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For $X$, a positive random variable, use Fubini's theorem applied to $\sigma$-finite measure to prove

\begin{align*} E(X)= \int_{[0, \infty)} P[X>t] dt \end{align*}

I found several references on the the site like this:

  1. Proving $EX=\int_{0}^{\infty}(1-F(x))\, dx$ when $X\geq0$
  2. Expression for $n$-th moment

but all of them assume that $X$ has density or discrete is the are different proof that does not use this assumption.

This is what I tried

\begin{align*} \int_{[0, \infty)} P[X>t] dt&= \int_{[0, \infty)} E(\mathsf{1}_{[X>t]}) dt=\int_{[0, \infty)} \int_{\Omega} \mathsf{1}_{[X>t]}dP dt\\ &=\int_{\Omega} \int_{[0, \infty)} \mathsf{1}_{[X>t]} dt dP=\int_{\Omega} \int_{[0, X)} 1 dt dP=\int_{\Omega} X dP=E[X] \end{align*} where the first step in second line is due to Fubinni's theorem

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marked as duplicate by Did probability-theory Dec 5 '14 at 7:45

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    $\begingroup$ Try to rewrite $P[X>t]$ as an integral of $\chi_{\{X>t\}}$, then Fubini's theorem will apply naturally. Also observe that $$\int_0^\infty \chi_{\{X>t\}}(x) dt = \int_0^{X(x)} 1 dt.$$ $\endgroup$ – Xiao Dec 2 '14 at 23:17
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    $\begingroup$ Dear, @Xiao I just made editions is this what you meant? $\endgroup$ – Boby Dec 2 '14 at 23:39
  • $\begingroup$ Yes, that is what Xiao meant. $\endgroup$ – copper.hat Dec 2 '14 at 23:46
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    $\begingroup$ @Boby Nicely done. $\endgroup$ – Xiao Dec 2 '14 at 23:47
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    $\begingroup$ That's okay, glad to help! $\endgroup$ – Xiao Dec 2 '14 at 23:52