# The Unapologetic Mathematician

## 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.

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Posted by John Armstrong | Analysis, Measure Theory

## 2 Comments	»

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 | October 27, 2010 | Reply

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