# How to prove that $\mathbf{E}\eta = \int_{0}^{\infty}\mathbf{P}(\{\eta > z\})dz$ for arbitrary non-negative random variable?

Let $\eta$ be non-negative (a.c.) continuous random variable. How to prove that $\mathbf{E}\eta = \int_{0}^{\infty}\mathbf{P}(\{\eta > z\})dz$ for arbitrary non-negative random variable?

If $\eta$ is discrete random variable it's quite simple. I have an idea that we can prove it easily for random variable $$\xi = \sum_{i=1}^{n}\xi_i\mathbf{1}_{\Delta_i}$$ that can take only finite set of values , where $\mathbf{1}_{\mathbf{X}}$ is an indicator function of a set $\mathbf{X}$. Since for every measurable function $\xi$ there is an approximating sequence $\{\xi_i\}_{i\ge1}$ of the functions of such kind that tends to $\xi$ pointwise, I would like to make a passage to the limit to prove this statement for an arbitrary continuous non-negative random variable? How can one do that rigorously?

• Have you considered Fubini theorem ? – Gabriel Romon Apr 26 '17 at 19:06
• @LeGrandDODOM I can't clearly understand what use it could be here? We don't have any measure products. Could you explain? – Illia Yurtsiv Apr 26 '17 at 19:09
• Note that $P(\eta >z)=P_\eta((z,\infty))=\int 1_{(z,\infty)}(u)dP_\eta (u)$. Hence $\int_{0}^{\infty}\int 1_{(z,\infty)}(u)dP_\eta (u)dz=\int \int1_{(0,u)}(z)dz dP_\eta(u)=\int u dP_\eta(u) = E(\nu)$ – Gabriel Romon Apr 26 '17 at 19:17
• @LeGrandDODOM, Thanks a lot! – Illia Yurtsiv Apr 26 '17 at 19:19

Do the following. Let $\mathbb{P}$ be the underlying probability measure. Then, \begin{align} \int_0^{\infty}\mathbb{P}(X(\omega)\geq x)dx & = \int_0^{\infty}\mathbb{E}\left[\mathcal{X}_{X(\omega)\geq x}\right]dx \\ &= \int_0^{\infty}\int_{\Omega}\mathcal{X}_{X(\omega)\geq x}d\mathbb{P}(w)dx\\ & =\int_{\Omega}\int_{0}^{X(\omega)}dxd\mathbb{P}(\omega) \\ & = \int_{\Omega}X(\omega)d\mathbb{P}(\omega) = \mathbb{E}[X]. \end{align}