A measure on a finite sigma-algebra is a finite weighted sum over its atoms

For $x\in X$ define $C_x:=\bigcap \{A\in \mathcal{A}:x\in A\}$. The intersected family is nonempty because it contains $X$, and it is finite, so $C_x\in \mathcal{A}$ and $x\in C_x$.

Enumerate the finite family of atoms by a natural number. Applying finite choice to that enumeration selects one representative $x_C\in C$ for every atom; when $X=\mathrm{\varnothing }$, the family and the choice function are empty.

If $y\in C_x$, then $x$ and $y$ belong to exactly the same members of $\mathcal{A}$: otherwise the complement of a measurable set separating them would contradict $y\in C_x$. Hence $C_y=C_x$, while $y\mathrm{\notin }{C}_x$ gives $C_y\cap C_x=\mathrm{\varnothing }$; the distinct $C_x$ form a finite measurable partition of $X$.

Every $C_x$ is an atom, and if $A\in \mathcal{A}$ and $x\in A$, then $C_x\subseteq A$ by definition; conversely, if $C$ is an atom and $x\in C$, then the nonempty measurable set $C_x\subseteq C$ forces $C_x=C$. Consequently every atom is one of the partition blocks and every measurable $A$ is the disjoint union of the atoms it contains.

For measurable $A$, finite additivity and step 3.1 give $\mu (A)={\sum }_{C\subseteq A}\mu (C)$.

For each atom $C$, the term $\mu (C){\delta }_{x_C}(A)$ equals $\mu (C)$ when $C\subseteq A$ and $0$ otherwise; this remains true when $\mu (C)=+\mathrm{\infty }$ because [L3] uses the null/non-null infinite branch.

Summing step 4.2 over the atoms and comparing with step 4.1 proves the representation. For $X=\mathrm{\varnothing }$ both sides are the zero measure and the sum is empty.