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So I'm trying to figure this out, it's really basic but I can't seem to find the full justification anywhere.

Let $\xi$ be a random variable from $\Omega$ to $\mathbb{R}$ which has an absolutely continuous (wrt Lebesgue measure $\lambda$) induced probability measure on $\mathbb{R}$. The underlying sample space $\Omega$ has $\mu$ a probability measure.

By Radon-Nikodym I know for any Borel set $A \subseteq \mathbb{R}$ there exists $f$ a density such that $P(A) = \int_{A} f(x) \, d\lambda(x)$.

By the change of variables formula for induced measures I know $\mathbb{E}\xi = \int_{\Omega} \xi(\omega) \, d\mu(\omega) = \int_{\mathbb{R}} x \, dP(x)$.

My question is, how do I use this information to justify $\mathbb{E}\xi = \int_{\mathbb{R}} x f(x) \, d\lambda(x)$? And is this the correct way to understand the rigorous construction of these concepts?

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The usual notation for a probability measure on $(\Omega,\mathcal F)$ is $P$ and one rather uses $\mu$ to denote the distribution of a random variable $X:\Omega\to\mathbb R$, that is, a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$. With these conventions, recall that $\mu$ is defined as the unique measure such that, for every $B$ in $\mathcal B(\mathbb R)$, $\mu(B)=P(X^{-1}(B))$ (since the hypothesis that $X$ is a random variable ensures exactly that $A=X^{-1}(B)$ is in $\mathcal F$, hence that $P(A)$ is well defined).

Thus, the definition of $\mu$ is to ask that, for every $u=\mathbf 1_B$, $$ \int_\Omega u(X) \mathrm dP=\int\limits_{\mathbb R}u(x)\mathrm d\mu(x)\tag{$\ast$}. $$ It is a standard result, explained in every decent introductory textbook on probability theory (with measure) that $(\ast)$ holds in fact for every measurable function $u:\mathbb R\to\mathbb R$ which is integrable with respect to $\mu$.

In the proof usually presented, one considers the class $C$ of measurable functions $u:\mathbb R\to\mathbb R$ such that $(\ast)$ holds. One first notes that $C$ contains every $\mathbf 1_B$ with $B$ in $\mathcal F$, then, by linearity, that $C$ contains every simple nonnegative function, then, by supremum, that $C$ contains every positive integrable function, and finally, again by linearity, that $C$ contains every integrable function.

Your post is concerned with the case where $u:x\mapsto x$. One sees that $(\ast)$ holds as soon as one side of the identity is finite, that is, as soon as $X$ is integrable for $P$ or, equivalently, as soon as $u:x\mapsto x$ is integrable for $\mu$. Thus, when $(\ast)$ holds and when $\mu$ has density $f$ with respect to $\lambda$, $$ \mathrm E(X)=\int_\Omega X(\omega) \mathrm dP(\omega)=\int\limits_{\mathbb R}x\mathrm d\mu(x)=\int\limits_{\mathbb R}xf(x)\mathrm d\lambda(x). $$

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I haven't fully thought this through, but here's the first thing that comes to my mind.

Expected value is defined only if its positive and negative parts are both finite, just as Lebesgue integrals are defined that way. So let's work first with nonnegative random variables.

Suppose $\Pr(X\ge 0)=1$. If $\Pr(X\ge a)=b$ then however $\mathbb{E}(X)$ is defined, it should be $\ge a b$. This must hold for all values of $a$ and $b$ for which the inequality is true. So I'd think about whether these infinitely many constraints are enough to determine the value of $\mathbb{E}(X)$. And then I'd notice the similarity between that and the way the Lebesgue integral of a nonnegative function is defined.

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