The finite union bound

For $i<n$, put $B_i=A_i\setminus {\bigcup }_{j<i}{A}_j$. By [L4], every $B_i$ is measurable. They are pairwise disjoint and satisfy $B_i\subseteq A_i$. Their union equals ${\bigcup }_{i<n}{A}_i$: a point in the latter has a least index $i<n$ at which it belongs to $A_i$, and then belongs to $B_i$; the reverse inclusion follows from $B_i\subseteq A_i$.

By [L1], $\mathbb{P}$ is a measure. Extend the disjoint family by $B_i=\mathrm{\varnothing }$ for $i\ge n$. Countable additivity and $\mathbb{P}(\mathrm{\varnothing })=0$ from [L3] give $\mathbb{P}({\bigcup }_{i<n}{A}_i)={\sum }_{i<n}\mathbb{P}(B_i)$. By [L2], each $\mathbb{P}(B_i)\le \mathbb{P}(A_i)$, so summing these finitely many inequalities proves $\mathbb{P}({\bigcup }_{i<n}{A}_i)\le {\sum }_{i<n}\mathbb{P}(A_i)$.

If $n=0$, the union is empty and the sum is the empty sum, so both sides are $0$ by [L3]. If $n=1$, the bound is the equality $\mathbb{P}(A_0)=\mathbb{P}(A_0)$. These arguments impose no finiteness assumption on $\mathrm{\Omega }$.