Define a measure in the following way... 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.
 A: 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.
A: 
$\textbf{Lemma}$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 $$ \sum_{x \in X}f(x)w(x)\leq \int_X f d\mu$$
