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$X$ be a random variable and $P$ is a probability measure on $\Omega$. Is it true that $$\displaystyle{\int_{\Omega}}X^+\,\mathrm{d}P = \displaystyle{\int _{0}^{\infty}}P(X^+ > t)\,\mathrm{d}t\,, $$

where $X^+(\omega)=\max\{X(\omega),0\}\,$?

I was thinking along the guidelines of Fubini's theorem as probability is a $\sigma$-finite measure. The left hand side of the integral is actually area under a curve taking vertical strips and the right hand side is calculating the same area taking horizontal strips. So by Fubini's theorem they must be equal.

But how do I rigorously show this ?

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    $\begingroup$ Exactly as you described it - maybe use Tonelli's theorem for the case when the LHS/RHS do not converge. $\endgroup$ Mar 2, 2021 at 14:18

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First note that $\displaystyle{\int _{0}^{\infty}}P(X^+ > t)dt =\displaystyle{\int _{0}^{\infty}}\int 1(X^+ > t)\, dP\,dt $, where $1(A)$ is the indicator function of the event $A$. Then use Tonelli's theorem to interchange the integral and get $$ \displaystyle{\int _{0}^{\infty}}\int 1(X^+ > t)\, dP\,dt =\int \displaystyle{\int _{0}^{\infty}}1(X^+ > t)\, dt\,dP = \int \displaystyle{\int _{0}^{X^+}}\, dt\,dP =\int {X^+}\,dP. $$

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  • $\begingroup$ So if I am working with a random variable whose expectation is finite then this is basically by Fubini's theorem right ? $\endgroup$ Mar 2, 2021 at 15:03
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    $\begingroup$ Correct. Generally speaking, you do not need to bother about the finiteness when the integrand is non-negative. You can just use Tonelli. $\endgroup$ Mar 2, 2021 at 15:05
  • $\begingroup$ Thank you so much $\endgroup$ Mar 2, 2021 at 15:11

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