Intro: I must take a small detour here which is only relevant if you do not know the book itself and care about my background. I am working with Königsberger Analysis I (can be found on Springerlink). Currently I am in Chapter 11 which focusses on integration.

Königsberger takes the following approach to introduce Integral Calculus:

  1. He defines step functions $\varphi: [a,b] \to \mathbb{C}$ such that for all $x \in (x_{k-1},x_k)$ the step function $\varphi$ is constant $c_k$ and then defines the Integral of step functions: $$\int_a^b \varphi(x)dx := \sum_{k=1}^n c_k \Delta x_k $$
  2. He defines regulated functions $f:I \to \mathbb{C}$ on an Intervall $I$ such that for all $x \in (a.b)$ the left-sided limit and the right-sided limit exists.
  3. He introduces the reader to the 'approximation theorem'

Approximation theorem: A function $f$ on a compact Intervall $[a,b]$ is a regulated function if and only if for all $\epsilon > 0$ there exists a step function $\varphi $ such that $|f(x)-\varphi(x)| \leq \epsilon$ for all $x \in [a,b]$

  1. Corollary to the theorem he shows that for regulated functions $f$ the following limit always exists and defines this as the integral of $f$ over $[a,b]$ $$ \int_a^b f(x) dx := \lim_{n \to \infty} \int_a^b \varphi_n(x)dx $$

My problem: I believe to understand the above topics and associated proofs 'quite well'. However in the last section of the chapter Königsberger tries to make the connection of regulated functions to the Riemann Sum with the following theorem:

Theorem: Let $f: [a,b] \to \mathbb{C}$ be a regulated function. Then for all $\epsilon > 0$ there exists a $\delta >0$ such that for all partitions $Z$ of $[a,b]$ with $\max \Delta x_k \leq \delta$ and arbitrary $\xi_k \in [x_{k-1},x_k]$ the following holds: $$ \left| \sum_{k=1}^n f( \xi_k) \Delta x_k - \int_a^b f(x) dx \right| \leq \epsilon $$

I struggle with the first part of the proof. (Page 216)

Proof: Königsberger suggest to first verify the theorem for step functions rather than regulated functions and then use the approximation theorem.

He says that showing it for step functions should be done by induction after the number of "jump points" $m$ (translator suggests saltus and jump discontinuity) and that it is very easy.

I only care about the first induction steps $m=0$ and $m=1$ which Königsberger describes as trivial, for $m=1$ he mentions to choose $\delta:= \epsilon / 4 \| \varphi\|$ and my problem is I really don't see how he obtains these things. After this I can complete the proof on my own because he's very thorough from that point on.

  • $\begingroup$ Just wondering, should it be $\Bbb R$ instead of $\Bbb C$? $\endgroup$ – Mark Fantini Jul 14 '14 at 23:02
  • $\begingroup$ Thanks for your comment @Fantini. In Königsberger's Analysis he always discusses functions of the form $f: I \to \mathbb{C}$. In an earlier chapter (C9, Differentiation) he mentions that he always takes $I \subset \mathbb{R}$ because for $I \subset \mathbb{C}$ we'd need complex analysis methods. He then mentions that all his theorems work for complex valued functions. $\endgroup$ – Spaced Jul 14 '14 at 23:10

$m=0$: For a constant step function $\varphi:[a,b]\rightarrow \mathbb{C}$, where $\exists c\in \mathbb{C}\space\forall x\in[a,b]: \varphi(x)=c.$ We have: $$ \left| \sum_{k=1}^n \varphi( \xi_k) \Delta x_k - \int_a^b \varphi(x)\space dx \right|$$ $$= \left| c \sum_{k=1}^n \Delta x_k -c(b-a) \right|=0$$ since the sum collapses to $(b-a).$

$m=1:$ For a step function $\varphi:[a,b]\rightarrow \mathbb{C}$ with one jump point $y \in [a,b]$ : $\exists c_{1},c_{2}\in \mathbb{C}\space\forall x\in[a,b]: (x<y \Rightarrow \varphi(x)=c_{1} \space \wedge \space x>y \Rightarrow \varphi(x)=c_{2}).$ Now the $y$ will lie in one of the $\Delta x_k$, say in $\Delta x_j$ i.e. between $x_{j-1}$ and $x_{j}$. $$ \left| \sum_{k=1}^n \varphi( \xi_k) \Delta x_k - \int_a^b \varphi(x) \space dx \right|$$ $$=\left| \sum_{k=1}^n \varphi( \xi_k) \Delta x_k - \sum_{k=1}^n\int_{x_k}^{x_{k+1}} \varphi(x) \space dx \right|$$ $$=\left| \sum_{k=1}^n \left(\varphi( \xi_k) \Delta x_k - \int_{x_{k-1}}^{x_{k}} \varphi(x) \space dx \right) \right|$$ For each $k$ except for $k=j$ this will be $0$. $$=\left| \varphi(\xi_j)(x_{j-1}-x_{j})\space- \space c_{1}(y-x_{j-1})\space-\space c_{2} (x_{j}-y)\right|$$ $$=\left| (\varphi(\xi_j)-c_{1})(y-x_{j-1})\space+ \space (\varphi(\xi_j)-c_{2})(x_{j}-y)\right|$$ $$\leq \left| (\varphi(\xi_j)-c_{1})\right|(y-x_{j-1})\space+ \space \left|(\varphi(\xi_j)-c_{2})\right|(x_{j}-y)$$ $$\leq (\left| \varphi(\xi_j)\right|+\left|c_{1}\right|)(y-x_{j-1})\space+ \space (\left|\varphi(\xi_j)\right|+\left|c_{2}\right|)(x_{j}-y)$$ $$\leq (\left| \varphi(\xi_j)\right|+\left|c_{1}\right|+\left|\varphi(\xi_j)\right|+\left|c_{2}\right|)\Delta x_j$$ $$\leq (\left| \varphi(\xi_j)\right|+\left|c_{1}\right|+\left|\varphi(\xi_j)\right|+\left|c_{2}\right|)\space\frac{\epsilon} {4 \| \varphi\|} \leq \epsilon.$$

  • $\begingroup$ Great answer, thanks for your detail rich work, I appreciate it a lot. $\endgroup$ – Spaced Jul 15 '14 at 12:35
  • $\begingroup$ You are very welcome! $\endgroup$ – Vincent Pfenninger Jul 15 '14 at 13:14

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