Hi

I need to show that $\displaystyle \int_{\Omega}\int_{0}^{\infty} I_{\left[0,X(\omega)\right]} \, dx \, dP(\omega) $, where $\displaystyle I_{\left[a,b\right]} $ is the indicator function, i.e. $\displaystyle I_{\left[a,b\right]}(x)=1 $ if $\displaystyle x \in \left[a,b\right]$ and zero otherwise, is equal to both

$\displaystyle E(X)$ and $\displaystyle \int_{0}^{\infty}(1-F(x)) \, dx $ , where $\displaystyle F(x)$ is the cdf of $\displaystyle X$.

Hence deducing that

$\displaystyle E(X)=\int_{0}^{\infty}(1-F(x)) \, dx $

I tried:

$\displaystyle \int_{\Omega}\int_{0}^{\infty} I_{\left[0,X(\omega)\right]} \, dx \, dP(\omega) = \int_{\Omega} \int_{0}^{X(\omega)} \, dx \, dP(\omega) = \int_{\Omega} X(\omega) \, dP(\omega) = E(X) $, since this is the books' definition of the expected value of $\displaystyle X$.

I´m not sure how to go about finding the other result.

Thanks