Riemann Integral for Step Functions

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However, to be sure that the integral is well-defined we need additionally that it is independent of the special choice of a decomposition.

Proof: Let \[\begin{aligned} Z_1:\quad&a=x_0<x_1<x_2<\ldots<x_{n-1}<x_n=b,\\ Z_2:\quad&a=y_0<y_1<y_2<\ldots<y_{m-1}<y_m=b \end{aligned}\] be two decompositions of \([a,b]\) such that \(\phi\) is constant on \((x_{i-1},x_i)\) and \(\phi\) is constant on \((y_{j-1},y_j)\) for all \(i=1,\ldots,n\), \(j=1,\ldots,m\). We distinguish between two cases:

1st Case: \(Z_2\) is a refinement of \(Z_1\). This means that for all \(i\in\{1,\ldots,n\}\) there exists some \(j(i)\in\{1,\ldots,m\}\) with \(x_i=y_{j(i)}\). Then for \(i\) holds \[x_i=y_{j(i)}<y_{j(i)+1}<\ldots<y_{j(i+1)}=x_{i+1}\] and \(\phi(x)=d_k=c_i\) for \(x\in~(y_{k-1},y_k)\), \(k\in\{j(i-1)+1,\ldots,j(i)\}\). This implies \[\begin{aligned} \sum_{j=1}^{m}d_j(y_{j}-y_{j-1})=&\sum_{i=1}^{n}\sum_{k=j(i-1)+1}^{j(i)}c_i(y_{k}-y_{k-1})\\ =&\sum_{i=1}^{n}c_i\sum_{k=j(i-1)+1}^{j(i)}(y_{k}-y_{k-1})\\ =&\sum_{i=1}^{n}c_i(y_{j(i)}-y_{j(i-1)})=\sum_{i=1}^{n}c_i(x_i-x_{i-1}). \end{aligned}\]

2nd Case: There exists a common refinement \(Z_3\) of \(Z_1\) and \(Z_2\), i.e., \(Z_3=Z_1\cup Z_2\). Then we have (by using the 1st case) \[\sum_{Z_1}...\;=\sum_{Z_3}\ldots\;=\sum_{Z_2}\ldots\;.\] \(\Box\)