Is Jost's version of the Lebesgue integral well-defined? In Postmodern Analysis by Jurgen Jost, the Lebesgue integral of a step function is defined as follows:
Suppose we have a step function $t:W\subset\mathbb{R}^d\to \mathbb{R}$ defined on a cube $W\subset\mathbb{R}^d$ given by 
$$ t = \sum_j c_j \mathbb{1}_{W_j} \quad (*)$$
where $c_j$ is a real number and $W_j$ is a cube with side-length $\ell_j>0$ for $j=1,2,...k$ and if $j\neq j'$, then $\text{int}(W_j)\cap \text{int}(W_{j'})=\varnothing$ and $\bigcup_j W_j = W$ (i.e. the cubes $\{W_j\}$ are almost disjoint and partition W, and $t$ is constant on the interior of each cube). Then we define the integral of $t$ to be
$$ \int_{\mathbb{R}^d} t = \sum_{j=1}^k c_j\ell_j^d. $$
I want to show that this definition is indepenedent of the collection of cubes $\{W_j\}_{j=1}^k$, as it is obvious that the function $t$ can have many representations of the form $(*)$. Although it is "geometrically obvious in my minds eye", I am having difficulty proving this fact. Any help is appreciated. 
What I have tried: Given two collections of cubes $\{W_j\}$ and $\{B_i\}$  where we can write 
$$ t = \sum_i a_i\mathbb{1}_{B_i} = \sum_j c_j\mathbb{1}_{W_j} $$
we need to show 
$$ \sum_i a_i k^d_i = \sum_j c_j \ell_j^d $$
where $k_i$ is the side length of the cube $B_i$ and $\ell_j$ is the side length of the cube $W_j$.
I am not sure how to formally prove this equality, in general. The other idea I have is to find some kind of canonical representation of $(*)$.
I feel as though I am missing something, because the author says this fact is obvious and gives no proof or justification. I don't want to rely on faith.
 A: (I use the superscript as just another index.) In dimension 1, what you can do is you form a list $A=\{a_1,\dots,a_n\}, a_i< a_{i+1},$ of the endpoints  of intervals in both collections $\mathcal W = \{W_i\}_i$ and $\mathcal B= \{B_j\}_j$ (note $|A|<\infty$), and then consider the collection of intervals
$\mathcal C = \{ [a_n,a_{n+1}) : n=1,\dots,n-1 \}$. Up to points in the boundary which we safely ignore, every set in $\mathcal W$ and $\mathcal B$ can be created as a union of intervals in $\mathcal C$. 
In dimension 2, you can form an increasing finite list $A^1=\{a^1_1,\dots,a^1_{n^1}\}$ of all possible $x$ coordinates of endpoints of both collections $\mathcal {W,B}$. Do the same for $y$-coordinates $A^2 =\{a^2_1,\dots,a^2_{n^2}\}$. Now form the collection of cubes
$$\mathcal C = \left\{ [a^1_{k^1},a^1_{k^1+1}) \times [a^2_{k^2},a^2_{k^2+1}) : \substack{k^1=1,\dots,n^1-1\\ k^2 = 1,\dots,n^2-1}\right\}$$

It should be clear how the construction goes for dimension $3,4,\dots,n$， and you should be able to formalise with induction.
A: Assume we have collections of cubes $\{W_j\}_{j=1}^k$ and $\{B_i\}_{i=1}^m$ where each collection is almost disjoint and covers the cube $W$. Also assume that $t:W\subset\mathbb{R}^d\to \mathbb{R}$ is constant on the open cubes $W_j^{\mathrm{o}}$ and $B_i^{\,\mathrm{o}}$ for $j=1,2,...,k$ and $i=1,2,...,m$. 
We want to show that 
$$ \sum_i t(B_i^{\,\mathrm{o}})\text{vol}(B_i^{\,\mathrm{o}}) = \sum_j t(W_j^{\mathrm{o}})\text{vol}(W_j^{\mathrm{o}}) \qquad (i)$$
where we slightly abuse notation to write $t(B_i^{\,\mathrm{o}})$ for the constant value of $t$ on the open cube $B_i^{\,\mathrm{o}}$ and similarly for $t(W_j^{\mathrm{o}})$.
