Partitioning non atomic finite measure space

Question

Let $(X,\Sigma, \mu)$ be a non-atomic, complete and finite measure space.

I would like to know if the following is true:

For every $\varepsilon \in (0, \mu(X))$ there are finitely many sets $X_1, \ldots , X_N$ such that:

1. $X_i \in \Sigma$ for every $i = 1, \ldots , N$ and $\bigcup_{i=1}^N X_i = X$;

2. $X_i \cap X_j = \emptyset$ for every $i \ne j$;

3. $\mu(X_i) \le \varepsilon$ for every $i=1, \ldots ,N$;

4. $\mu(X_i) \le \mu(X_i^c)$ for every $i = 1,\ldots , N$.

I got confused by the fourth property: I think it is not a problem to construct a finite partition into sets of arbitarily small measure but I am confused whether I can assume that $\mu(X_i) \le \mu(X_i^c)$ or not.

Answer 1

By the result in the linked question there exists a partition $(X_i)_{i \leq N}$ of measurable sets such that $\mu(X_i) \leq \min\{\epsilon,\tfrac{1}{2} \mu(X)\}$ for all $i=1,\ldots,N$. The sets $(X_i)_{i \leq N}$ then clearly satisfy property 1-3. Moreover, $$\mu(X_i^c) = \mu(X)-\underbrace{\mu(X_i)}_{\leq \mu(X)/2} \geq \frac{1}{2} \mu(X) \geq \mu(X_i),$$ i.e. property 4 holds.