In this answer (to my own question), I show how the elementary definition of an expectation of a continuous random variable is consistent with the abstract measure theoretic formulation, and also mention the appearance of the Radon-Nikodym derivative. I heavily use notation from David William's \textit{Probability with Martingales} [PW], and also restate his Lemma from Section 6.12 along with results from Section 5.14 in [PW] with detailed proof.

In this answer (to my own question), I show how the elementary definition of an expectation of a continuous random variable is consistent with the abstract measure theoretic formulation, and also mention the appearance of the Radon-Nikodym derivative. I heavily use notation from David William's Probability with Martingales [PW], and also restate his Lemma from Section 6.12 along with results from Section 5.14 in [PW] with detailed proof.

In this answer (to my own question), I show how the elementary definition of an expectation of a continuous random variable is consistent with the abstract measure theoretic formulation, and also mention the appearance of the Radon-Nikodym derivative. I use notation from David William's Probability with Martingales [PW], and also restate his Lemma in Section 6.12 of [PW] (as my Proposition 1, Proposition 3) and add a detailed proof. Proposition 2 is sketched out in Section 5.14 of [PW] but in a more general context.

• Let $(\Omega,\F,\P)$ be a probability space, where $\Omega$ is a non-empty set, $\F$ is a $\sigma$-algebra on $\Omega$, and $\P\colon\F\to \R$ is a probability measure.
• Let $\B$ be the Borel $\sigma$-algebra on $\R$.
• Let $m\F$ be the set of random variables, i.e., $$m\F=\{X\colon\Omega\to\R:X\text{ is }\F/\B\text{ measurable}\}.$$
• Let $m\B$ be the set of real-valued measurable functions on $\R$, i.e. $$m\B=\{h\colon\R\to\R:h\text{ is }\B/\B\text{ measurable}\}.$$

Before continuing, fix an $X\in m\F$.

• Let $\Lambda_X$ be the law of $X$, which is the probability measure $\Lambda_X\colon\B\to\R$ such that $\Lambda_X(B)=\P(X\in B)$ for all $B\in\B$.
• Use $\E$ to denote integration of elements of $m\F$ with respect to the measure $\P$, e.g. $\E[X]=\int_{\Omega} X \d \P$.
• Let $\mathbf{L}$ be the integration operator (analagous to $\E$) of elements of $m\B$ with respect to $\Lambda_X$. That is, $\mathbf{L}(h)=\int_{\R} h\d\Lambda_X$ for each $h\in m\B$, where integration is with respect to $\Lambda_X$ instead of $\P$.
• Set $\L^1(\Omega,\F,\P) = \{X \in m\F:\E|X|<\infty\},$ the set of integrable random variables.
• Let $\L^1(\R,\B,\Lambda_{X})= \{h\in m\B: \int_{\R}|h|\d\Lambda_X<\infty \}$, the set of integrable functions on $\R$.
• Let $\mathrm{Leb}$ denote the Lebesgue measure, and we write $\int_{\R} h \d\mathrm{Leb}$ or $\int_{\R}h(x)\d x$ to denote the Lebesgue integral of $h\in m\B$.

Note that we are using two different notions of integral for $\E$ and $\Lambda_X$. That is, just as we define integration of $\F/\B$ measurable functions $X\colon\Omega\to\R$ with respect to the probability measure $\P$, we may define integration of $\B/\B$ measurable functions $h\colon\R\to\R$ with respect to the probability measure $\Lambda_X$. Instead of $\int_{\Omega} X\d \P$, we write $\int_{\R} h \d \Lambda_X$. Observe that $\mathbf{L}(h)$ relates to $X$ precisely through the measure $\Lambda_X$ we are integrating with respect to, while $\P$ has no connection a priori to $X$ in $\E[h(X)]$.

Our goal is to establish that the $\mathbf{L}$ and $\E$ operators agree (Proposition 1), and in the special case that the random variable admits a density, $\mathbf{L}$ has the standard elementary form $\int_{\R}xf(x)\d x$.

Proposition 1: Suppose $h\in m\B$. Then $h(X)\in\L^1(\Omega,\F,\P)$ iff $h\in\L^1(\R,\B,\Lambda_X)$ and $\E h(X)= \mathbf{L}(h)$.

