Hello,

This limit holds only if at least one of the has a finite measure. If they're all of infinite measure, I think it should be easy to finish your proof by using (*) below. So now assume we can use this formula.Suppose that is continuous from above, that is, if with , then I have .

Since is finite-additive, the first axiom of a measure :I will show that is a measure.

is verified.

And the positivity too.

And we want to prove that for any pairwise disjoint sequence , we have :Let be disjoint sets.

Claim:

(*)

The backslash is not \ which gives a space in latex, but \backslashLet , note that and

I guess you're being confused with below and above...Now, since is continuous from above, I have

The above continuity deals with an intersection, not an union, and with (which you actually have)

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To sum up... (not too formally) :

We know that , (continuous from above)

We know that pairwise disjoint, (finite additivity)

We want to prove that pairwise disjoint,

With the information we have, this is equivalent to proving

Then I recommend you read this : http://www.mathhelpforum.com/math-he...ection-ei.html (the second post).