Define a measure in the following way...

Question

Suppose $X$ is a set, $S$ is the $\sigma$-algebra of all subsets of $X$, and $\omega: X \to [0, \infty]$ is a function. Define a measure $\mu$ on $(X, S)$ by

$$ \mu(E) = \sum_{x \in E} w(x)$$

for $E \subset X$. Prove that if $f: X \to [0, \infty]$ is a function, then

$$\int f \, \mathrm{d}\mu = \sum_{x\in X}w(x)f(x), $$

where the infinite sums above are defined as the supremum of all sums over finite subsets of $X$.

I was thinking to show it first works for indicator functions, then simple functions by linearity, and then taking simple functions converging to f from below and using MCT to get the final result. We've used a similar idea on other questions but I'm getting hung up on the details here. Thank you in advance for any help.

Answer 1

$\textbf{Lemma}$Let $I=\bigcup_{k \in S}J_k$ where $S$ is a nonempty set of indices. and $J_k$'s are nonempty and pairwise disjoint.Then for every $w:I \to [0,+\infty]$ we have that $\sum_{i \in I}w(i)=\sum_{k \in S}(\sum_{i \in J_k}w(i))$

Now let $\phi=\sum_{j=1}^nk_j1_{E_j}$ a simple functions with its standard representation, i.e, all $k_i$ are distinct and the sets $E_j$ are pairwise disjoint.

Then $$\int_X \phi d\mu=\sum_{j=1}^nk_j \mu(E_j)=\sum_{j=1}^nk_j(\sum_{x \in E_j}w(x))$$ $$=\sum_{j=1}^n(\sum_{x \in E_j}k_jw(x))=\sum_{j=1}^n(\sum_{x \in E_j}\phi(x)w(x))$$ $$=^{Lemma}\sum_{x\in X}\phi(x)w(x)$$

Now by definition of the integral of non-negative functions let $\phi_n$ an increasing sequence of simple non-negative functions such that $\phi_n \to f$ and $\int_X\phi_nd\mu \to \int_Xfd\mu$

Then $\int_X\phi_n d\mu=\sum_{x \in X}\phi_n(x)w(x) \leq \sum_{x \in X}f(x)w(x)$

So by taking the limit we have that $\int_Xf d\mu \leq \sum_{x \in X}f(x)w(x)$

Now let $F \subset X$ finite.

Then $ \sum_{x \in F}\phi_n(x)w(x) \leq \sum_{x \in X}\phi_n(x)w(x)=\int_X\phi_nd\mu$

Using the fact that $\lim_n \sum_{x \in F}\phi_n(x)w(x)= \sum_{x \in F}f(x)w(x)$

we find,by taking limits, that $ \sum_{x \in F}f(x)w(x) \leq \int_X f(x)d\mu$

Taking the supremum over all finite subsets of $X$ we have that $$ \sum_{x \in X}f(x)w(x)\leq \int_X f d\mu$$

Answer 2

We have that for $w:X\to [0,\infty ]$ $$ \mu (E):=\sup\left\{\sum_{x\in A}w(x):A\subset E\,\land\, \#A<\aleph _0\right\}\tag1 $$ for any $E\in \wp (X)$. If $X$ is countable or there is some $\mu (A)\inf_{x\in A}f=\infty $ the statement to be proved is clear, so assume that $X$ is uncountable and that $\mu (A)\inf_{x\in A}f<\infty$ for all $A\subset X$ . Recall that $$ \int f\,\mathrm d \mu :=\sup\left\{\sum_{j=1}^N\mu \left(A_j\right)\inf_{x\in A_j}f:X=\bigcup_{j=1}^NA_j\right\}\tag2 $$ where the $A_j$ defines an arbitrary partition of $X$ (the $A_j$ are mutually disjoint). And we need to show that $\mathrm{(2)}$ is equivalent to $$ \sup\left\{\sum_{x\in A}w(x)f(x):A\subset X\,\land\, \#A<\aleph _0\right\}\tag3 $$ Now note that $$ \sum_{j=1}^N w(x_j)f(x_j)=\sum_{j=1}^N\mu (\{x_j\})\inf_{x\in \{x_j\}}f\tag4 $$ so, completing the list $\{x_1\},\ldots ,\{x_n\}$ to a partition of $X$ adding $H:=X\setminus \{x_1,\ldots x_n\}$ we have that

$$ \sum_{j=1}^N w(x_j)f(x_j)\le\sum_{j=1}^N\mu (\{x_j\})\inf_{x\in \{x_j\}}f+\mu(H)\inf_{x\in H}f\tag5 $$ for any chosen finite list $x_1,\ldots ,x_n$ contained in $X$, so $\rm (3)\leqslant (2)$. By the other side we have that if $\mu (A)<\infty $ then, by $\rm (1)$, for any $\epsilon >0$ there is a finite $C\subset A$ such that

$$ \mu (A)<\epsilon +\sum_{x\in C}w(x)\tag6 $$ Hence $$ \begin{align*} \sum_{j=1}^N\mu \left(A_j\right)\inf_{x\in A_j}f&<\sum_{j=1}^N\left(\epsilon _j+\sum_{x\in C_j}w(x)\right)\inf_{x\in A_j}f\\ &<\epsilon +\sum_{j=1}^N\sum_{x\in C_j}w(x)f(x) \end{align*}\tag7 $$ where $\epsilon $ is arbitrarily small for suitable choosing of the $\epsilon _j$ according to the values of $\inf_{x\in A_j} f$, thus we can conclude easily from $\rm (7)$ that $\rm (2)\leqslant (3)$, finishing the proof.