Proof. First, suppose $h=1_B$ for $B\in\B$. Then $$\E h(X)=\E[1_{\{X\in B\}}] =P(X\in B),$$ and $$\mathbf{L}(h) =\int_{\R}1_B \d\Lambda_X = \int_B \d\Lambda_X = \Lambda_X(B) :=P(X\in B).$$

Next, let $h$ be a non-negative function in $m\B$. Then there is a sequence $(h_n)$ of non-negative simple functions in $m\B$ such that $h_n\uparrow h$. By the Monotone Convergence Theorem applied to each distinct measure space, $$\E[h(X)]=\lim_n \E[h_n(X)] = \lim_n \mathbf{L}(h_n)=\mathbf{L}(h),$$ since $\E[h_n(X)]=\mathbf{L}(h_n)$ for each $n$.

Next, let's obtain a familiar expression for $\mathbf{L}(h)$ when $X$ admits a PDF $f_X$, i.e., $$\Lambda_X(B)=P(X\in B)=\int_B f_X\d \mathrm{Leb} = \int_B f_X(x)\d x.$$ In the notation of [PW], this means $\Lambda_X$ has measure $f_X\mathrm{Leb}$, $\Lambda_X$ has density $f_X$ relative to $\mathrm{Leb}$, and we write $\frac{\d \Lambda_X}{\d \mathrm{Leb}}=f_X$. The PDF $f_X$ is called the Radon-Nikodym derivative of $\Lambda_X$ relative to $\mathrm{Leb}$ on $(\R,\B)$. Our precise formulation is below, but more suggestively, we demonstrate that $$\int_{\R} h(x)f_X(x)\d x=\int_{\R} h f_X\d\mathrm{Leb} = \int_{\R} h \frac{\d \Lambda_X}{\d \mathrm{Leb}}\d\mathrm{Leb} = \int_{\R} h\d\Lambda_X.$$

Suppose $h\in\L^1(\R,\B,\Lambda_X)$ with $h\ge0$. Then there is an increasing sequence $(h_n)$ of non-negative simple functions such that $h_n\uparrow h$. By the Monotone Convergence Theorem, \begin{align*} \int_{\R}h\d\Lambda_X =\lim_n\int_{\R}h_n\d\Lambda_X=\lim_n\int_{\R} h_nf_X\d\mathrm{Leb}= \int_{\R} hf_X\d\mathrm{Leb}, \end{align*} yielding the claim. For general $h\in\L^1(\R,\B,\Lambda_X)$, writing $h=h^{+}-h^{-1}$ finishes the proof.

In this answer (to my own question), I show how the elementary definition of an expectation of a continuous random variable is consistent with the abstract measure theoretic formulation, and also mention the appearance of the Radon-Nikodym derivative. I heavily use notation from David William's \textit{Probability with Martingales} [PW], and also restate his Lemma from Section 6.12 along with results from Section 5.14 in [PW] with detailed proof.

• Let $(\Omega,\F,\P)$ be a probability space, where $\Omega$ is a non-empty set, $\F$ is a $\sigma$-algebra on $\Omega$, and $\P\colon\F\to \R$ is a probability measure.
• Let $\B$ be the Borel $\sigma$-algebra on $\R$.
• Let $m\F$ be the set of random variables, i.e., $$m\F=\{X\colon\Omega\to\R:X\text{ is }\F/\B\text{ measurable}\}.$$
• Let $m\B$ be the set of real-valued measurable functions on $\R$, i.e. $$m\B=\{h\colon\R\to\R:h\text{ is }\B/\B\text{ measurable}\}.$$
• Let $\Lambda_X$ be the law of $X$, which is the probability measure $\Lambda_X\colon\B\to\R$ such that $\Lambda_X(B)=\P(X\in B)$ for all $B\in\B$.
• Use $\E$ to denote integration of elements of $m\F$ with respect to the measure $\P$, e.g. $\E[X]=\int_{\Omega} X \d \P$.
• Let $\mathbf{L}_X$ be the integration operator (analagous to $\E$) of elements of $m\B$ with respect to $\Lambda_X$. That is, $\mathbf{L}_X(h)=\int_{\R} h\d\Lambda_X$ for each $h\in m\B$, where integration is with respect to $\Lambda_X$ instead of $\P$.
• Set $\L^1(\Omega,\F,\P) = \{X \in m\F:\E|X|<\infty\},$ the set of integrable random variables with respect to $\P$.
• Let $\L^1(\R,\B,\Lambda_{X})= \{h\in m\B: \int_{\R}|h|\d\Lambda_X<\infty \}$, the set of integrable functions on $\R$ with respect to $\Lambda_X$.
• Let $\mathrm{Leb}$ denote the Lebesgue measure, and we write $\int_{\R} h \d\mathrm{Leb}$ or $\int_{\R}h(x)\d x$ to denote the Lebesgue integral of $h\in m\B$.

