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Let $X$ denote a measurable space, that is, a set equipped with a $\sigma$-algebra $\Sigma(X)$. Let $M(X)$ denote the space of real-valued measures over $X$. This is a vector space over the real numbers, since the sum of two measures is again a measure, as is a scalar multiple of a measure. I would like to know the most general setting for which $M(X)$ is a measurable vector space.

Does $M(X)$ admit a canonical choice of $\sigma$-algebra, turning it into a measurable space?

If the answer is "no", then what about the setting where $X$ is a localizable measurable space?

If the answer is again "no", then what is the most general setting so that $M(X)$ is admits a canonical measurable structure?

Most generally, what is the largest subcategory $\mathcal C$ of $\mathbf{Meas}$ so that $M : \mathcal C \to \mathcal C$ is an endofunctor?

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I guess you want to add some property to your "canonical" measurable structure, otherwise you can always use e.g. the set of all parts of $M(X)$ as $\sigma$-algebra. – Benoît Kloeckner Dec 24 '12 at 15:02
Since you seem to be excited about the approach emphasized by Dmitri Pavlov, I'd like to point out to you that a lot of what DP is talking about in his answers is discussed in volume 3 of Fremlin's measure theory: (mostly in the notes and exercises). Instead of working with von Neumann algebras, Fremlin emphasizes Boolean and measure algebras and Riesz spaces where the translation to/from classical measure theory, probability theory and ergodic theory seems to become clearer, at least to someone with a more classical analytic slant like me. – Theo Buehler Dec 26 '12 at 15:10
@Theo, thank you! – Tom LaGatta Dec 26 '12 at 18:45
up vote 5 down vote accepted

Let $(X,\Sigma)$ be the measurable space. I think the sigma-algebra on $\mathcal M$ that you want is this. The least sigma-algebra so that for all $A \in \Sigma$, the map $\mu \mapsto \mu(A)$ is measurable.

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Thanks, Gerald! This is absolutely right. – Tom LaGatta Dec 24 '12 at 18:50
I think it is worth pointing out that the corresponding endofunctor takes a measurable map $f:X\to Y$ to the map $\mu\mapsto \mu\circ f^{-1}$ from $M(X)$ to $M(Y)$. – Michael Greinecker Dec 25 '12 at 15:04
@Gerald Edgar or @Michael Greinecker, I have another question for you folks. In the same vein, one may also consider the minimal topology so that the evaluation functions $\mu \mapsto \mu(A)$ are continuous; this topology generates a Borel $\sigma$-algebra on the space. How is this $\sigma$-algebra related to the one that Gerald proposed? – Tom LaGatta Dec 26 '12 at 2:45
@Tom: that uninteresting toplogy is discrete. So of course its Borel sigma-algebra is the power set. – Gerald Edgar Dec 26 '12 at 14:23
@Gerald, that's good to hear. Thank you very much for your answers. – Tom LaGatta Dec 26 '12 at 18:45

First try it with $X$ a point. Then the space of measures is $\mathbb R^+$.

If you want the measure to be bounded, you give up at this point. I think the right thing to do here is to look at the space of probability measures. Then on a finite set of points you get a simplex, which has a canonical measure. However, once you set $X = \mathbb N$ (with the discrete $\sigma$-algebra, which is also the unique $\sigma$-\algebra that separates the points), it is clear that there cannot be a natural probability measure on the space of probability measures, because it would have to be invariant under the permutation action, so the expected size of each point must be equal, which is clearly impossible.

So you have to give up boundedness -some sets of measures will have infinite measure. For a single point, Lebesque measure is the obvious choice. For a finite set of points, Lebesgue measure still makes sense. For an infinite set of points, I have no idea what to do - I'm not even sure which sets of measures should have infinite measure.

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We're not asking for a measure on $\mathcal{M}(X)$, are we? Just a $\sigma$-algebra. – Nate Eldredge Dec 24 '12 at 6:19
And $\mathcal{M}(X)$ here includes all real-valued (i.e. signed) measures on $X$; they need not be positive. So if $X$ is a point then $\mathcal{M}(X)$ is $\mathbb{R}$. – Nate Eldredge Dec 24 '12 at 6:23
Thanks Will. Like Nate says, I'm looking at not-necessarily-bounded, real-valued measures on $X$. As you point out, the space of measures is invariably too big for there to be a canonical choice of a measure. Nonetheless, I'm hoping that there is a suitably general assumption on $X$ to guarantee a canonical choice of $\sigma$-algebra on $M(X)$. – Tom LaGatta Dec 24 '12 at 8:19
Oh, I completely misread the question. I think Gerald Edgar's answer is correct. – Will Sawin Dec 24 '12 at 16:09

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