Karhunen–Loève theorem expansion random variables

[cianfa72]

TL;DR Summary

About the integral involved in the definition of Karhunen–Loève theorem expansion random variable coefficients.

Hi,
in the Karhunen–Loève theorem's statement the random variables in the expansion are given by $$Z_k = \int_a^b X_te_k(t) \: dt$$
$X_t$ is a zero-mean square-integrable stochastic process defined over a probability space $(\Omega, F, P)$ and indexed over a closed and bounded interval $[a, b]$, with continuous covariance function $K_X(s, t)$.

$e_k(t)$ are elements of an orthonormal basis on $L^2([a, b])$ formed by the eigenfunctions of the linear operator $T_{K_{X}}$ with respective eigenvalues $\lambda_k$. As explained here $L^2([a, b])$ is the Hilbert space of measurable square functions defined on the space $([a,b], \mathcal B([a,b], P)$ that have finite measure. Note that $\mathcal B([a,b])$ is the Borel $\sigma$-algebra on the closed set $[a,b]$ and $P$ is the Lebesgue measure on it.

I'm confused about the meaning of the integral involved in the definition of random variables $Z_k$

Which kind of integral is this ?

Ps. I asked the same question on MSE without luck.

 

[cianfa72]

It seems that integrals used in the definition of expansion coefficients are just integrals over the realizations of the stochastic process in the interval $[a,b]$.

Every coefficient is thus a random variable defined on the same $(\Omega, F, P)$ probability space.

However the above definition requires that each function product of stochastic process's realizations (sample paths) times basis elements $e_k(t)$ has to be integrable over the closed bounded set $[a,b]$ -- see also this lecture.

In particular each of the above function products has to be a measurable function w.r.t. the Borel $\sigma$-algebra on $[a,b]$. Thus their integrals are actually Lebesgue integrals, I believe.