If $(E,\mathcal E,\mu)$ is a finite measure space, $f\in\mathcal L^1(\mu)$ is nonnegative and $c\ge0$, can we somehow pull $c$ out in $\int\min(f,c)\:{\rm d}\mu$?
We may clearly write $$\int\min(f,c)\:{\rm d}\mu=\frac12\left(\int f\:{\rm d}\mu+c\mu(E)-\int|f-c|\:{\rm d}\mu\right),\tag1$$ but then the question is if we can simplify $\int|f-c|\:{\rm d}\mu$.
You can write $$ \int |f-c|\,\mathrm d \mu =\int_{f>c}(f-c)\,\mathrm d \mu -\int_{f<c}(f-c)\,\mathrm d \mu \\ =c(\mu (f<c)-\mu (f>c))+\int_{f>c}f\,\mathrm d \mu -\int_{f<c}f\,\mathrm d \mu $$ I dont know if this is a "simplification", it would depend about what you want to do.