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I have difficulty interpreting the following question:

If the simple function $\phi$ in $M^{+}(X,\mathcal{A})$ has the (not necessarily standard) representation $$\phi=\displaystyle\sum_{k=1}^{n}b_k\chi_{F_k}$$ where $b_k \in \mathbb{R}$ and $F_k \in \mathcal{A}$, show that

$$\int \phi \mathrm{d}\mu = \displaystyle\sum_{k=1}^{n}b_k\mu({F_k}).$$

We know that a simple function $\phi$ is of $\bf{standard \;\;representation}$ if

$$\phi=\displaystyle\sum_{k=1}^{n}a_k\chi_{E_k}$$ where $a_k$ are distinct and $E_j$ are disjoint nonempty subsets of $X$. Moreover, its integral is of the form $$\int \phi \mathrm{d}\mu = \displaystyle\sum_{k=1}^{n}a_k\mu({E_k}).$$

Is the question stating that the integral will hold for $\phi$ where $b_k$ are not distinct and where $F_k$ are not disjoint? I would like to say that this will hold when $b_k$ are not distinct, but I do not think this will hold if $F_k$ are not disjoint sets.

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See Lemma 11.4 in Aliprantis-Border, p.405 - they give similar proof as Davide. (This page does not display for me in Google books, but maybe you'll have better luck.) – Martin Sleziak Apr 1 '12 at 10:07
(Originally an answer; transforming into comment now that I have enough reputation.) It also holds when the $F_k$ are not disjoint; of course in that case we must have $a_k \neq b_k$. For example $\phi = \chi_{[0,1]}$ can have non-standard representations $\chi_{[0,1/2)} + \chi_{[1/2, 1]}$ (disjoint $F_k$ but non-distinct $b_k$'s) and $\frac{1}{3}\chi_{[0,1]} + \frac{2}{3}\chi_{[0,1]}$ (distinct $b_k$'s but overlapping $F_k$). – lazyhaze Apr 1 '12 at 20:05
up vote 3 down vote accepted

In fact, it holds even if the $F_j$ are not disjoint and even if the $b_j$ are not distinct. First, we show that if $\{S_1,\ldots,S_N\}$ is a partition of $X$ with $S_j$ measurable and $(b_1,\ldots,b_N)\in\mathbb R^N$, not necessary disjoint, we have $$\int \sum_{j=1}^Nb_j\chi_{S_j}d\mu=\sum_{j=1}^Nb_j\mu(S_j).$$ Indeed, let $\sum_{k=1}^na_k\chi_{T_k}$ the standard representation of $\sum_{j=1}^Nb_j\chi_{S_j}$. We have \begin{align*} \sum_{j=1}^Nb_j\mu(S_j)&=\sum_{j=1}^Nb_j\mu(S_j\cap\bigcup_{k=1}^nT_k)\\ &=\sum_{j=1}^N\sum_{k=1}^nb_j\mu(S_j\cap T_k)\\ &=\sum_{j=1}^N\sum_{k=1}^na_k\mu(S_j\cap T_k)\\ &=\sum_{k=1}^na_k\mu(T_k), \end{align*} since if $S_j\cap T_k$ is not empty, we take $x$ in it and its value for the two functions is respectively $b_j$ and $a_k$.

Now we can show that for two simple functions, the integral of the sum is the sum of the integral. Let $f=\sum_{j=1}^{n_1}c_j^{(1)}\chi_{S_j^{(1)}}$ and $g=\sum_{k=1}^{n_2}c_k^{(2)}\chi_{S_j^{(2)}}$ written in their standard representation. Let $I_1:=\{1,\ldots,n_1\}$ and $I_2:=\{1,\ldots,n_2\}$. Then $f+g$ can be written as $$\sum_{j\in I_1}\sum_{k\in I_2}(c_j^{(1)}+c_k^{(2)}) \chi_{S_j^{(1)}}\chi_{S_k^{(2)}}.$$ It may not be the standard representation but it doesn't matter thanks to the previous result. We have \begin{align*} \int (f+g)d\mu&=\sum_{j\in I_1}\sum_{k\in I_2}(c_j^{(1)}+c_k^{(2)})\mu(S_j^{(1)}\cap S_k^{(2)})\\ &=\sum_{j\in I_1}\sum_{k\in I_2}c_j^{(1)}\mu(S_j^{(1)}\cap S_k^{(2)})+\sum_{j\in I_1}\sum_{k\in I_2}c_k^{(2)}\mu(S_j^{(1)}\cap S_k^{(2)})\\ &=\sum_{j\in I_1}c_j^{(1)}\mu(S_j^{(1)})+\sum_{k\in I_2}c_k^{(2)}\mu(S_k^{(2)}), \end{align*} since $\{S_j^{(1)}\}_{j=1}^{n_1}$ and $\{S_k^{(2)}\}_{k=1}^{n_2}$ are partitions of $X$.

Now to see the result, just use an induction.

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