I have managed to do that in the following way: since you pick $\displaystyle \epsilon$, then for $\displaystyle X_1,\ldots X_M,$ these are all null so you can define sequences $\displaystyle \{ I_{n_r} \}$ with $\displaystyle \displaystyle\sum_{r=0}^{\infty}m(I_{n_r}) < \frac{\epsilon}{M+1}$. Hence we define a sequence $\displaystyle \{ I_r\}$ by putting all the sequence elements of $\displaystyle \{I_{1_r}\}$ up to $\displaystyle \{ I_{M_r}\}$ in the sequence, ie the sequence $\displaystyle \{ I_{1_1}, I_{2_1},\ldots I_{M_1}, I_{2_1},I_{2_2},\ldots I_{M_2}, \ldots \}$. Of course $\displaystyle \displaystyle\bigcup_{n=1}^{M} X_n \subseteq \displaystyle\bigcup_{r=1}^{\infty} I_r$, and we have that $\displaystyle \displaystyle\sum_{n=1}^{\infty}m(I_r) < M\frac{\epsilon}{M+1} < \epsilon$. Therefore $\displaystyle \display\bigcup_{n=1}^{M} X_n$ is null for all $\displaystyle M\in\mathbb{N}$.

But I am confused about the infinite union? Can I deduce it from the above?

what about choosing $\displaystyle \epsilon$, then for each $\displaystyle X_n$, choose a sequence $\displaystyle \{ I_{n_r}\}$ such that $\displaystyle \displaystyle\sum_{r=1}^{\infty} m(I_{n_r}) < \frac{\epsilon}{2^{n+1}}. $Then I would have to define a sequence $\displaystyle \{ I_r\}$ which somehow contains infinitely many of the above sequences, so that I get $\displaystyle \displaystyle\sum_{r=1}^{\infty} I_r < \epsilon$. But can I construct a sequence like that?

Any help would be appreciated, thank you