Confused with Bounded and Dominated convergence theorem I am confused with  bounded and dominated convergence theorem.
I have a random variable $Z$  which is $\mathscr{G}$-measurable  and bounded almost sure, ie, there is a positive number $M$ such that $|Z| \leq M $ a.s.Then I have to show that for the conditional expectation $$Y = \mathbb{E}[X|\mathscr{G}]$$ i have :
$$\mathbb{E}[YZ] = \mathbb{E}[XZ]\quad\text{(1)}\quad$$
and more general if $Z$ is $\mathscr{G}$-measurable and $$\mathbb{E}[|ZX|]< \infty$$ then the equality to be proven is still true.
My answer according to David William (Probability with Martingales) page 90 is that :
If $Z$ is the indicator of a set in $\mathscr{G}$, then (1) is true by definition of the conditional expectation $Y$. Linearity then shows that (1) holds for $Z \in \mathscr{G}$ . Next, Monotone Convergence Theorem shows that (1) is true for $Z \in \mathscr{G}$  with the understanding that both sides might be infinite.
or
We know that for integrable r.v.'s $X$, the defining equation for its conditional expectation $Y=E[X|\mathscr{G}]$ given some sub-$\sigma$-field $\mathscr{G}$ is
$$E[X\cdot1_A]=E[Y\cdot1_A],\ \forall A\in\mathscr{G}\quad(1)$$
Changing to:
$$E[XZ]=E[YZ],\ \forall\text{ nonnegative, bounded,} \mathscr{G}\text{-measurable r.v. }Z\quad(2)$$
Now $(2)\implies(1)$ is obvious. For the other direction. My proof is as follows:
Since $Z$ is $\mathscr{G}$-measurable, nonnegative and bounded, we can find an increasing sequence of simple functions $z_n\geq0$ such that $z_n\nearrow Z$ a.s. Moreover, $z_n$ has the form
$$z_n=\sum_{j=1}^{nk}a_j\cdot1_{A_{kj}},\text{ where}\ A_{kj}\in\mathscr{G}$$
Clearly, $E[X z_k]=E[Y z_k]$ by $(1)$. Let $k\to\infty$ and by dominated convergence we have $E[XZ]=E[YZ]$.
But how this can be proven with dominated convergence theorem?
Any help?
 A: Just as you did for a bounded $Z$, you can define a sequence $(z_n)$ of simple functions such that $z_n \nearrow Z$ a.s. You only need $Z$ to be $\mathscr{G}$-measurable for such a sequence to exist, the main difference being that if $Z$ is unbounded then the convergence is not uniform (see e.g. this question with the linked notes for more details).
We now have the following inequality : $\forall n\ge 0$, $|z_nY| = |z_n||Y| \le |Z||Y|$. And we can see that $|ZY|$ is integrable, indeed :
$$\begin{align}
|Y| &\le \mathbb{E}[|X|\,\,|\mathscr{G}] \quad\text{by Jensen's inequality} \\
\implies |YZ| &\le |Z|\mathbb{E}[|X|\,\,|\mathscr{G}] \\
\implies \mathbb{E}[|YZ|] &\le \mathbb{E}\left[|Z|\mathbb{E}[|X|\,\,|\mathscr{G}]\right] \\
&= \mathbb{E}\left[\mathbb{E}[|XZ|\,\,|\mathscr{G}]\right] \quad\text{by $\mathscr{G} $- measurability of $|Z|$}\\
&= \mathbb{E}[|XZ|] < \infty
\end{align}$$
So we have shown that the sequence $(z_nY)$ is dominated by an integrable function.
On top of that, it also holds that $\mathbb{E}[z_nY] = \mathbb{E}[z_nX] \, \, \forall n\ge 0$, by definition and linearity of conditional expectation.
Therefore, by dominated convergence theorem we can conclude that
$$\lim_{n\to \infty} \mathbb{E}[z_nY] = \mathbb{E}[ZY] =\lim_{n\to \infty} \mathbb{E}[z_nX] = \mathbb{E}[ZX] $$
