# Definition of a probability measure via expected value

I have seen quite a few times doing something like this:

I have a probability space $$(\Omega, \mathbb{P})$$ where $$\mathbb{P}$$ has to be defined.

I would expect them do define it on the measurable sets: given $$E\subset\Omega$$, $$\mathbb{P}(E)=...$$

Instead, they bring in something someone called "test functions". It works like this: for every test function $$f$$ (if I understood well, they usually request $$f$$ to be continuous), then

$$\begin{equation} \mathbb{E}(f(X))=... \tag{1} \end{equation}$$

where $$X$$ is a random variable on $$\Omega$$ with distribution $$\mathbb{P}$$.

I have two issues with this approach:

1. If I'm free enough to choose $$f$$, then if $$f={1_E}$$ charcteristic function, I get that $$\mathbb{E}(f(X))=\int 1_E(x)d\mathbb{P}= \int_E d\mathbb{P} = \mathbb{P}(E)$$ What's the need for the definition (1), then?

2. If $$f$$ has to be continuous, then I can't choose it characteristic. So I don't even see how the (1) definition even works.

• Well, it's hard to say much without an example of what you are talking about, but I'd assume it's because they can't explicitly define the probability of each set in the $\sigma$ algebra. That is, they know enough about the distribution to compute certain things about it, and maybe even enough to prove that the thing actually is a probability distribution, but the computations are too difficult to make it practical to compute the probabilities attached to some specific sets. – lulu Jan 13 at 18:00
• Addressing question #2 in your original post, characteristic functions aren't continuous but they can be approximated by continuous functions. So usually if the measure $\mathbb{P}$ is defined explicitly for continuous test functions, it can extend to be defined on characteristic functions by taking a limit – Adam Jan 13 at 21:15