How can I say a set has measure $1$? Suppose $(\Omega,\mathscr{E},\mathbb{P})$ is a measure space such that $\mathbb{P}(\Omega)=1$.
Suppose $A_i \in \mathscr{E}$ for $1 \leq i \leq n$ where $n \in \mathbb{N}$.
Suppose $\mathbb{P}(\bigcup_{1 \leq i \leq n} A_i)=1$.
Can I conclude that there exists $1 \leq i \leq n$ such that $\mathbb{P}(A_i)=1$? If not, how can I found a counterexample?
I know the inclusion-exclusion principle, but I do not know if we can use it here and how.
In addition, are there conditions (i.e.  $(\Omega,\mathscr{E},\mathbb{P})$ has not atoms) such that the statement holds?

No. There seems to be no inclusion hypothesis in your statement.
Consider $\Omega=[0,1]$ with the Borel $\sigma$-algebra generated by the open sets and the the usual probability $dx$. Then you can have
$$A_j = \left(\frac{j-1}{n},\frac{j}{n}\right)\,.$$
Clearly the union has probability one but they are all disjoint and with probability $1/n$.
