This is part of the proof for the following result from Rene Schilling's Measures, Integrals and Martingales.
Let $(X,\mathcal{A},\mu)$ be $\sigma$-finite and let $\phi:[0,\infty) \to [0,\infty)$ with $\phi(0)=0$ be increasing and continuously differentiable. Then $$\int \phi \circ u d\mu = \int_0^\infty \phi'(s)\mu\{u\ge s\}ds$$ holds for all $u\in \mathcal{M}^+(\mathcal{A})$; the right-hand side is an improper Riemann integral.
The heart of the proof lies in proving the identity $\int \phi \circ ud\mu = \int_{(0,\infty)} \mu\{\phi \circ u\ge t\}\lambda^1(dt)=\int_0^\infty \mu\{\phi \circ u\ge t\}dt$, where the LHS is a Lebesgue integral and the RHS is an improper Riemann integral. We know that we can do this if $t\mapsto \mu\{\phi \circ u\ge t\}$ is Lebesgue a.e. continuous and bounded. Boundedness is not a problem since we may consider $k\wedge \mu\{\phi \circ u\ge t\}\mathbb{1}_[k^{-1},k](t),k\in \mathbb{N}$, and let $k\to \infty$ using Monotone Convergence Theorem.
Question: I don't see how the argument given in the bolded sentence above makes sense. If we replace $\mu\{\phi \circ u\ge t\}$ as above, then using Monotone Convergence Theorem, we have $$\int \phi \circ u d\mu = \lim_{k\to \infty} \int_{[k^{-1},k]} k \wedge \mu\{\phi \circ u\ge t\},$$ where the limit is equal to, since the integrand is bounded and Riemann integrable, $$\lim_{k\to \infty} \int_{k^{-1}}^k k\wedge \mu\{\phi \circ u\ge t\}.$$ But the improper integral we want is $$\lim_{k\to \infty} \int_{k^{-1}}^k \mu\{\phi \circ u\ge t\}$$without the $k$ in the integrand. I don't see how these two integrals are equal. In the case where $\phi \circ u \in L^1(\mu)$, we have $\mu\{\phi \circ u\ge t\}$ bounded on bounded intervals by the Markov inequality so we need not consider such approximations. But the theorem in the book states explicitly that the relation holds for all positive measurable $u$. So I must make sense of this identity, but I can't figure it out. I would greatly appreciate any help.
