I've come across at least three different definitions of the upper integral of a $[0,\infty]$-valued function with respect to an outer measure $\mu$.
- $\int_X^\ast f d\mu = \int_0^\infty \mu(\{f(x)>t\}) dt$ (also known as the Choquet-integral)
- $\int_X^\ast f d\mu = \inf\{\sum_{i\in\mathbb{N}} c_i \mu(E_i) \mid E_i\subseteq X, c_i\geq 0, f\leq \sum c_i \chi_{E_i}\}$
- $\int_X^\ast f d\mu = \inf\{\int_X \phi d\mu \mid f\leq\phi, \phi \;\mu-\text{measurable}\}$
Denoting these three right hand values with $I_1, I_2, I_3$ respectively, I can show $I_1\leq I_3$ and $I_2\leq I_3$ and equality if $f$ is measurable, but I am not sure whether any one of these is a strict inequality or what the precise relationship between $I_1$ and $I_2$ is.
If they indeed turn out to be different in general, is there a nice condition under which they coincide, $\mu$ being finite for example or $\mu$ being a metric outer measure?
Okay, first wrinkle: The Choquet integral is not subadditive in general. It is subadditive iff $\mu$ is submodular, i.e. if $\mu(A\cup B)+\mu(A\cap B) \leq \mu(A)+\mu(B)$. The other two notions of upper integrals are always subadditive. Thus they cannot be equal in general.
But if that is the case, then $I_1\leq I_2$.
EDIT: Because it caused confusion, here are my proofs:
1.) $I_1(f)$ is monotone in $f$ and for measurable $\phi$ one can apply Fubini-Tonelli: $$I_1(\phi) = \int_0^\infty \mu(\{\phi(x)>t\}) dt = \int_0^\infty \int_X \chi_{\phi(x)>t} d\mu\;dt = \int_X \int_0^\infty \chi_{\phi(x)>t} dt\;d\mu = \int_X \lambda([0,\phi(x)[) d\mu = \int_X \phi(x) d\mu$$ Therefore for all measurable $\phi$ with $f\leq\phi$ we find $I_1(f)\leq I_1(\phi)=\int_X\phi d\mu$. Taking the infimum over all such $\phi$ we find $I_1(f)\leq I_3(f)$.
2.) $I_2(f)$ is also monotone in $f$. Every measurable $\phi$ can be expressed as $\sum_{i\in\mathbb{N}} c_i \chi_{E_i}$ with $E_i$ even measurable. This can be seen by using dyadic expansion:
Start with $\phi_0 := \sum_{n=1}^\infty \chi_{\{\phi\leq n\}}$. This is a function with $\phi-1\leq\phi_0\leq\phi$. Then if we have given a $\phi_k$ with $\phi-2^{-k}\leq\phi_k\leq\phi$ then recursively set $\phi_{k+1} := \phi_k + 2^{-(k+1)} \chi_{\{\phi-\phi_k \geq 2^{-(k+1)}\}}$. This is then also a function with $\phi-2^{-(k+1)} \leq \phi_{k+1} \leq \phi$. The limit $\lim_{k\to\infty} \phi_k$ is then an infinite series of the form $\sum_{i\in\mathbb{N}} c_i \chi_{E_i}$ and exactly equal to $\phi$.
Therefore for all measurable $\phi$ the integral $\int_X \phi d\mu$ is actually contained in the set over which the infimum in $I_2$ is taken. Therefore $I_2(f)\leq I_2(\phi)\leq I_3(\phi) = \int_X \phi d\mu$ for all measurable $\phi$ with $f\leq \phi$. Taking the infimum over all such $\phi$ we arrive at $I_2(f)\leq I_3(f)$.