Definition of closed Ideal in the space of finite measures.

Question

Let $M(\mathbb{R})$ be the space of finite signed measure on $(\mathbb{R})$. What is the norm defined on this space? From theorem 1.3.5 in Rudin Fourier analysis on groups. Interscience Tracts in Pure and Applies Mathematics, It is proved that $L^{1}(\mathbb{R})$ is a closed ideal in the space $M(\mathbb{R})$. What is the definition of closed ideal means here?

Answer 1

A fairly standard norm in this case would be the total variation; in particular, for $\nu \in M(\mathbb{R})$, we define $$\| \nu\| = \lvert \nu \rvert (\mathbb{R}).$$ If I'm not mistaken, a closed ideal is simply an ideal (very much like those from ring theory) which is closed in the topology induced by the norm on $M(\mathbb{R})$.

Edit: To flesh out a few more details, first note that we can embed $L^1(\mathbb{R}) \hookrightarrow M(\mathbb{R})$ via the identification $f \mapsto \nu$, where $d\nu = fdm$ (or, equivalently, $\nu(E) = \int_E f \ dm$), where $m$ denotes Lebesgue measure. As Rudin notes in his text, multiplication of measures is defined by convolution. In particular, for $\lambda, \mu \in M(\mathbb{R})$, we define $$(\lambda * \mu)(E) = (\lambda \times \mu)(E_2),$$ where, in our case, $E_2 = \{ (x,y) \in \mathbb{R}^2 : x+y \in E \}$. We are almost ready to show that $L^1$ is an ideal of $M$. We just need to think about how to show that $\mu * \nu \in L^1$ for $\mu \in M$ and $\nu \in L^1$. Recall that the Radon-Nikodym theorem implies that if $\mu \in M$ is absolutely continuous with respect to $m$, then $d\mu = g dm$ for some $g \in L^1(\mathbb{R})$.

We can now show that $L^1$ is an ideal of $M$ by taking arbitrary $\nu "\in" L^1$ (ie, $d\nu = fdm$ for $f \in L^1$) and showing that $\mu * \nu \ll m$ for all $\mu \in M$ (see Rudin for the proof). This, of course, implies (again by Radon-Nikodym) that $d(\mu * \nu) = gdm$ for some $L^1$ function $g$. Therefore, $\mu * \nu \in L^1$ and so $L^1$ is an ideal of $M$. That $L^1$ is complete implies that $L^1$ is closed in $M$ and therefore $L^1$ is in fact a closed ideal of $M$.