Equality of expected value using Fubini's theorem

Question

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, $\alpha > 0$ and $X: \Omega \to \mathbb{R}$ a non-negative real-valued random variable. I need to prove that

$$\int_{[0,\infty)}\alpha^{-2}\mathbb{E}(\min(X, \alpha^2))\mathrm{d}\lambda^1(\alpha) = 2\mathbb{E}(\sqrt{X}),$$

where $\lambda^1$ denotes the lebesgue measure. In a first step I proved that

$$\mathbb{E}(\min(X, \alpha^2)) = \int_{[0,\alpha^2]}\mathbb{P}(X \geq t)\mathrm{d}\lambda^1(t).$$

Using this, we get

$$\int_{[0,\infty)}\alpha^{-2}\mathbb{E}(\min(X, \alpha^2))\mathrm{d}\lambda^1(\alpha) = \int_{[0,\infty)}\int_{[0,\alpha^2]}\alpha^{-2}\mathbb{P}(X \geq t)\mathrm{d}\lambda^1(t)\mathrm{d}\lambda^1(\alpha)$$

and here is where I get stuck. This smells like Fubini but I don't see how to reach

$$2\mathbb{E}(\sqrt{X}) = 2\int_{\Omega}\sqrt{X}\mathrm{d}\mathbb{P}$$ from here. I would appreciate any help.

Answer 1

Hint: \begin{align} \int_0^\infty \int_0^{\alpha^2}\ldots \mathrm d\lambda(t)\mathrm d\lambda(\alpha) &=\int_0^\infty \int_0^{\infty}\ldots \, \mathbf 1_{t\le \alpha^2}\mathrm d\lambda(t)\mathrm d\lambda(\alpha)\\ &= \int_0^\infty \int_0^{\infty}\ldots \, \mathbf 1_{t\le \alpha^2}\mathrm d\lambda(\alpha)\mathrm d\lambda(t)\\ &= \int_0^\infty \int_0^{\infty}\ldots \, \mathbf 1_{\sqrt t\le \alpha}\mathrm d\lambda(\alpha)\mathrm d\lambda(t)\\ &=\int_0^\infty \int_{\sqrt{t}}^\infty \ldots \mathrm d\lambda(\alpha)\mathrm d\lambda(t) \end{align}