## The Banach Space of Totally Finite Signed Measures

Today we consider what happens when we’re working over a -algebra — so the whole space is measurable — and we restrict our attention to totally finite signed measures. These form a vector space, since the sum of two finite signed measures is again a finite signed measure, as is any scalar multiple (positive *or* negative) of a finite signed measure.

Now, it so happens that we can define a norm on this space. Indeed, taking the Jordan decomposition, we must have both and , and thus . We define , and use this as our norm. It’s straightforward to verify that , and that implies that is the zero measure. The triangle inequality takes a bit more work. We take a Hahn decomposition for and write

So we know that this defines a norm on our space.

But is this space, as asserted, a Banach space? Well, let’s say that is a Cauchy sequence of finite signed measures so that given any we have for all sufficiently large and . But this is larger than any , which itself is greater than . If is a positive measurable set then this shows that is kept small, and we find similar control over the measures of negative measurable sets. And so the sequence is always Cauchy, and hence convergent. It’s straightforward to show that the limiting set function will be a signed measure, and that we will have control over . And so the space of totally finite signed measures is indeed a Banach space.

I believe there’s a mistake in line 3 of the proof of the Triangle Inequality. Second term should actually be -(\mu + \nu) (X \cap B).

Comment by SW | October 27, 2010 |

I think you’re right, thanks. Does it look better now?

Comment by John Armstrong | October 27, 2010 |