First a Definition:
Let $I$ be a bounded interval, let $P$ be a partition of $I$ Let $f:I \rightarrow\mathbb R$ which is piecewise constant with respect to $P$. Then we define the piece wise constant integral $p.c.\int_{[P]}f$ with respect to the partition $P$ by the formula: $p.c.\int_{[P]}f = \sum_{J\in P}c_J |J|$ where for each $J$ in $P$, we let $c_J$ be the constant value of $f$ on $J$
Now he then goes on to state a proposition
Let $I$ be a bounded interval, and let $f:I \rightarrow\mathbb R$ be a function. Suppose $P$ and $P'$ are two partitions of $I$ in which $f$ is piecewise constant with respect to. Then we have $p.c.\int_{[P]}f = p.c.\int_{[P']}f$
Ive been stuck on one part of Taos solution for such a long time now posted here (problem 6): https://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/HW9.pdf
At the end of the solution how on earth would we prove this equality: $\sum_{J\in P}c_J \sum_{I\in P', I\subset J}|I|=p.c.\int_{[P']}f$ All he does is simply say "and this last sum is $p.c.\int_{[P']}f$" But in the verification process I want to prove this and don't know how.
for added clarity some definitions from the book:
Definition (Summation over finite sets) Let $X$ be a finite set with $n$ elements let $f:X \rightarrow\mathbb R$ be a function. We define $\displaystyle \sum_{x \in X} f(x)$ as follows. Select any bijection $g$ from {${i \in N : 1 \leq i \leq n}$} to $X$ we then define: $\displaystyle \sum_{x \in X} f(x)$ $:=$ $\sum_{i=1}^{n} f(g(i))$
Definition (Partions) Let $I$ be a bounded interval. A Partition of $I$ is a finite set P of bounded intervals contained in $I$, such that every $x$ in $I$ lies in exactly one of the bounded intervals $J$ in P