Exercise 1.1.20 (Piecewise constant integral)

Question

Let $[a,b]$ be an interval, let $f:[a,b]	o\mathbb{R}$ be a piecewise constant function. Let $I_1,\ldots,I_n$ be the partition of $[a,b]$ into finitely many intervals, such that $f$ is equal to a constant $c_i$ on each of the intervals $I_i$. Show that the expression
$$\sum_{i=1}^{n}c_i m(I)$$
is independent of the choice of partition used to demonstrate the piecewise constant nature of $f$.

Elu Thingol · Accepted Answer

Let $P = \{I_1,\ldots,I_n\}$ and $P' =\{ J_1,\ldots,J_n\}$ be partitions of $[a,b]$ into intervals such that $f$ is piecewise constant with respect to both of them $\forall I \in P: f(I) = c_I$ and $\forall J \in P': f(J) = c_J$. The idea of the proof is to compare both partitions with the common refinement $P\# P' = \left\{ I \cap J : I \in P, J \in P'\right\}$. This is possible, since $f$ is piecewise constant with respect to $P\# P'$. (For any $x \in I \cap J$ we have $f(x) = c_I = c_J$ by definition.) Notice that 
\begin{align*}
\sum_{I \in P} c_I \cdot m(I) &= \sum_{I \in P} c_I \cdot m\left( \bigcup\left\{ I \cap J: J \in P'\right\} \right)\[5pt] 
&= \sum_{I \in P} c_I \cdot \sum_{J \in P'} m\left(I \cap J\right)\[5pt]
&= \sum_{I \in P} \sum_{J \in P'} c_I \cdot m\left(I \cap J\right)\[5pt]
&= \sum_{J \in P'} \sum_{I \in P} c_J \cdot m\left(I \cap J\right)\[5pt]
&= \sum_{J \in P'} c_J \cdot \sum_{I \in P}  m\left(I \cap J\right)\[5pt]
&= \sum_{J \in P'} c_J \cdot m\left( \bigcup\left\{ I \cap J: I \in P\right\} \right) \[5pt] 
&= \sum_{J \in P'} c_J \cdot m\left(J \right)
\end{align*}
where we have used the fact that $\bigcup_{J \in P'} I \cap J = I$ is a disjoint partition of an interval $I$.

Exercise 1.1.20 (Piecewise constant integral)

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Exercise 1.1.20 (Piecewise constant integral)

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