measure theory problem

I am having trouble following a proof in the book "Mathematical
Analysis" by Andrew Browder. It is theorem 9.13 on page 207 stating:

Let A be a sigma algebra of subsets of the set X, and let M be the
collection of measures with domain A. Suppose that M has the property:
for any mu_1, mu_2 in M there exists a mu_3 in M such that mu_1 =<
mu_3, and mu_2 =< mu_3. Where less than or equal to (=<) for measures
is defined to be mu =< nu if mu(E) =< nu(E) for all E in the sigma
algebra. Finally define nu(E) = sup{mu(E): mu in M}, then nu is a
measure on A.

The proof proceeds as follows:

nu is well-defined (because these are positive extended reals with an
upper bound so a sup exists even if its infinity), nonnegative set
function with domain A. Must show countable additivity. Let {B_n} be a
disjoint sequence in A. Then for each mu in A,

mu(U_{n=1}^infinity B_n) = sum_{n=1}^infinity mu(B_n) =<
sum_{n=1}^infinity nu(B_n).

No problems so far.
Then they conclude that nu(U_{n=1}^infinity B_n) =< sum_{n=1}^infinity
nu(B_n).

Thats what I'm not getting. I figured that since nu(E) >= mu(E) for all
mu hence nu(U_{n=1}^infinity B_n) >= mu(U_{n=1}^infinity B_n), since
U_{n=1}^infinity B_n is in A. Thus leaving it incomparable with the
infinite sum of nu's above.

Thanks for any help.

.

Relevant Pages

  • Re: measure theory problem
    ... > upper bound so a sup exists even if its infinity), nonnegative set ... > infinite sum of nu's above. ... You haven't used the fact that the sup is the *least* upper bound. ...
    (sci.math)
  • Re: measure theory problem
    ... Let A be a sigma algebra of subsets of the set X, ... Let be a disjoint sequence in A. Then for each mu in A, ... U_^infinity B_n is in A. Thus leaving it incomparable with the ... infinite sum of nu's above. ...
    (sci.math)