This is an exercise from the book Probability Theory by Heinz Bauer:
Let $(\Omega,\mathcal{A},P)$ be the probability space having $\Omega:=[0,1]$, $\mathcal{A}:=\Omega \cap \mathcal{B}$, $P:=\lambda_{\Omega}$ (Lebesgue measure on $\Omega$) and let $f:\mathcal{A}\to \Omega$ be the mapping
$$f(A)= \begin{cases} \sup A \hspace{0.3cm}\text{if}\, A\neq \emptyset\\ 0\hspace{1.1cm}\text{if}\, A= \emptyset \end{cases} \hspace{2cm} (A\in \mathcal{A})$$
and $Q:\Omega\times\mathcal{A}\to\mathbb{R}$ the mapping
$$Q(\omega,A):=1_{A}(\omega)+1_{\{f(A)\}}(\omega) \hspace{2cm} (A\in \mathcal{A} \text{ , } \omega \in \Omega)$$
Show that for each $A\in \mathcal{A}$, $\omega\mapsto Q(\omega,A)$ is a version of the conditional probability $P(A|\mathcal{A}):=E(1_A |\mathcal{A})$, although there is no $P$-null set $N\in \mathcal{A}$ such that $A\mapsto Q(\omega,A)$ is a probability measure on $\mathcal{A}$ for each $\omega\in \complement N$. Is there a $P$-null set $N\in \mathcal{A}$ for which this map is finitely additive on $\mathcal{A}$?
Attempt:
Let $A\in \mathcal{A}$. Since $1_A$ is $\mathcal{A}$ measurable we have $E(1_A |\mathcal{A})=1_A$ $P$-a.s., and since $P\{f(A)\}=0$ we see that $\omega\mapsto Q(\omega,A)$ is indeed a version of $P(A|\mathcal{A})$.
If $N\in \mathcal{A}$ is $P$-null set and we let $A:=\{\omega\}\in \mathcal{A}$ for some $\omega\in \complement N$, then we see that $Q(\omega,A)=2$, so $A\mapsto Q(\omega,A)$ cannot be a probability measure.
If $N\in \mathcal{A}$ is $P$-null set and $\omega\in \complement N$ there are two posssbilities:
$\omega=0$, in which case $A:=\emptyset=:B$ are two disjoint sets in $\mathcal{A}$ such that $Q(\omega,A\cup B)=1\neq 2 =Q(\omega,A)+Q(\omega,B)$.
$\omega>0 $, in which case we let $$A:= \{\omega-1/n: n\in\mathbb{N} \text{ is even and } \omega-1/n \in \Omega \}$$ $$B:= \{\omega-1/n: n\in\mathbb{N} \text{ is odd and } \omega-1/n \in \Omega \}$$
Then $A,B$ are disjoints sets in $\mathcal{A}$ with $f(A)=f(B)=f(A\cup B)=\omega$, and so $Q(\omega,A\cup B)=1\neq 2 =Q(\omega,A)+Q(\omega,B)$.
Hence there is no $P$-null set outside of which $A\mapsto Q(\omega,A)$ is finitely additive.
Am I missing something?