These are not well-defined up to null sets. Indeed, even if all $A_i$ are null so the supremum is supposed to be $[\varnothing]$, you can easily arrange it so that $\bigcup_{i \in I} A_i$ is not null - after all, every subset of $[0, 1]$ is the union of all its singleton subsets, which are all null.
However, $\mathcal{B}([0, 1])/\text{Null}$ is nevertheless complete. The point here is that your definition of supremum does work when $I$ is countable, and an arbitrary supremum can actually be reduced to a countable one in $\mathcal{B}([0, 1])/\text{Null}$. Indeed, let $\{[A_i]\}_{i \in I} \subset \mathcal{B}([0, 1])/\text{Null}$. Let,
$$r = \sup_{F \subset I \text{ finite subset}} m\left(\bigcup_{i \in F} A_i\right)$$
where $m$ is the Lebesgue measure. Then $r \leq 1 < \infty$, so for any $n \in \mathbb{N}$, there exists $F_n \subset I$ finite s.t.,
$$m\left(\bigcup_{i \in F_n} A_i\right) > r - \frac{1}{n}$$
Now, I claim that $\bigcup_{n \in \mathbb{N}} \bigcup_{i \in F_n} A_i$ contains all $A_i$ up to null sets, from which it is not hard to see,
$$\sup_{i \in I} [A_i] = \left[\bigcup_{n \in \mathbb{N}} \bigcup_{i \in F_n} A_i\right]$$
To prove the claim, fix $i_0 \in I$. Let $n \in \mathbb{N}$,
$$\begin{split}
m\left(A_{i_0} \setminus \left(\bigcup_{n \in \mathbb{N}} \bigcup_{i \in F_n} A_i\right)\right) &\leq m\left(\left(\bigcup_{i \in F_n \cup \{i_0\}} A_i\right) \setminus \left(\bigcup_{i \in F_n} A_i\right)\right)\\
&= m\left(\bigcup_{i \in F_n \cup \{i_0\}} A_i\right) - m\left(\bigcup_{i \in F_n} A_i\right)\\
&\leq r - m\left(\bigcup_{i \in F_n} A_i\right)\\
&< \frac{1}{n}
\end{split}$$
As this holds for all $n \in \mathbb{N}$, letting $n \to \infty$ yields,
$$m\left(A_{i_0} \setminus \left(\bigcup_{n \in \mathbb{N}} \bigcup_{i \in F_n} A_i\right)\right) = 0$$
i.e., $A_{i_0} \subset \bigcup_{n \in \mathbb{N}} \bigcup_{i \in F_n} A_i$ up to a null set, as claimed.
