Show that $\int_\mathbb{R^n} F(|g(x)|) dx = - \int_0^{\infty} F(\alpha) d \lambda(\alpha) = \int_0^{\infty} F'(\alpha) \lambda(\alpha) d\alpha$

Question

So, I've been tasked with proving the following:

Show that if $F$ is a non-negative differentiable function defined on $[0, \infty)$ with $F(0) = 0$ and $g$ is a measurable function on $\mathbb{R^n}$ with $0 < \int_{\mathbb{R^n}} F(|g(x)|) dx < \infty$ then $\int_\mathbb{R^n} F(|g(x)|) dx = - \int_0^{\infty} F(\alpha) d \lambda(\alpha) = \int_0^{\infty} F'(\alpha) \lambda(\alpha) d\alpha$

Here, $\lambda(\alpha) = \mu\{x : |g(x)| > \alpha\}$

I've been able to prove the left and right-most sides are equal:

Observe that $$ \begin{align} \int_{\mathbb{R^n}} F(|g(x)|) dx &= \int_{\mathbb{R^n}} \int_0^{|g(x)|} F'(\alpha) d\alpha dx \\ &= \int_{\mathbb{R^n}} \int_0^{\infty} 1_{(\alpha, \infty)} |g(x)| F'(\alpha) d\alpha dx \\ &= \int_0^{\infty} \int_\mathbb{R^n} 1_{(\alpha, \infty)} |g(x)| F'(\alpha) dx d\alpha \\ &= \int_0^\infty F'(\alpha) \lambda(\alpha) d\alpha \end{align}$$

However, now I'm stuck at showing why the middle equality is true. Any help would be greatly appreciated!

Answer 1

Use integration by parts with $u = \lambda(\alpha)$, $du = d\lambda(\alpha)$, $dv = F'(\alpha) \, d\alpha$, $v = F(\alpha)$

\begin{align*} \int_0^\infty u \, dv &= \left[ uv \right]_0^\infty - \int_0^\infty v \, du \\ \int_0^\infty F'(\alpha) \lambda(\alpha) \, d\alpha &= \left[ F(\alpha) \lambda(\alpha) \right]_0^\infty - \int_0^\infty F(\alpha) \, d\lambda(\alpha) \\ \end{align*}

Since $F(0) = 0$ and $\lim\limits_{\alpha \to \infty} \lambda(\alpha) = 0$ then $\left[ F(\alpha) \lambda(\alpha) \right]_0^\infty = 0$ so that

\begin{align*} \int_0^\infty F'(\alpha) \lambda(\alpha) \, d\alpha &= - \int_0^\infty F(\alpha) \, d\lambda(\alpha) \\ \end{align*}

and we are done.