In the probability book of Bauer it is claimed that for nonegative X and Y, or integrable X and Y we have $$ (1) \quad X=Y \, a.s. \Rightarrow \mathbb{E}(X \mid \mathcal{A}) = \mathbb{E}(Y \mid \mathcal{A}) \, a.s. $$ and $$ (2) \quad X \leq Y \, a.s. \Rightarrow \mathbb{E}(X \mid \mathcal{A}) \leq \mathbb{E}(Y \mid \mathcal{A}) \, a.s..$$ I agree with this observation.
Now Bauer argues that if $(X_n)$ is a monotonously increasing sequence, then duo to $(1)$ and $(2)$ also $\mathbb{E}(X_n \mid \mathcal{A})$ can be seen as an increasing sequence.
I guess the term "can be seen as" refers to the fact that the conditional expectation is only almost surely given. But why do we need to use $(1)$ and $(2)$ to see that $\mathbb{E}(X_n \mid \mathcal{A})$ can be seen as an increasing sequence ? Isn't $(2)$ sufficient? Usually when I read something like this I am missing something important.
I think for the claim it is not necessary to use $(1)$ additionaly to $(2)$.
However, if $(X_n)$ is only almost surely monotonously increasing, then one needs to use $(1)$ and $(2)$, because $(2)$ alone finds for each $n$ only a representative with $$ \mathbb{E}(X_n \mid \mathcal{A}) \leq \mathbb{E}( X_{n+1} \mid \mathcal{A}) $$ but the sequence of those representative $(\mathbb{E}(X_n \mid \mathcal{A}))_{n\in \mathbb{N}}$ does not have to be monoton increasing. In this case one would have to use $(1)$ together with the fact that the a countable infinite union of zero sets is again a zero set.