Difference between a measure and a premeasure

Solution 1:

Let $X$ be a non-empty set and $\mathscr A$ an algebra on it. A premeasure on a $\mathscr A$ is a function $\lambda:\mathscr A\to[0,\infty]$ such that

• $\lambda(\varnothing)=0$; and
• if $A_1,A_2,\ldots$ is a countable collection of disjoint sets in $\mathscr A$ and if their union is contained in $\mathscr A$, then $$\lambda\left(\bigcup_{n=1}^{\infty} A_n\right)=\sum_{n=1}^{\infty}\lambda(A_n).$$

If $\mathscr B$ is a $\sigma$-algebra on $X$, then a measure on $\mathscr B$ is a function $\mu:\mathscr B\to[0,\infty]$ such that

• $\mu(\varnothing)=0$; and
• if $B_1,B_2,\ldots$ is a countable collection of disjoint sets in $\mathscr B$ (and, since $\mathscr B$ is a $\sigma$-algebra, their union is already contained in $\mathscr B$, so this condition need not be prescribed explicitly in this case), then $$\mu\left(\bigcup_{n=1}^{\infty} B_n\right)=\sum_{n=1}^{\infty}\mu(B_n).$$

Basically, yes, the main difference is that of the domains, viz., a premeasure is defined on an algebra and a measure is defined on a $\sigma$-algebra. There is another subtle difference, though: while both concepts are required to satisfy $\sigma$-additivity, in the case of a premeasure this makes sense only for countable collections of disjoint sets of an algebra whose unions, too, are actually in the algebra.