TITLE: Boundedness in the functional monotone class theorem
QUESTION [5 upvotes]: Let

$\Omega$ be a set
$\mathcal M\subseteq 2^\Omega$ be a $\pi$-system with $\Omega\in\mathcal M$
$\mathcal H$ be a set of functions $\Omega\to\mathbb R$

Suppose $\mathcal H$ satisfies the following conditions 

$A\in\mathcal M\Rightarrow 1_A\in\mathcal H$
$f,g\in\mathcal H\Rightarrow\alpha f+g\in\mathcal H$, for all $\alpha\in\mathbb R$
If $f:\Omega\to\mathbb R$ is bounded and there is a sequence $\left\{f_n\right\}\subseteq H$ with $f_n\ge 0$ and $f_n\uparrow f$, then $f\in\mathcal H$

Then, the montone class theorem states, that $\mathcal H$ contains all bounded $\sigma(\mathcal M)$-measurable functions $\Omega\to\mathbb R$.
Why is boundedness crucial? Let's step through a possible proof: Let $$\mathcal D:=\left\{A\subseteq\Omega:1_A\in\mathcal H\right\}\;.$$

Claim: $\mathcal H$ is a $\lambda$-system

Proof:

$\Omega\in\mathcal M$ and (1.) $\Rightarrow$ $\Omega\in\mathcal D$
Let $A\in\mathcal D\Rightarrow 1_{A^c}=1_\Omega-1_A\in\mathcal H$ by (1.) and (2.)$\Rightarrow A^c\in\mathcal D$
Let $\left\{A_n\right\}\subseteq\mathcal D$ be pairwise disjoint, then $$1_{A^n}=\sum_{m\le n}1_{A_m}\in\mathcal H\;,$$ for $A^n:=\biguplus_{m\le n}A_m$, by (2.) and therefore $$1_A=\lim_{n\to\infty}1_{A^n}\in\mathcal H$$ by (3.) (since $1_A$ is bounded)

Now, since $\mathcal M$ is a $\pi$-system, $$\sigma(\mathcal M)=\delta(\mathcal M)\subseteq\mathcal D\;,$$
by (1.) and the $\pi$-$\lambda$-theorem. So, by (2.), all simple $\sigma(\mathcal M)$-measurable functions are contained in $\mathcal H$.
Until now, the whole argumentation has not made use of the boundedness restriction in (3.).
Let $f:\Omega\to\mathbb R$ be $\sigma(\mathcal M)$-measurable. Then, it's positive and negative parts $f^+$ and $f^-$ can be written as the monotone limit of non-negative $\sigma(\mathcal M)$-measurable simple functions $f_n^+$ and $f_n^-$, respectively. Since $f_n^\pm\in\mathcal H$, we can conclude $$f=f^+-f^-=\lim_{n\to\infty}f_n^+-\lim_{n\to\infty}f_n^-\in\mathcal H\;,\tag{1}$$ even without the boundedness restriction in (3.) as long as the pointwise limits exist (in $\overline{\mathbb R}$).
In the sense of existence in $\overline{\mathbb R}$, the limits exist as long there sum is well-defined, i.e. as long as there can't be a $\infty+(-\infty)$ term. So, it seems like we could loosen the boundedness in (3.) and only force $$\inf_\Omega f>-\infty\;.$$ Do I miss something?

REPLY [7 votes]: Probably not an answer (I didn't read the question completely), but too long for a comment 

The class of bounded measurable functions satisfies all of 1-3.  The boundedness in 3 is in fact not a "restriction"! With the word "bounded" this assumption is weaker. (3 would be void with $f=0$, which is "very restrictive").
If you drop the boundedness in 3, then you can prove that $\mathcal H$ contains all measurable functions.
So why the boundedness is there? Because this theorem is used usually to do something with expectations, and the boundedness is a convenient condition to derive convergence of expectations from that of functions.