\end{gather} It is easily seen that they satisfy the "step function approximation" and as $0\le \int_a^bh-g\le \epsilon$ (uniform continuity) by a previous theorem $f$ is integrable.

The book then goes on to generalise by discussing regulated functions. I would like however to see a self contained proof similar to the previous one if $f$ is monotone. This question is reduced to the two questions asked in the beggining of the post

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2 Answers 2

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Let $\varepsilon>0$ and $N_\varepsilon$ the smallest $n \in \mathbb{N}$ such that $$ \frac{1}{n}(b-a)(f(b)-f(a)) \le \varepsilon $$ For $n \ge \max\{2,N_\varepsilon\}$ consider the following partition of $[a,b]$: $$ \mathcal{P}=\{a=x_0<x_1<\ldots<x_n=b\}, \ x_i=a+i\frac{b-a}{n}\ 0 \le i \le n. $$ Set $$ A_i=\begin{cases} [x_0,x_1] & \text{ for } i=0\\ (x_i,x_{i+1}]& \text{ for } 1 \le i \le n-1 \end{cases} $$ Since $f$ is strictly increasing for each $i \in \{0,\ldots,n-1\}$ we have $$ f(x_i) \le f(x) \le f(x_{i+1}) \quad \forall\ x_i \le x \le x_{i+1}. $$ Setting $$ h=\sum_{i=0}^{n-1}f(x_{i+1})\chi_{A_i},\ g=\sum_{i=0}^{n-1}f(x_i)\chi_{A_i} $$ we have $$ g(x)\le f(x) \le h(x) \quad \forall\ x \in [a,b], $$ and $$ h(x)-g(x)=\sum_{i=0}^{n-1}\Big(f(x_{i+1})-f(x_i)\Big)\chi_{A_i}(x)>0 \quad \forall\ x \in [a,b]. $$ In addition \begin{eqnarray} \int_a^b(h-g)&=&\sum_{i=0}^{n-1}(f(x_{i+1})-f(x_i))\int_a^b\chi_{A_i}=\sum_{i=0}^{n-1}(f(x_{i+1})-f(x_i))(x_{i+1}-x_i)\\ &=&\frac{b-a}{n}\sum_{i=0}^{n-1}(f(x_{i+1})-f(x_i))=\frac{1}{n}(b-a)(f(b)-f(a))\\ &\le&\frac{1}{N_\varepsilon}(b-a)(f(b)-f(a))\le \varepsilon. \end{eqnarray}

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    $\begingroup$ You didn't have to include so many details but thank you sincerely $\endgroup$
    – Optional
    Jan 22, 2013 at 11:19
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Let $P$ be a finite partition of $[f(a),f(b)]$ into subintervals of length at most $\epsilon$. For $A\in P$, write $m_A<M_A$ for the endpoints of $A$ and define $$g=\sum_{A\in P}m_A\chi_{f^{-1}(A)}\qquad\mbox{and}\qquad h=\sum_{A\in P} M_A\chi_{f^{-1}(A)}.$$

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  • $\begingroup$ Thanks for the answer. Can you suggest such a partition $P$ (that is preferably simple)? $\endgroup$
    – Optional
    Jan 21, 2013 at 19:19
  • $\begingroup$ Sure. Pick $n\ge1$ large enough that $\delta:=(f(b)-f(a))/n\le \epsilon$, and take $P=\{A_1,A_2,\dots,A_n\}$ where $A_1=[f(a),f(a)+\delta)$, $A_2=[f(a)+\delta,f(a)+2\delta)$, ... , $A_{n-1}=[f(a)+(n-2)\delta,f(a)+(n-1)\delta)$, $A_n=[f(a)+(n-1)\delta,f(a)+n\delta]$. $\endgroup$
    – user108903
    Jan 21, 2013 at 19:23
  • $\begingroup$ Yeah, thats what I had in mind. The whole process of partitioning the range instead of the domain is a little complicated. Can we partition the domain instead? $\endgroup$
    – Optional
    Jan 21, 2013 at 19:28
  • $\begingroup$ Well you are partitioning the domain (even into intervals), but you're using a partition of the range to find the partition of the domain. $\endgroup$
    – user108903
    Jan 21, 2013 at 19:29
  • $\begingroup$ But how do we know $f^{-1}(A)$ is an interval and so the two functions are integrable? $\endgroup$
    – Optional
    Jan 21, 2013 at 19:31

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