# Defining the expectation of a measurable function with respect to a (non-probability) measure

The typical definition of expectation requires a probability space and a random variable

Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a probability space, $$(\mathsf{X}, \mathcal{X})$$ be a measurable space and $$\mathrm{X}:\Omega\to\mathsf{X}$$ be a random variable. The expectation of $$\mathrm{X}$$ is defined as $$\mathbb{E}[X] := \int_\Omega X \,\,d\mathbb{P}.$$

Can we relax this? Meaning, can I sensibly define the expectation of a measurable function, given a measure space $$(\Omega, \mathcal{F}, \mathbb{M})$$, where $$\mathbb{M}$$ is now simply a measure?

Notes:

1. I am using the following classical distintion between a random variable and a measurable function: every random variable is a measurable function, but not any measurable function is a random variable. For a measurable function to be a random variable, we need the existance of a probability measure on $$(\Omega, \mathcal{F})$$.
2. In theory, I know this is generalisable because we typically work with Lebesgue integrals and the Lebesgue measure is not a probability measure, in general. However, what condition on $$\mathbb{M}$$ do I need? Intuitively, my guess is that it needs to be a $$\sigma$$-finite measure.

# Summary

Let $$(\Omega, \mathsf{F}, \mathbb{M})$$ be any measure space, and let $$(\mathsf{X}, \|\cdot\|_\mathsf{X})$$ be any Banach space, where $$\|\cdot\|_{\mathsf{X}}$$ is the norm on $$\mathsf{X}$$. Let $$\mathcal{B}(\mathsf{X})$$ be the Borel sigma algebra on $$\mathsf{X}$$ that has been generated by the topology $$\tau_\mathsf{X}$$ which has been induced by the metric $$d(x, y) := \|x - y\|_{\mathsf{X}}$$ which has itself been induced by the norm $$\|\cdot\|_{\mathsf{X}}$$. Then we can define the $$L^p$$ spaces $$L^p(\Omega, \mathcal{F}, \mathbb{M}) := \left\{f:\Omega\to\mathsf{X}\left| f \text{ is } \mathcal{F}\text{-measurable and } \left(\int \|f(\omega)\|_{\mathsf{X}}^p \, \mathbb{M}(d\omega)\right)^{1/p} < \infty \right.\right\}, \qquad p\geq 1.$$

These exist both for finite and infinite-dimensional Banach spaces $$\mathsf{X}$$. The larger $$p$$, the more structure we are giving. Meaning that under mild regularity conditions, the spaces are nested $$L^q \subset L^p$$ for $$q > p$$. In particular, for $$p = 1$$ all functions in $$L^1$$ are Bochner integrable.

An integral is called expectation when $$\mathbb{M}$$ is a probability measure (or finite, I suppose, since by rescaling they are equivalent). Then expectations exist when the measurable function $$f$$ is in $$L^1$$.

• How are you defining expectation if $\mathbb{M}(\Omega) =\int_\Omega 1 \,\,d\mathbb{M} \not = 1$ ? Commented Feb 19 at 17:35
• Nobody should call such an object an "expectation", but any mesure theory textbook worth its salt should discuss abstract integration (which is what it seems you want). Commented Feb 19 at 17:36
• @BrianMoehring What I know is that we can talk about expectation when $$\int_{\Omega} |X| \, d\mathbb{M} < \infty$$ but this only works when $(\mathsf{X}, \mathcal{X}) = (\mathbb{R}, \mathcal{B}(\mathbb{R}))$, since the absolute value function $|\cdot|$ is only defined there. I don't know how to generalise this. My guess is that if $(\mathsf{X}, d)$ is a metric space, then one could require $$\int_{\Omega} \sqrt{d(X, X)} \, d\mathbb{M} < \infty.$$ I am literally just guessing now. Commented Feb 19 at 17:40
• @BrianMoehring I am confused as to why nobody should call it expectation. If we use the Lebesgue measure and the law of the unconscious statistician, we can write $$\mathbb{E}[f\circ X] = \int f(x) p(x) dx$$ where $dx$ is the Lebesgue measure and $p(x)$ is the Radon-Nikodym derivative of the distribution of $X$ with respect to $dx$. Commented Feb 19 at 17:53
• Indeed that's not general (but I find it interesting because it recasts some problems in a probabilistic setting - and makes integrals expectations and vice versa). That's why I just left a comment :) Commented Feb 19 at 19:10

