Nonnegative scalar multiples and countable weighted sums of measures are measures

For $c=0$, the set function $c\mu$ is the zero measure.

For $0<c<+\mathrm{\infty }$, $(c\mu )(\mathrm{\varnothing })=0$; for disjoint $(E_j)$, multiplying the finite partial-sum identities by $c$ and taking their supremum gives $c\mu ({\bigcup }_j{E}_j)={\sum }_{j}c\mu (E_j)$, whether the common value is finite or $+\mathrm{\infty }$.

For $c=+\mathrm{\infty }$, a disjoint union has $\mu$-measure zero exactly when every member has $\mu$-measure zero: this follows directly from countable additivity and nonnegativity. Hence the infinite branch takes value $0$ on the union exactly when every term value is $0$, and otherwise both it and the series of term values are $+\mathrm{\infty }$; this branch is a measure without forming $0\cdot (+\mathrm{\infty })$.

Steps 1.1, 1.2 and 1.3 prove that every scalar multiple $c_k{\mu }_k$ is a measure for all possible coefficients.

Put $\nu (E)={\sum }_k(c_k{\mu }_k)(E)$. Then $\nu (\mathrm{\varnothing })=0$.

If $(E_j)$ is disjoint, then step 2.1 and Tonelli give $\nu ({\bigcup }_j{E}_j)={\sum }_k{\sum }_j(c_k{\mu }_k)(E_j)={\sum }_j{\sum }_k(c_k{\mu }_k)(E_j)={\sum }_j\nu (E_j)$.

Steps 3.1 and 3.2 prove that the countable weighted sum is a measure; the same proof for a finite index range, or zero coefficients thereafter, gives every finite weighted sum, including the empty zero measure.