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The statement, as in Lieb-Loss Analysis, for the Layer cake representation is:

Let $\nu$ be Borel measure on $[0,\infty)$ such that $\phi(t) := \nu([0,t))$ is finite for each $t > 0$. Now let $(\Omega,\Sigma,\mu)$ be a sigma-finite measure space and $f$ any nonnegative measurable function on $\Omega$. Then $$\int_\Omega\phi(f(x))\mu(dx) = \int_0^\infty \mu( f > t)\nu(dt).$$

For instance if $\phi(t) = t$ then we get just the integral of $f$. If $\phi(t) = \sin(t)$ then we get a $\cos(t)$ on the right side (avoiding all the signed measures details). This examples lead me to think that we can rewrite the equality as $$\int_\Omega \phi(f(x))\mu(dx) = \int_0^\infty \mu(f>t)\phi'(t)dt.$$In some way I could see $d\nu(t) = d\phi(t) = \phi'(t)dt$ but perhaps I am abusing notation.

Thanks in advance.

user57
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1 Answers1

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Based on your question, I believe you're interested only in situations in which $\phi'$ exists (we can easily come up with some examples of $\nu$ for which $\phi'$ does not exist). If we assume $\phi'$ is continuous (that is, $\phi$ is continuously differentiable), then $\phi'$ is the Radon-Nikodym derivative of $\nu$ with respect to Lebesgue measure $\lambda$. See this answer for a discussion of equivalence of usual derivative $\phi'$ and Radon-Nikodym derivative of $\nu$ with respect to $\lambda$.

In this case, we get that for any measurable $A$, $\nu(A)=\int_A \phi'(t)dt$ (defining property of Radon-Nikodym derivative). This easily extends to $\int_0^\infty s(t)\nu(dt)=\int_0^\infty s(t)\phi'(t)dt$ for any measurable simple function $s$. We deduce that for a measurable $g$ on $[0,\infty)$, $\int_0^\infty g(t)\nu(dt)=\int_0^\infty g(t)\phi'(t)dt$ whenever $g$ is either non-negative or $\int_0^\infty |g(t)|\nu(dt)<\infty$.