Let $\{A_n\}_n$ be a sequence of sets of a $\delta$-ring $\mathfrak{M}$ of measurable sets with finite Lebesgue measure. Let us suppose that $$\forall\varepsilon>0\quad\exists N\in\mathbb{N}^+:\forall m,n\geq N\quad\mu(A_n\triangle A_m)<\varepsilon$$ I suspect that it is possible to show that there is a constant, independent from $M$, such that $\mu(\bigcup_{n=1}^M A_n)$ is bounded, but I am not able to prove it. Since $d(A,B)=\mu(A\triangle B)$ can define a metric and since a Cauchy sequence is bounded, I see that $\exists A\in\mathfrak{M}\quad\exists K\geq0:\forall n\in\mathbb{N}^+\quad\mu(A_n\triangle A)\leq K$, but I am not sure whether that can be useful to prove that $\exists K'\geq0:\forall M\in\mathbb{N}\quad\mu(\bigcup_{n=1}^M A_n)\leq K'$....
Has anybody any idea about how to do so, if it is possible? I thank you all for any help!
If I understand your question correctly, then what you want to show is not true.
Take $A_n = [n, n + 1/n]$ and let $\mu$ be Lebesgue measure. Then $\mu (A_n) \to 0$, which easily implies your "Cauchy" property, but
$$ \mu(\bigcup_{n=1}^M A_n) = \sum_{n=1}^M 1/n \to \infty$$ for $M \to \infty$.