Alternative proof of Monotone Convergence Theorem Let $(X, \mathfrak{A}, \mu)$ be a measure space. Show that the Monotone Convergence Theorem holds if $f$, $f_{1}$, $f_{2}$, ... are real-valued measurable functions and $f_{1}$ is integrable. Further,
(1) $f_{1} \leq f_{2}...$
and 
(2) f = $\lim_{n} f_{n}$ hold almost everywhere. 
My try:
Assume (1) and (2) hold everywhere.
From the monotonicity of the integral we have
$\int f_{1} d\mu \leq \int f_{2} d\mu \leq$ .. $\leq \int f d\mu$.
We need to show the reverse inequality. 
Let $g^{+}_{n,k}$  and $g^{-}_{n,k}$ be two sequences of positive valued simple functions such that
$\lim_{k} g^{+}_{n,k} - g^{-}_{n,k} = f_{n}$ and $g^{+}_{n,k} - g^{-}_{n,k} \leq f_{n}$ for all $k$.
Define $h_{n}^{+}$ and $h_{n}^{-}$ as $h^{+}_{n} = \max_{k}(g_{k,n}^{+}$) and $h^{-}_{n} = \max_{k}(-g^{-}_{k,n})$, then both $h_{n}^{+}$ and $h^{-}_{n}$ are non-decreasing measurable simple functions and $\lim_{n} h^{+}_{n} + h^{-}_{n } = f^{+} - f^{-} = f$. Also $h^{+}_{n} \leq f^{+}_{n}$ and $h^{-}_{n} \leq -f^{-}_{n}.$ Since $h^{+}_{n}$ and $h^{-}_{n}$ are simple functions, it follows 
$\lim_n \int h_{n}^{+} d\mu = \int f^{+} d\mu$ and $ \lim_{n} \int h^{-}_{n} d\mu = -\int f^{-} d\mu$
and
$\lim_{n} \int h^{+}_{n} + h^{-}_{n} d\mu = \int f d\mu$.
Since $\int f^{+} d\mu = \lim_{n} h^{+}_{n} d\mu \leq \lim_{n} \int f^{+}_{n}d\mu$ and similary $-\int f^{-}d\mu = \lim_{n} \int h^{-}_{n} d\mu \leq \lim_{n} -\int f^{-}_{n}d\mu$
$\int f d\mu = \lim_{n} \int h^{+}_{n} + h^{-}_{n} d\mu \leq \lim_{n} \int f^{+}_{n} - f^{-}_{n} d\mu = \int f_{n}d\mu$.
Assume now (1) and (2) hold almost everywhere, let $N$ be the set consisting all $x\in X$ for which at least one of (1) and (2) fails. 
Then the functions $f\chi_{n^{C}}$ and $f_{n}\chi_{N^{C}}$ fulfills (1) and (2) and thus
$\int f\chi_{N^{C}} d\mu = \lim_{n} \int f_{n}\chi_{N^{C}}d\mu$ and from this we can conclude that
$\int f d\mu = \lim_{n} \int f_{n}d\mu$.
This is very similar to a proof for MCT in the book Measure Theory by Donald Cohn, however, the book is for positive valued functions only. 
My question is whether this is correct or if I have missed something. 
Secondly, why is the constraint that $f_{1}$ is integrable necessary?
 A: Some comments while I'm reading your proof and the original from Donald Cohn's (that you cited). These might sound critical, but I'm just trying to raise your awareness in details. 


*

*The verse "We need to show the reverse inequality" is a bit puzzling. If, e.g. $(X,\mathcal{M},\mu)=([0,1],\text{Leb}([0,1]),m_{1})$, where $m_{1}$ is the Lebesgue measure, and we choose $f_{n}\equiv 1-\frac{1}{n}$ for all $n\in\mathbb{N}$, then $\int f_{n}\,dm_{1}<\int f_{n+1}\,dm_{1}$ for all $n\in\mathbb{N}$. So there is really no reason why the reverse inequality should hold, and I doubt this is what you meant. You want to show that $\lim_{n} \int f_{n}\,d\mu\geq \int f\,d\mu$.

*The way you defined the functions $h_{n}^{+}$ and $h_{n}^{-}$ does not work, i.e. defining them as maximums over an infinite set. Are you sure these maximums exist, or should you replace it with a supremum? And if you take supremum, what would be the result? Would these functions, as a supremum of simple functions, be simple functions? In any case, what you want to do is to define $h_{n}^{+}=\max\{g^{+}_{1,n},...,g^{+}_{n,n}\}$ instead, and $h_{n}^{-}$ similarly (but instead minimums). 
I am going to assume from now on that the functions $h_{n}^{+}$ and $h_{n}^{-}$ are defined as above. Also it is worth to note that you have not defined $f^{+}$ or $f^{-}$ before you start using these notations.


*

*Note that the sequence $(h_{n}^{-})$ does not consist of non-negative simple functions, so you can't use Proposition 2.3.3 as it stands in the next step. Something analogous could be argumented though.

*And here comes the situation where some level of integrability from $f_{n}$ would be required:
\begin{equation*}
\lim_{n}\int h_{n}^{+}+h_{n}^{-}\,d\mu=\int f\,d\mu.
\end{equation*}
What if $\int |f_{n}|\,d\mu=\infty$ for all $n$? Would it be possible that one of the sides would yield $\infty-\infty$ in the steps where you proved this equality, which would not be defined?
Here's a suggestion for an alternative approach to this result, using the MCT for non-negative functions as Donald Cohn has proven it. I don't think you have to make the proof from scratch again. 
Take $f_{1}\leq f_{2}\leq ...\leq f$ as before and assume that $f_{1}$ is integrable. Then $0\leq f_{2}-f_{1}\leq f_{3}-f_{1}\leq ... \leq f-f_{1}$ is a non-negative sequence, and thus
\begin{equation*}
\lim_{n}\int f_{n}\,d\mu-\int f_{1}\,d\mu=\lim_{n}\int f_{n}-f_{1}\,d\mu\overset{MCT}{=}\int f-f_{1}\,d\mu=\int f\,d\mu-\int f_{1}\,d\mu.
\end{equation*}
Since $-\infty<\int f_{1}\,d\mu<\infty$, then this implies $\lim_{n}\int f_{n}\,d\mu=\int f\,d\mu$.
Edit: Note that the integrability of $f_{1}$ is necessary. Let $(X,\mathcal{M},\mu)=(\mathbb{R},\text{Leb}(\mathbb{R}),m_{1})$ and $f_{n}=-\chi_{[n,\infty)}$ for all $n\in\mathbb{N}$. Now $f_{1}\leq f_{2}\leq ...\leq f=0$, but $\lim_{n}\int f_{n}\,dm_{1}=-\infty\neq 0=\int f\,dm_{1}$.
