## Monotonicity of Measures

First, we make a couple of definitions. A set function on a collection of sets is “monotone” if whenever we have and in with , then . We say that is “subtractive” if whenever further and , then .

Now I say that any measure on an algebra is both monotone and subtractive. Indeed, since is an algebra, then is guaranteed to be in as well, and it’s disjoint from . And so we see that

Since is non-negative, we must have , and so is monotone. And if is finite, then we can subtract it from both sides of this equation to show that is subtractive as well.

Next, say is a measure on an algebra . If we have a set and a finite or (countably) infinite sequence of sets so that , then we have the inequality . To show this, we’ll invoke a very useful trick: if is a sequence of sets in an algebra , then there is a *pairwise disjoint* sequence of sets so that each and . Indeed, we define

That is, is the same as , and after that point each is everything in that hasn’t been covered already.

So we start with , and come up with a new sequence . The (disjoint) union of the is, like the union of the . Then, by the countable additivity of , we have . But the monotonicity says that , and also that , and so .

On the other hand, if is a pairwise disjoint and we have , then . Indeed, if is a finite sequence, then the union is in . Monotonicity and finite additivity then tell us that

However if is an infinite sequence, then what we just said applies to any finite subsequence. Then the sum is the limit of the sums . Each of these is less than or equal to , and so their limit must be as well.

[...] in , and if at least one of the has finite measure, then . Indeed, if has finite measure then by monotonicity, and thus the limit must have finite measure as well. Now is an increasing sequence, and we [...]

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[...] is, . Thus we must have . On the other hand, if is contained in the union of a sequence , then monotonicity tells us that , and thus . That is, must be equal to for sets ; as a set function, indeed [...]

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[...] it’s a Borel set. Further, monotonicity tells us [...]

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[...] set and pick another one . Then we see that , and so our hypothesis tells us that . Since , the subtractivity of tells us that , and we conclude that . That is, every measurable set that fits into must have [...]

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[...] We can rewrite this as . If is integrable, then and both have finite measure, and so is subtractive. Thus we can [...]

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[...] course, just as for measures, signed measures are finitely additive (which we used above) and thus subtractive. It won’t be monotone, in general, though; Given a set and a subset we can [...]

Pingback by Signed Measures « The Unapologetic Mathematician | June 22, 2010 |