(related to Proposition: Riemann Integral for Step Functions)
Let \(\phi\in T[a,b]\) be a step function with respect to the partition \(a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b\). We want to show that the definition of the Riemann integral \[\int_a^b\phi(x)dx:=\sum_{i=1}^nc_i(x_i-x_{i-1})\] does not depend on the specific choice of the partition \(a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b\), i.e. for any such partition, the above sum takes the same value.
Assume
\[\begin{array}{rcl} S&:=&a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b\\ T&:=&a=t_0 < t_1 < \ldots < t_{m-1} < x_m=b\\ \end{array} \]
are two different partitions of the closed interval \([a,b]\) such that
\[\begin{array}{rcl} \phi (x)=c_i\quad&\quad&\text{for all } x\in(x_{i-1},x_i),\quad i=1,\ldots,n,\\ \phi (x)=d_j\quad&\quad&\text{for all } x\in(t_{j-1},t_j),\quad j=1,\ldots,m.\\ \end{array} \]
and assume that \(S\subset T\). This means that for each \(i\) there is an index \(j_i\) such that \(x_i=t_{j_i}\). Then we have
\[x_{i-1}=t_{j_{i-1}} < t_{j_{i-1}+ 1} < t_{j_{i-1}+ 2} < \ldots < t_{j_i}=x_i\quad\quad\text{for }1\le i\le n.\] and \[c_i=d_j\quad\quad\text{for }j_{i-1} < i \le j_i,\quad 1\le i\le n.\] It follows \[\sum_{i=1}^{n}c_i(x_i-x_{i-1})=\sum_{i=1}^n\sum_{k=j_{i-1}+1}^{j_i}d_k(t_k-t_{k-1}).\]
In other words, the Riemann integrals with respect to the partitions \(S\) and \(T\) equal each other.
Now assume arbitrary partitions \(S\) and \(T\) and that \(U=S\cup T\). Then we have \(S\subset U\) and \(T\subset U\). According to case 1, the Riemann integrals with respect to the partitions \(S\) and \(U\) equal each other, as well as the Riemann integrals with respect to the partitions \(T\) and \(U\) equal each other. Thus, the Riemann integrals with respect to the partitions \(S\) and \(T\) also equal each other. In other words, the Riemann integral of the step function does not depend on the specific choice of the partition.