I am new to measure theory and am wondering:

Is the only difference between a measure and a premeasure the fact that measures are defined on $\sigma$-algebras and premeasures are defined only on algebras?

If there are more fundamental differences than this please elaborate on them.


1 Answer 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .