# Expectation value - does it need to be on a Banach space?

I came across the following definition of a exception value of a random variable $X$ in (Lord, Powell and Shardlow, 2014;pg139; emphasis not mine):

Let $X$ be a Banach space-valued random variable on the probability space $(\Omega,\mathcal{F},\Bbb{P})$. If $X$ is integrable, the expectation of the $X$ is $$\Bbb{E}[X]:= \int_\Omega X(\omega)d\Bbb{P}(\omega), \tag{4.1}$$

Why is there a resection to Banach space-random variables here? Why won't only old measurable space do?

• It's Banach-space valued, it's still defined on a normal probability space. Commented Aug 7, 2017 at 15:03
• So we think of this $X$ as a "random element" of the Banach space. Just as an ordinary random variable is a "random real number". Commented Aug 7, 2017 at 15:04
• @1524 I know, but from the definitions I have seen random variables go from measurable spaces to probability spaces. Commented Aug 7, 2017 at 15:05
• to define a $L_1$ function and thus expectation you need a norm. Commented Aug 7, 2017 at 15:10
• Sorry, misunderstood. You use the borel sigma algebra on the Banach space Commented Aug 7, 2017 at 15:34

There are two objects being considered here - the sample space $\Omega$, and an unnamed state space which I shall call $E$, which in this case is assumed to be a Banach space. (It is fairly unhelpful that the author does not give the state space a name, and probably contributes to your confusion.) Our random variable is $X:\Omega\to E$.
(Edit, to make something clear from the comments: whenever you have a topological space, you also have a measurable space by considering the Borel $\sigma$-field. Unless otherwise specified, you should always assume this is the $\sigma$-field considered when we have a topological space being considered in measure/probability theory.)
In this case, we are looking at the expectation of a random variable, which we know is simply the integral of the measurable function $X$. But what is the integral? Any definition of an integral requires linear combinations and limits, so at the least we will require $E$ to be a topological vector space. Assuming $E$ is Banach simply makes things a little cleaner - the topology of a normed vector space is easier to understand, and making it complete (i.e. Banach) merely guarantees that the space possesses enough limits to avoid running into problems.