Measures with Compact Support in Complex Analysis: Finiteness Assumptions

[lark]

I was reading in a book, says \mu is a measure with compact support K in C, meaning \mu(U)=0 for U\cap K=0..
Is \mu(K) assumed to be finite in this case?
It doesn't say in the book, but they make a statement which is true if that's so. Is there usually some assumption about measures being finite on compact sets?
I know complex measures are assumed to be finite. But in C you would usually be integrating over a positive measure (which could be infinite normally?)
thanks
Laura

 

[HallsofIvy]

Yes, a measure of a compact set is finite.

 

[jostpuur]

Yes, a measure of a compact set is finite.

For sure? I see no reason for a measure of a compact set to be finite in general. Are you assuming something?

 

[lark]

Yes, a measure of a compact set is finite.

That would normally be assumed for a measure? This is some arbitrary measure, of course, not necessarily Lebesgue measure. All that Rudin's RACA says is that a measure's range is
[0,\infty].
Laura

 

[morphism]

No, it's not usually assumed that all measures are finite on compact sets. On the other hand, most classes of measures that pop up in practice are assumed to have this property. For instance, most authors have this requirement in the definition of a Borel measure. One reason for doing so is that the measures one gets from the Riesz representation theorem are finite on compact sets.

It doesn't say in the book, but they make a statement which is true if that's so.

Can you elaborate on this?

 

[lark]

Can you elaborate on this?

Something about differentiating inside the integral sign. It's easy to see if the measure is finite.
Laura

 

[lark]

Lemma 2.2 in Lang's Real and Functional Analysis is about differentiating under the integral sign, and it doesn't assume the measure's finite on compact sets. But it makes a lot of assumptions to make up for that. The book's in books.google.com.
So I guess, the book I'm reading does assume that measures are finite on compact sets, because they aren't making those other assumptions.
Laura

A few years ago I found a mistake in one of the problems in Lang's Algebra (I'm sure there are a lot of mistakes!) and I sent him a counterexample. I got a note back saying the problem would be taken out. But I was just thrilled and delighted to find out recently that he published my counterexample in the new edition, in place of the original problem! I wanted to send him a thank you note. But then I found out Lang had died ... and I couldn't send a thank you note ...