Is the conditional expectation of a partition really only defined up to $\mathbb{P}$ uniqueness?

Question

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and let $(B_i)_i$ be a finite or countable partition of $\Omega$, that is, $\Omega = \cup_{i}B_i$ and $B_i\cap B_j=\emptyset$ for all $i\neq j$. Then, let $\mathcal{C}\subseteq \mathcal{A}$ be $\sigma$-algebra generated by that partition. It can be shown that this $\sigma$-algebra is defined by $\mathcal{C}=\{\cup_{i\in I}B_i:I\subseteq \mathbb{N}\}$ (in the countable case). Now according to this answer this $\sigma$ algebra does not contain any non-empty nullsets. I.e., the empty set is the only null set in that $\sigma$-algebra.

Now the conditional expectation is given by $$\mathbb{E}[X|\mathcal{C}]=\Sigma_{i\in I} ,円\mathbb{E}[X|B_i],円\chi_{B_i}$$ $\mathbb{P}$-a.s., according to e.g. the books by Resnick and Bauer. However, I do not understand why this formula would only hold $\mathbb{P}$-a.s. In the case of conditional expectation, the exception null-set on which the different versions may differ is a set in $\mathcal{C}$. Since, however, this $\sigma$-algebra only contains the empty set as null set, the different versions of the conditional expectation can only differ on the empty set, hence they have to be equal everywhere. Where is my error of thinking?

• Kavi Rama Murthy: If 0ドル<P(B_i)<1$ for all $i$ and $\mathcal C$ is generated by this partition, then, yes, what you say is correct.
• Ramiro: The answer you mention is for a question where it is assumed that 0ドル<\mathbb{P}(B_i)<1$ for all $B_i$. In this case, your reasoning is correct. However, in the general case we may have some $B_i$ with $\mathbb{P}(B_i) =0$. Then, $\mathcal{C}$ contains other nullsets (not only the empty set) and the formula holds $\mathbb{P}$-a.s.