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This answer says that if $X$ is a random variable and $X_+ = \mathrm{max}(0, X)$, then $X_+ = \int_0^\infty I_{\{X > x\}}\mathrm{d}x$. I'd like to know how to derive this starting with $A \in \mathcal{S} \implies \int_S 1_A\mathrm{d}\mu = \mu(A)$ (from "Desired Properties").

My thinking is that for a probability space $(\Omega, F, \mu)$, $\{X > t\} \in F \implies \int_\Omega I_{\{X > t\}}\mathrm{d}\mu = \mu(\{X > t\})$ but it's the wrong variable of integration. Am I on the right track?

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  • $\begingroup$ Does max(0, X) mean max{0, X}, i.e., the larger of X and 0 ? $\endgroup$
    – Daniel Asimov
    Commented Mar 25, 2023 at 1:01
  • $\begingroup$ Equality $X_+ = \int_0^\infty I_{\{X > x\}}\mathrm{d}x$ is obvious and is an equality between random variables. It cannot follow from equalities between expectations. Moreover, Mathoverflow is devoted to questions about research. $\endgroup$
    – Christophe Leuridan
    Commented Mar 25, 2023 at 9:22
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    $\begingroup$ Just to amplify what Christophe said: $y_+=\int_0^\infty I_{\{y>x\}}\mathrm{d}x$ is simply true for any real number $y$. There is nothing special involving random variables going on. $\endgroup$
    – James Martin
    Commented Mar 25, 2023 at 11:43

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The layer cake representation of a non-negative measurable function, $X$, is applied in the proof of proposition 2.1 here.

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