The LOTUS as stated is something like this (for continuous case):
Let $X$ be a continuous real-valued random variable that follows some density function $f(x)$. Let $g(X)$ be some real-valued measurable function. Then we have
$$\mathbb{E}[g(X)] = \int_\mathbb{R}g(x)f(x)\ \mathrm{d}x.$$
My question is in the domain that $g$ is defined on. When we have $g(X)$ as a function of a random variable, what is its domain? My intuition is $\mathbb{R}$ because (implicitly perhaps) $X$ as a random variable is some measurable function from $\Omega \to\mathbb{R}$. So in a sense, $g:\mathbb{R}\to\mathbb{R}$ and law is essentially show that
$$\int_\Omega g(X(\omega))\ \mathrm{d}\mathbb{P} = \int_\mathbb{R} g(x)f(x)\ \mathrm{d}m$$
where $m$ is the Lebesgue measure on $\mathbb{R}$. Is that a correct interpretation?
If the above interpretation is correct, then would the following also be for LOTUS of two random variables. Suppose $X,Y$ are independent real-valued continuous random variables (real valued measurable functions from the same $\Omega$) with $f_X(x),\ f_Y(y)$ as their densities. Let $h(X,Y)$ be some real-valued measurable function. By the LOTUS, we must then have
$$\mathbb{E}[h(X,Y)] = \int_{\mathbb{R}^2} h(x,y) f_X(x) f_Y(y) \mathrm{d}m_2$$
where $m_2$ is the Lebesgue measure on $\mathbb{R}^2$. If my interpretation in the previous part is correct, then the domain of $h$ should be $h:\mathbb{R}^2\to\mathbb{R}$ and the law shows that
$$\int_\Omega h(X(\omega),Y(\omega)) \mathrm{d}\mathbb{P} = \int_{\mathbb{R}^2} h(x,y) f_X(x) f_Y(y) \mathrm{d}m_2.$$
Is this a correct understanding / extension?
Lastly, I'm looking to find a measure-theoretic proof of this law so I can better grasp its derivation (in contrast to a lot of the proofs that assume some monotonic behavior of $g$). Does anybody have a reference / could provide one? Thanks!
Best Answer
Your understanding is correct. The proof is completely straight-forward. Let $\mu$ be a measure on $\Omega$ and let $X:\Omega\to E$ be a measurable map into some target space. Define $X^*\mu$ to be the measure on $E$ given by
$$ X^*\mu(A)=\mu(\{X\in A\}) $$
Let $g=\sum_{i=1}^n c_i1_{A_i}$ be a positive, simple function. Then,
\begin{align} \int_{\Omega} g(X)\textrm{d}\mu&=\int_{\{X\not \in \cup_{i=1}^n A_i\}} g(X)\textrm{d}\mu+\sum_{i=1}^n \int_{\{X\in A_i\}} g(X)\textrm{d}\mu\\ &=\sum_{i=1}^n c_i \mu(\{X\in A_i\})\\ &=\sum_{i=1}^n c_i X^*\mu(A_i)\\ &=\int_{E} g\textrm{d} X^*\mu \end{align} Thus, applying monotone convergence, we get the equality for all positive, measurable $g$. Finally, for general $g\in L^1(X^*\mu),$ we get the equality by decomposing into positive and negative parts.