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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.

Can somebody help or provide some reference about this approach?

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    $\begingroup$ 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. $\endgroup$ – lulu Jan 13 at 18:00
  • $\begingroup$ Thank you for your answer. I think I found what I was looking for though. It seems to me that the answer must be looked into the distribution theory. $\endgroup$ – roddik Jan 13 at 18:23
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    $\begingroup$ You don't need to go into distribution theory. Instead of using test functions to define probability, just use the characteristic function. This is how Kolmogorov did it in the 1930s (if i recall the approximate date correctly) $\endgroup$ – nomen Jan 13 at 19:13
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    $\begingroup$ 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 $\endgroup$ – Adam Jan 13 at 21:15
  • $\begingroup$ Adam what's the most general setting for it to be possible? I mean it's just that continuous functions always appropximate characteristic functions, and then I can use dominated convergence? $\endgroup$ – roddik Jan 13 at 21:29
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https://en.wikipedia.org/wiki/Riesz%E2%80%93Markov%E2%80%93Kakutani_representation_theorem Allows you to define a linear funtional on a suitable space of continuous functions, and deduce that it must arise from integration against a measure.

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  • $\begingroup$ Thank you for your answer, you must be right. Do you think one can see Riesz theorem as a particular case of distribution theory here? $\endgroup$ – roddik Jan 13 at 23:36
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    $\begingroup$ Yes. Note that not all distributions arise from measures. $\endgroup$ – Yuval Peres Jan 13 at 23:50

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