Note that we are using two different notions of integral for $\E$ and $\Lambda_X$. That is, just as we define integration of $\F/\B$ measurable functions $X\colon\Omega\to\R$ with respect to the probability measure $\P$, we may define integration of $\B/\B$ measurable functions $h\colon\R\to\R$ with respect to the probability measure $\Lambda_X$. Instead of $\int_{\Omega} X\d \P$, we write $\int_{\R} h \d \Lambda_X$. Observe that $\mathbf{L}_X(h)$ relates to $X$ precisely through the measure $\Lambda_X$ we are integrating with respect to, while $\P$ has no connection a priori to $X$ in $\E[h(X)]$.

Our goal is to establish that the $\mathbf{L}_X$ and $\E$ operators agree (Proposition 1), i.e., $\int_{\R} h\d\Lambda_X = \int_{\Omega}h(X)\d\P$, and in the special case that the random variable admits a density, $\mathbf{L}_X$ has the standard elementary form $\int_{\R}xf(x)\d x$ (Proposition 3).

Proposition 1: Suppose $h\in m\B$. Then $h(X)\in\L^1(\Omega,\F,\P)$ iff $h\in\L^1(\R,\B,\Lambda_X)$ and $\E h(X)= \mathbf{L}_X(h)$, i.e. $\int_{\R} h\d\Lambda_X = \int_{\Omega}h(X)\d\P$.

Proof. First, suppose $h=1_B$ for $B\in\B$. Then $$\E h(X)=\E[1_{\{X\in B\}}] =P(X\in B),$$ and $$\mathbf{L}_X(h) =\int_{\R}1_B \d\Lambda_X = \int_B \d\Lambda_X = \Lambda_X(B) :=P(X\in B).$$

Next, let $h$ be a non-negative function in $m\B$. Then there is a sequence $(h_n)$ of non-negative simple functions in $m\B$ such that $h_n\uparrow h$. By the Monotone Convergence Theorem applied to each distinct measure space, $$\E[h(X)]=\lim_n \E[h_n(X)] = \lim_n \mathbf{L}_X(h_n)=\mathbf{L}_X(h),$$ since $\E[h_n(X)]=\mathbf{L}_X(h_n)$ for each $n$.

Next, let's obtain a familiar expression for $\mathbf{L}_X(h)$ when $X$ admits a PDF $f_X$, i.e., $$\Lambda_X(B)=P(X\in B)=\int_B f_X\d \mathrm{Leb} = \int_B f_X(x)\d x.$$ In the notation of [PW], this means $\Lambda_X$ has measure $f_X\mathrm{Leb}$, $\Lambda_X$ has density $f_X$ relative to $\mathrm{Leb}$, and we write $\frac{\d \Lambda_X}{\d \mathrm{Leb}}=f_X$. The PDF $f_X$ is called the Radon-Nikodym derivative of $\Lambda_X$ relative to $\mathrm{Leb}$ on $(\R,\B)$. Our precise formulation is below, but more suggestively, we demonstrate that $$\int_{\R} h(x)f_X(x)\d x=\int_{\R} h f_X\d\mathrm{Leb} = \int_{\R} h \frac{\d \Lambda_X}{\d \mathrm{Leb}}\d\mathrm{Leb} = \int_{\R} h\d\Lambda_X.$$

Suppose $h\in\L^1(\R,\B,\Lambda_X)$ with $h\ge0$. Then there is an increasing sequence $(h_n)$ of non-negative simple functions such that $h_n\uparrow h$. By the Monotone Convergence Theorem applied to each distinct measure space, \begin{align*} \int_{\R}h\d\Lambda_X =\lim_n\int_{\R}h_n\d\Lambda_X=\lim_n\int_{\R} h_nf_X\d\mathrm{Leb}= \int_{\R} hf_X\d\mathrm{Leb}, \end{align*} yielding the claim. For general $h\in\L^1(\R,\B,\Lambda_X)$, writing $h=h^{+}-h^{-1}$ finishes the proof.