Exercise 1.1.21 (Basic properties of the piecewise constant integral)

In the previous proof we have demonstrated that if $P$ and $P^{\prime }$ are partitions of $[a,b]$ with respect to which $f$ and $g$ are piecewise constant respectively, then both are pieciwise constant with respect to the common refinement $\mathit{P\#}{P}^{\prime }$:

where we used the indicator $1_{\text{{⋅}}}:[a,b]\to \{0,1\}$ function to represent $f$ in an obvious manner.

(i)

The piecewise constant integral of the linear combination above is $$\begin{array}{llll}\hfill \text{pc}{\int \; <!--\; nolimits\; -->}_{[a,b]}\mathit{cf}+g& =\sum _{I\in \mathit{P\#}{P}^{\prime }}(c\cdot c_I+d_I)\cdot m(I)& \hfill & \\ \hfill & =c\sum _{I\in \mathit{P\#}{P}^{\prime }}{c}_I\cdot m(I)+\sum _{I\in \mathit{P\#}{P}^{\prime }}{d}_I\cdot m(I)& \hfill & \\ \hfill & =c\cdot \text{pc}{\int \; <!--\; nolimits\; -->}_{[a,b]}f+\text{pc}{\int \; <!--\; nolimits\; -->}_{[a,b]}g.& \hfill & \end{array}$$

(ii)

The monotonicity can be proven by observing that from $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in [a,b]:f(x)=\sum _{I\in \mathit{P\#}{P}^{\prime }}{c}_I\cdot 1_{\{x\in I\}}(x)\le \sum _{I\in \mathit{P\#}{P}^{\prime }}{d}_I\cdot 1_{\{x\in I\}}(x)=g(x)$$

follows that

$$\forall <!--\; FUNCTION\; APPLICATION\; -->I\in \mathit{P\#}{P}^{\prime }:c_I\le d_I.$$

The monotonicity is then obvious, by monotonicity of summation operation

$$\text{pc}{\int \; <!--\; nolimits\; -->}_{[a,b]}f(x)=\sum _{I\in \mathit{P\#}{P}^{\prime }}{c}_I\cdot m(I)\le \sum _{I\in \mathit{P\#}{P}^{\prime }}{d}_I\cdot m(I)=\text{pc}{\int \; <!--\; nolimits\; -->}_{[a,b]}g(x).$$

(iii)

Since $E$ is elementary, we can represent it as a union of intervals $E=I_1\cup \dots I_k$. But the complement $[a,b]∕E$ is also elementary; thus, $[a,b]∕E=I_{k+1},\dots ,I_n$. By definition $1_E$ is piecewise constant with respect to $P:=I_1\dots I_n$, and we have $$\begin{array}{llll}\hfill \text{pc}{\int \; <!--\; nolimits\; -->}_{[a,b]}{1}_E& =\sum _{I\in P}{c}_{I}m(I)& \hfill & \\ \hfill & =\sum _{i=1}^k{c}_{I_i}m(I_i)+\sum _{i=k+1}^n{c}_{I_i}m(I_i)& \hfill & \\ \hfill & =\sum _{i=1}^{k}1\cdot m(I_i)+\sum _{i=k+1}^{n}0\cdot m(I_i)& \hfill & \\ \hfill & =m\left(\bigcup _{i=1}^k{I}_i\right)=m(E)& \hfill & \end{array}$$

where in the last step we used the additivity property of the elementary measure.