Integrals wrt measure space of countable and co-countable sets I need help working on the exercise 1.6 (page 32 in the third edition) from the book "Real and complex analysis" by Rudin.
Let $X$ be an uncountable set and $$\Sigma = \big\{ E \in \wp(X) \, : \, |E| \leq \aleph_0 \ \mbox{ or } \ |E^c| \leq \aleph_0 \big\} \, . $$ I proved that $\Sigma$ is a $\sigma$-algebra for $X$, named the $\sigma$-algebra of countable and co-countable sets. Then I proved that $\Sigma$ is generated by the family $ \, \big\{ \{ x \} : x \in X \big\} \, $ of singleton subsets of $X$.
Let $Y$ be a Hausdorff topological space and $\mathfrak{B}$ be the Borel $\sigma$-algebra for $Y$. Then I proved that a function $ \ f:X \to Y \ $ is measuble with respect to the measurable spaces $(X, \Sigma)$ and $(Y, \mathfrak{B})$ if, and only if, there exists an unique $ \ y_f \in Y \ $ such that $ \ \big| f^{-1} [Y \setminus \{ y_f \}] \big| \leq \aleph_0 \, $. Thus clearly one has $ \ \big| f^{-1} [\{ y_f \}] \big| > \aleph_0 \, $.
Let $ \ Y = [0 , \infty] \ $ be the topological subspace of the extended real number line with the usual (order) topology and let $ \ \mu: \Sigma \to [0, \infty] \ $ be a function such that, $\forall E \in \Sigma$, $$\mu(E) = \left\{ \begin{array}{ll} 0 \, , & \mbox{if $ \ |E| \leq \aleph_0 \, $}; \\ 1 \, , & \mbox{if $ \ |E^c| \leq \aleph_0 \, $}.\end{array} \right.$$ Then $\mu$ is well-defined since $ \ |E| \leq \aleph_0 \ $ if, and only if, $ \ |E^c| > \aleph_0 \, $, $\forall E \in \Sigma \, $. Thereafter I proved that $\mu$ is a measure for $(X, \Sigma)$.
By the trivial characterization of simple functions, we have a bijection between the set of measurable simple functions of $ \, (X, \Sigma, \mu) \, $ and the set of finite partitions $\mathcal{P}$ of $X$ such that $ \ \mathcal{P} \subset \Sigma \ $ and one, and only one, of the elements of $\mathcal{P}$ is uncountable. Therefore, if $ \ \phi : X \to [0, \infty] \ $ is simple and measurable, it is easy to show that $ \ \int \phi \ d \mu = t_{\phi} \ $, where $ \ t_{\phi} \in [0 , \infty] \ $ is the only extended real number such that $ \ \big| \phi ^{-1} \big[ [0, \infty] \setminus \{ t_{\phi} \} \big] \big| \leq \aleph_0 \, $. Hence, I suspected that perhaps this could hold for any measurable function.
I am not supposed to use the monotone convergence theorem (MCT) but, for heuristic purposes, let $ \ f: X \to [0, \infty] \ $ be a measurable function and $ \ (\beta_n)_{n \in \mathbb{N}^*} \ $ be a pointwise non-decreasing sequence of $ \ [0, \infty]$-valued measurable functions such that $ \ \displaystyle \lim_{n \to \infty} \beta_n = f \ $ pointwise. By MCT one has $$ \int f \ d \mu = \int \lim_{n \to \infty} \beta_n \ d \mu = \lim_{n \to \infty} \int \beta_n \ d \mu = \lim_{n \to \infty} t_{\beta_n} \ . $$ That must hold for all non-decreasing sequences $(\beta_n)$ of simple measurable functions that converges to $f \, .$ So I suspected that $ \ \int f \ d \mu = $ $\displaystyle \lim_{n \to \infty} t_{\beta_n} = t_f \, $, but I am not sure, the last equality is only a guess. Is it valid? How can I prove that?
Thanks in advance.
 A: Finally I got it.
Let $ \ f: X \to [0, \infty] \ $ be a measurable function, $$S_f = \Big\{ \phi \in [0, \infty]^X \ : \ \phi \ \mbox{ is simple and measurable and } \ \phi \leq f \ \Big\} \, , $$ $$I_f = \Big\{ \textstyle \int \phi \ d \mu \in [0, \infty] \ : \ \phi \in S_f \ \Big\}$$ and $ \ E_f = f^{-1} [ \{ t_f \} ] \, $. Then $ \ \int f \ d \mu = sup(I_f) \, $, $ \, E_f \, $ and $ \, E_f^c \, $ are measurable, $ \, |E_f| > \aleph_0 \ $ and $ \ |E_f^c| \leq \aleph_0 \, $, where $ \ E_f^c = X \setminus E_f = f^{-1} \big[ [0, \infty] \setminus \{ t_f \} \big] \, $.
Let $ \ s \in I_f \ $. There exists $ \ \phi \in S_f \ $ such that $ \ s = \int \phi \ d \mu \, $. If we had $ \ E_f \cap \big( \phi^{-1} [ \{ t_{\phi} \} ] \big) = \varnothing \ $, then we would have $ \ \phi^{-1} [ \{ t_{\phi} \} ] \subset E_f^c \ \Rightarrow \ |E_f^c| \geq \big|\phi^{-1} [ \{ t_{\phi} \} ] \big| > \aleph_0 \ $, a contradiction. So there exists $ \ x_0 \in \ E_f \cap \big( \phi^{-1} [ \{ t_{\phi} \} ] \big) \ $ and we have that $ \ s = \int \phi \ d \mu = t_{\phi} = \phi(x_0) \leq f(x_0) = t_f \ $. It follows that $ \, t_f \, $ is an upper bound of $ \, I_f \, $.
Let $ \ \phi_0 = t_f \cdot \chi_{E_f} = 0 \cdot \chi_{E_f^c} + t_f \cdot \chi_{E_f} \ $. Then $ \ \phi_0 : X \to [0, \infty] \ $ is a simple and measurable function. Let $ \ x \in X \, $. If $ \ x \in E_f \ $, then $ \ \phi_0 (x) = t_f = f(x) \, $. If $ \ x \in E_f^c \ $, then $ \ \phi_0 (x) = 0 \leq f(x) \, .$ Since $x$ is arbitrary, we have $ \ \phi_0 \leq f \, .$ Therefore $ \ \phi_0 \in S_f \ $ and $ \ t_f = \int \phi_0 \ d \mu \in I_f \ $.
It follows that $ \ t_f = max(I_f) = sup(I_f) = \int f \ d \mu \ $.
