The Banach Space of Totally Finite Signed Measures

Today we consider what happens when we’re working over a $\sigma$ -algebra — so the whole space $X$ 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 $\mu^+(X)<\infty$ and $\mu^-(X)<\infty$ , and thus $\lvert\mu\rvert(X)<\infty$ . We define $\lVert\mu\rVert=\lvert\mu\rvert(X)$ , and use this as our norm. It’s straightforward to verify that $\lVert c\mu\rVert=\lvert c\rvert\lVert\mu\rVert$ , and that $\lVert\mu\rVert=0$ implies that $\mu$ is the zero measure. The triangle inequality takes a bit more work. We take a Hahn decomposition $X=A\uplus B$ for $\mu+\nu$ and write

$\displaystyle\begin{aligned}\lVert\mu+\nu\rVert&=\lvert\mu+\nu\rvert(X)\\&=(\mu+\nu)^+(X)+(\mu+\nu)^-(X)\\&=(\mu+\nu)(X\cap A)-(\mu+\nu)(X\cap B)\\&=\mu(A)+\nu(A)-\mu(B)-\nu(B)\\&\leq\lvert\mu\rvert(A)+\lvert\mu\rvert(B)+\lvert\nu\rvert(A)+\lvert\nu\rvert(B)\\&=\lvert\mu\rvert(X)+\lvert\nu\rvert(X)\\&=\lVert\mu\rVert+\lVert\nu\rVert\end{aligned}$

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 $\{\mu_n\}$ is a Cauchy sequence of finite signed measures so that given any $\epsilon>0$ we have $\lvert\mu_n-\mu_m\rvert(X)<\epsilon$ for all sufficiently large $m$ and $n$ . But this is larger than any $\lvert\mu_n-\mu_m\rvert(E)$ , which itself is greater than $\lvert\lvert\mu_n\rvert(E)-\lvert\mu_m\rvert(E)\rvert$ . If $E\subseteq A$ is a positive measurable set then this shows that $\lvert\mu_n(E)-\mu_m(E)\rvert$ is kept small, and we find similar control over the measures of negative measurable sets. And so the sequence $\{\mu_n(E)\}$ is always Cauchy, and hence convergent. It’s straightforward to show that the limiting set function $\mu$ will be a signed measure, and that we will have control over $\lVert\mu_n-\mu\rVert$ . And so the space of totally finite signed measures is indeed a Banach space.

June 28, 2010 - Posted by John Armstrong | Analysis, Measure Theory

4 Comments »

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 | Reply
- I think you’re right, thanks. Does it look better now?
  
  Comment by John Armstrong | October 27, 2010 | Reply
Hi John. Thanks for the blog. I was looking for a citable source for my thesis. Could you suggest a book where I can find this clearly stated and also find other signed measure related theory.

Comment by Sanket | April 13, 2019 | Reply
- I’m glad it’s helpful. Unfortunately, I don’t have a great answer for you.
  
  I was never really an analyst, and over the last ten years I’ve forgotten what further result I even intended this post to support. If I could remember, it might help pin down my own references. I’d been thinking it would come up in the Radon-Nikodym theorem for signed measures, but I don’t seem to have used the fact that they form a Banach space there.
  
  I think, in general, that I learned most of this from Rudin’s Real and Complex Analysis, but I can’t seem to find where he proves the Banach property. It’s possible that I just needed it to be a normed linear space, but saw that I could get Banach and went ahead to show that on my own; it’s kind of an exercise, really. If the Rudin reference helps, great, but past that I have no real idea what other citation to use.
  
  Comment by John Armstrong | April 13, 2019 | Reply

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