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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
up vote 5 down vote accepted

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 $b_k$.

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We do not need to use the dominated convergence theorem, nor the linearity of the integral. The simple functions I am talking about take just a finite number of real values (i.e. their image is finite). – Kiuhnm Jan 19 '13 at 19:46
@Kiuhnm: if their image is finite, doesn't the first part of my reply answers your question? – Ilya Jan 19 '13 at 19:48
I would like to prove it without using the linearity of the integral, i.e. just by using the definition of the integral of simple functions (see the latest revision of my question). To say it another way, this question is more about sum and set manipulations than about measure theory. – Kiuhnm Jan 19 '13 at 20:11
@Kiuhnm: I guess, now it answers your question better. – Ilya Jan 19 '13 at 20:34
Thank you; that's exactly what I was looking for. I tried to edit your post but I couldn't because I wanted to change just two characters. The limits of the last and the third from last summation might be wrong. – Kiuhnm Jan 20 '13 at 1:41

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