# Integral of simple functions in standard and non-standard representation

Some definitions

Let $(X,\mathbb X,\mu)$ be a measure space. A real-valued function is simple if it has only a finite number of values. A simple $\mathbb X$-measurable function $\varphi$ can be represented in the form $$\varphi=\sum_{j=1}^n a_n\chi_{E_j}$$ where $a_j\in\mathbb R$ and $\chi_{E_j}$ is the characteristic function of a set $E_j\in\mathbb X$. If we add the restriction that the $a_j$ be distinct and the $E_j$ form a partition of X, then the representation is unique and is called the standard representation of $\varphi$.

If $\varphi$ is a simple function in $M^+(X,\mathbb X)$ with the standard representation above, we define the integral of $\varphi$ with respect to $\mu$ to be the extended real number $$\int\varphi\,d\mu=\sum_{j=1}^n a_j\mu(E_j)$$

My question

If the simple function $\varphi\in M^+(X,\mathbb X)$ has the (not necessarily standard) representation $$\varphi=\sum_{k=1}^m b_k\chi_{F_k}$$ where $b_k\in\mathbb R$ and $F_k\in\mathbb X$, it can be shown that $$\int\varphi\,d\mu = \sum_{k=1}^m b_k\,\mu(F_k).$$ My problem is that I cannot find a clean yet rigorous step-by-step proof of that result.

My idea is to rewrite the function $\varphi$ as $\sum_{k=1}^n a_k\chi_{\phi^{-1}(a_k)}$ where $a_k$ is the sum of some $b_k$ terms and $\phi^{-1}(a_k)$ is the union of intersections of some $F_k$ terms. After some manipulations I note that I can put back together all the "pieces" and find $\sum_{k=1}^m b_k\,\mu(F_k)$. Unfortunately, that very last passage is left to the reader. Is there a way to make all the process explicit and yet clean and easy to follow?

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By non-standard you mean the non-standard analysis, like $m$ being an infinite number? – Ilya Jan 19 '13 at 18:21
@Ilya. I believe by standard he means that the sets $F_{k}$ make a partition of $X$. – T. Eskin Jan 19 '13 at 18:25
Yes, I meant what Thomas said. – Kiuhnm Jan 19 '13 at 19:34
does $\varphi\in \Bbb M^+$ mean that it is non-negative? – Ilya Jan 19 '13 at 20:09
Yes, it means that it is a non-negative measurable function. – Kiuhnm Jan 19 '13 at 20:16

Suppose $m$ is finite. Using the linearity of the integral, we have $$\int \left(\sum_{k=1}^m b_k1_{F_k}(x)\right)\mu(\mathrm dx) = \sum_{k=1}^m b_k\int1_{F_k}(x)\mu(\mathrm dx) = \sum_{k=1}^mb_k\mu(F_k)$$ regardless of the shape of the collection $F_k$.
If you don't want to use linearity, note that given some finite collection $\{F_k\}_{k=1}^m$ of measurable sets there is a unique coarsest partition $\mathscr G = \{G_i\}_{i=1}^n$ such that $F_k$ are unions of some elements in $\mathscr G$. Let $g:\{1,\dots,n\}\to2^{\{1,\dots,m\}}$ be the index function uniquely defined by $$k\in g(i)\quad\Leftrightarrow\quad G_i\subset F_k$$ for any $k\in\{1,\dots,m\}$ and any $i\in \{1,\dots,n\}$. Furthermore, note that for any $k\in \{1,\dots,m\}$ the inverse of $g$ satisfies $$\{i:k\in g(i)\} = \{i:G_i\subset F_k\}$$ is partition of $F_k$ in $\mathscr G$. In particular, $\mu(F_k) = \sum_{i:k\in g(i)}\mu(G_i).$ Then we have: $$\varphi(x) = \sum_{k=1}^m b_k1_{F_k}(x) = \sum_{i=1}^n\left(\sum_{k\in g(i)}b_k\right)1_{G_i}(x)$$ where the first equality is the definition of $\varphi$ and the latter function is standard simple one. Thus $$\int\varphi\;\mathrm d\mu = \sum_{i=1}^n\left(\sum_{k\in g(i)}b_k\right)\mu(G_i) = \sum_{k=1}^n b_k\left(\sum_{i:k\in g(i)}\mu(G_i)\right) = \sum_{k=1}^n b_k\mu(F_k)$$ where we passed from the summation over $G_i$ to the summation over $b_k$.