For any measure space $$(X,\mathfrak{M},\mu)$$, and any finite-dimensional normed vector space $$V$$, one has a very standard theory of integration for maps $$f:X\to V$$ which are $$\mathfrak{M}$$-$$\mathcal{B}(V)$$ measurable ($$\mathcal{B}(V)$$ being the Borel $$\sigma$$-algebra on $$V$$). Literally open up any measure theory book (Folland/Rudin/Stein-Shakarchi etc) and read one of the first few chapters (ok Stein does abstract measure theory in chapter 6 or somehting, but Folland and Rudin do it in Chapters 1/2). One can then talk all about the $$L^p(\mu;V)$$ spaces. Folland and Rudin may seem to describe things only for $$V=\Bbb{C}$$, but the case of finite-dimensional $$V$$ follows immediately.

For a general Banach space $$V$$ as the target, one has the notion of Bochner integral; see Lang or Amann-Escher for the details (Folland outlines the basics in an exercise, and Rudin talks about integration $$V$$-valued integrals when $$V$$ is some type of topological vector space in his functional analysis text… though briefly).

Note that for 99% of the basic theory, there is no reason at all to make any restrictions on the measure space on your domain. We don’t need finiteness, or $$\sigma$$-finiteness or semifiniteness or completeness or anything. For more complicated ‘structure theorems’ like the Radon-Nikodym theorem or the Lebesgue decomposition, or Fubini-Tonelli’s theorem you’re going to want $$\sigma$$-finite. But the finiteness assumption can easily be removed for MANY things.

So, for any $$f\in L^1(\mu; V)$$, we can talk about its integral $$\int_Xf\,d\mu$$; this is a vector in $$V$$. However, if $$\mu$$ is not a probability measure, one should NOT refer to this as an expectation value. Why? It’s just bad terminology. Also from an intuitive/linguistic perspective, ‘expectation’ is supposed to mean ‘the average value of something’, and average values are supposed to be the sum divided by the number of things (i.e this division is meant to provide a probability measure). Well, if you have an infinite measure space, then you’ll be dividing by $$\infty$$, which is just bad, or more concretely, there’s no natural way to convert an infinite measure $$\mu$$ into a probability measure. However, (as mentioned in the comments) if $$0<\mu(X)<\infty$$, then you can easily rescale to get a probability measure.

• Thank you! By the way, I have not forgotten about this answer, but I am trying to clear the ideas in my head :) Commented Feb 20 at 12:09
• Let's see if I understood, I am going to write a few comments. The most general theory of integration requires any measure space $(\Omega, \mathcal{F}, \mathbb{M})$ and any Banach space $(\mathsf{X}, \mathcal{X})$ with norm $\|\cdot\|_\mathsf{X}$. Then given a $\mathcal{F}$-measurable function $f:\Omega\to\mathsf{X}$, it's integral exists $$\int f \, d\mathbb{M} = \int_{\Omega} f(\omega) \mathbb{M}(d\omega).$$ Commented Feb 20 at 13:51
• The integral above, to my knowledge, is not generalisable to more general measurable spaces $(\mathsf{X}, \mathcal{X})$. Now, with the exact same set up as above and $p \geq 1$, one can define $L^p$ spaces $$L^p(\Omega, \mathcal{F}, \mathbb{M}) := \left\{f:\Omega\to\mathsf{X}\left| f \text{ is } \mathcal{F}\text{-measurable and } \left(\int \|f(\omega)\|^p \mathbb{M}(d\omega) \right)^{1/2} < \infty \right.\right\}.$$ This space contains fewer functions than in my earlier comment, because not all measurable functions satisfy the other condition. Commented Feb 20 at 13:51
• first comment: “given a $\mathcal{F}$-measurable function, its integral $\int f\,dM$” exists. NO. You need $f\in L^1$ to define the integral. Commented Feb 20 at 14:18
• second thing is, your target space cannot be an arbitrary measure space. You need more structure, like a normed vector space (ideally Banach, and for convenience, finite-dimensionality). Otherwise, integrals are pretty much meaningless (perhaps you could get away with some sort of abelian group structure or something, but I haven’t really seen such things). Commented Feb 20 at 14:20