# Topology on the space of universally integrable functions

Let $X$ be a compact space. Let us call a function $f:X\to {\mathbb C}$ universally integrable if it is integrable with respect to each regular Borel measure $\mu$ on $X$ (i.e. a positive functional on ${\mathcal C}(X)$, according to Riesz–Markov–Kakutani representation theorem). Let us denote by ${\mathcal U}(X)$ the space of all universally integrable functions on $X$.

My questions:

1. What is known about ${\mathcal U}(X)$?
2. Is it an algebra (with respect to the pointwise multiplication)?
3. Did anybody try to find a natural topology on ${\mathcal U}(X)$?

EDIT. This was put on hold as too broad, so I add some details.

1. I would be grateful for any references where what I am asking is discussed.

2. Nate Eldredge already answered the second of the three questions here, so there is no need to discuss this further.

3. As to the third one, the correct specification will be the following. As I wrote in comments to Eldredge’s answer, I believe, there is a topology on ${\mathcal U}(X)$ that turns it into an involutive stereotype algebra (see here, or here, or here) such that the involutive spectrum (see again here) of ${\mathcal U}(X)$ as a set is equal to $X$ (like in the case of ${\mathcal C}(X)$):
$$\text{Spec}\ {\mathcal U}(X)=X$$
Of course, such a topology on ${\mathcal U}(X)$ must be much weaker than the $C^*$-topology that Nate Eldredge suggests.

Am I right?

As a hypothesis, the following construction can be discussed: to each regular Borel measure $\mu$ on $X$ one can associate a mapping $\varPhi_\mu:{\mathcal U}(X)\to L(\mu)$. So ${\mathcal U}(X)$ can be naturally embedded into the direct product of the spaces $L(\mu)$.
$${\mathcal U}(X)\to \prod_{\mu} L(\mu).$$
I am now thinking about the topology generated on ${\mathcal U}(X)$ by this embedding (i.e. the initial topology generated by the mappings $\varPhi_\mu$). Is it possible that this is what I need?

#### Solutions Collecting From Web of "Topology on the space of universally integrable functions"

Let $M(X)$ denote the set of all regular Borel measures on $X$. (Note in particular that, by your definition, only finite measures are included.)

For each $\mu \in M(X)$, let $\mathcal{B}_\mu$ denote the completion of the Borel $\sigma$-algebra $\mathcal{B}$ with respect to the measure $\mu$ (that is, $A \in \mathcal{B}_\mu$ iff there exist Borel sets $B_1, B_2$ with $B_1 \subset A \subset B_2$ and $\mu(B_1) = \mu(B_2)$. It’s well known that $\mathcal{B}_\mu$ is a $\sigma$-algebra. So let $\mathcal{B}_u = \bigcap_{\mu \in M(X)} \mathcal{B}_\mu$, which is also a $\sigma$-algebra. (In the context of Polish spaces, this is the $\sigma$-algebra of universally measurable sets.) We can then say:

A function $f$ is universally integrable iff it is bounded and $\mathcal{B}_u$-measurable.

Clearly any bounded $\mathcal{B}_u$-measurable function is universally integrable. Conversely, if $f$ is to be integrable with respect to $\mu$, it must be $\mathcal{B}_\mu$-measurable, so every universally integrable function is $\mathcal{B}_u$-measurable. Now I claim any universally integrable function must be bounded. To prove the contrapositive, suppose $f$ is unbounded, so that for each $n$ there exists $x_n \in X$ with $|f_n(x)| \ge 2^n$. Let $\mu$ be the probability measure which assigns measure $2^{-n}$ to the point $x_n$. Then $\int |f|\,d\mu = \infty$, so $f$ is not universally integrable.

In particular, $\mathcal{U}(X)$ is an algebra, since a sum or product of bounded $\mathcal{B}_u$-measurable functions is again bounded and $\mathcal{B}_u$-measurable.

A simple and natural topology on $\mathcal{U}(X)$ is the one induced by the uniform norm $\|f\| = \sup_{x \in X} |f(x)|$. This makes $\mathcal{U}(X)$ into a Banach space, and moreover, a $C^*$ algebra.