This is from Asymptotic Statistics (van der Vaart) chapter 11 (problem 4).
Let $X$ and $Y$ be random variable defined on the same probability space.
Define $E[X|Y]$ as the measurable function $g_0$ such that $E(X-g_0(Y))^2 \leq E(X-g(Y))^2$ for all measurable $g$. In other words, $E[X|Y]$ is the projection of $X$ onto the linear space of measurable functions of $Y$.
The above definition of $E[X|Y]$ is well defined if $X, Y$ and the space of measurable $g$ are all square integrable.
Now consider $Z \geq 0$ also defined on the same probability space, but not necessarily square integrable. Define $E[Z|Y] = \lim_{M\rightarrow \infty} E[\min(M,Z)|Y]$
Show that this limit is well defined (exists almost surely) and that the definition coincides with the previous one of $E[Z] < \infty$.
What I've tried
Rewriting the expression above: $$E[Z|Y] = \lim_{M\rightarrow \infty} E[\min(M,Z)|Y] = \lim_{M\rightarrow \infty} E[Z|Y]P(X\leq M | Y) + MP(X>M|Y)$$
But it's not clear to me that this limit exists almost surely since $E[Z|Y]$ is not well defined. For the second part I want to show that this expression minimizes $E(Z-g(Y))^2$ for all measurable $g$ which is straightforward if $E[Z|Y]$ is well defined since in that case $\lim_{M\rightarrow \infty} E[Z|Y]P(X\leq M | Y) + MP(X>M|Y) = E[Z|Y]$.
Hint : show that, if $X, X'$ are square integrable random variables such that $X \leq X'$ almost surely, then $E[X | Y] \leq E[X' | Y]$ almost surely.