I tried to solve an exercise involving conditional expectations, and in doing so some question's popped up in my mind. First the exercise:

$|Z| \le c \textrm{ P.-a.s.} \Rightarrow |E\{ Z | \mathcal G \}| \le c \textrm{ P.-a.s.}$

At first sight I thought it is quite trivial, its just monotonicity, but in my notes and also on Wikipedia it is written in the form $Z \le Y$ implies $E\{Z|\mathcal G\} \le E\{Y|\mathcal G\}$, i.e. without the ''P.-a.s''-part. But here my first question, as the conditional expectation is just almost sure uniquely determined (i.e. two version differ on a set of $P$-measure zero), every such equation involving conditional expectations (i.e. also all the other listed on wikipedia) should be just read $P.-a.s.$, i.e. they just hold with probability one, otherwise it would not make any sense, or? If this is agreed upon, then indeed the exercise is just a trivial consequence of monotonicity.

But I am not sure about the above points, and because I do not think the exercise is trivial I tried to solve it in some other way, and here could you please take a look and say if I am correct or even point out if I am overcomplicating things? My solution:

By presupposition we have $P(\{ \omega : |Z(\omega)| > c \}) = 0$, now define an auxiliary random variable with $\hat Z(\omega) := 0$ on the above set of measure zero, and which agrees with $Z$ everywhere else. So then $|\hat Z| \le c$ everywhere (not just P.-a.s.), so we can apply monotonicity to get $|E\{\hat Z|\mathcal G\}| \le c$. Now let $G \in \mathcal G$ and apply the defining equation of conditional expectation $$ \int_G E\{ \hat Z | \mathcal G \} dP = \int_G \hat Z(\omega) dP = \int_G Z(\omega) dP $$ as $\hat Z$ and $Z$ just disagree on a set of measure zero the last equation is valid. But this shows that $E\{ \hat Z | \mathcal G \}$ is a version of the conditional expectation for $Z$, and so every other version $E\{Z|\mathcal G\}$ disagrees with this version just on a set of measure zero, i.e. we have $E\{ \hat Z | \mathcal G \} = E\{ Z |\mathcal G \}$ with probability one, which implies $E\{Z|\mathcal G\} \le c$ P.-a.s. $\square$

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    $\begingroup$ You are right, formally every statement about conditional expectation holds only almost surely. Indeed, nothing prevents you from making this expectation being $>c$ over some set of measure zero, so that the inequality will hold strictly almost surely. Monotonicity itself only requires $Z \leq Y$ almost surely, so that's the only thing you need for the proof. $\endgroup$ – Ilya May 29 '15 at 22:17

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