The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Analysis, Measure Theory


  1. 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

  2. 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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