Notice that the sets of values 
$$ \{ t(W_1^{\mathrm{o}}), t(W_2^{\mathrm{o}}),...,t(W_k^{\mathrm{o}}) \} \quad\text{and}\quad \{ t(B_1^{\,\mathrm{o}}), t(B_2^{\,\mathrm{o}}),...,t(B_m^{\,\mathrm{o}}) \} $$
must match, so after discarding repetitions, we may write these sets as
$$ \{t(W_j^{\mathrm{o}})\}_{j=1}^k=\{t(B_i^{\,\mathrm{o}})\}_{i=1}^m= \{c_1,c_2,...,c_n\} $$
where all of the $c_r$'s are distinct real numbers. Think of $\{c_r\}_{r=1}^n$ as the collection of distinct values that the function $t$ realizes (this is true up to a set of measure zero — can only be false for a point on the boundary of some cube in the collections).  
Now define index sets
$$ J_r:=\{j:W_j^{\mathrm{o}}\subseteq t^{-1}(\{c_r\})\} $$
$$ I_r:=\{i:B_i^{\,\mathrm{o}}\subseteq t^{-1}(\{c_r\})\} $$
for each $r=1,2,...,n$ so that $(i)$ is equivalent to
$$ \sum_{r=1}^n c_r \sum_{i\in I_r} \text{vol}(B_i^{\,\mathrm{o}}) = \sum_{r=1}^n c_r \sum_{j\in J_r} \text{vol}(W_j^{\mathrm{o}}) \qquad (ii)$$
because we must have $\bigcup_r J_r = \{1,2,...,k\}$ and $\bigcup_r I_r = \{1,2,...,m\}$. 
However $(ii)$ is equivalent to 
$$ \sum_{r=1}^n c_r \left( \sum_{i\in I_r} \text{vol}(B_i^{\,\mathrm{o}}) - \sum_{j\in J_r} \text{vol}(W_j^{\mathrm{o}}) \right) = 0. \qquad (iii) $$
I claim that for each $r\in \{1,2,...,n\}$ we have 
$$ \sum_{i\in I_r} \text{vol}(B_i^{\,\mathrm{o}}) - \sum_{j\in J_r} \text{vol}(W_j^{\mathrm{o}}) = 0. \qquad (\star) $$
In words, $(\star)$ says "whether we partition the graph of $t$ with one collection of cubes or another, the volume of points mapping to a particular value must be the same". This conforms with 'the picture in my mind'. 
Fix $r\in\{1,2,...,n\}$ and let 
$$ C := (\partial B_1)\cup\cdots\cup(\partial B_m)\cup(\partial W_1)\cup\cdots \cup(\partial W_k) $$
be the set consisting of all boundary components of the cubes in both collections. As the boundary of a (d-dimensional) cube has (d-dimensional) measure zero, we know that $C$ is a set of measure zero. Now we claim 
$$ \{x\in B_i^{\,\mathrm{o}}: i\in I_r\}\setminus C = \{ y\in W_j^{\mathrm{o}}:j\in J_r \}\setminus C \quad (\star')$$
which will prove $(\star)$ by simply computing the volume of each side and subtracting (each side is a finite union of disjoint open cubes). 
Notice that each collection of open cubes $\{W_j^{\mathrm{o}}\}_{j=1}^k$ and $\{B_i^{\,\mathrm{o}}\}_{i=1}^m$ covers $W\setminus C$. Thus
$$ t^{-1}(\{c_r\})\setminus C = \{x\in W\setminus C : t(x) = c_r\} = \{ x \in W_j^{\mathrm{o}} : t(W_j^{\mathrm{o}}) = c_r \}\setminus C = \{x\in W_j^{\mathrm{o}}:j\in J_r\}\setminus C $$
and similarly, we have 
$$ t^{-1}(\{c_r\})\setminus C = \{x\in W\setminus C : t(x) = \{x\in B_i^{\,\mathrm{o}}:i\in I_r\}\setminus C $$.
Thus $(\star')$ has been proven, justifying our claim $(\star)$, so we see that $(iii)$ is true, and hence $(i)$ is true, which completes the proof of the general case. 
