Radon-Nikodyn derivative identity

Question

Let $\nu$ and $\mu$ be two $\sigma$-finite measure defined on the measure space $(X,\mathcal{M})$. Suppose $\nu \ll \mu$ and let $w$ be the Radon-Nikodyn derivative of $\nu$ with respect to $\mu$, that is $w=\frac{d \nu}{d \mu}$. Prove that for every non negative measurable function $g$ one has: $$ \int_{X} g \, d \nu = \int_{X} g w \, d \mu $$ The only idea I have is to use a sort of "density" argument: first prove the assertion for constant and simple functions, then approximate the measurable function $g$ with a (monotone) sequence of simple functions. Is there a simpler way to do this (provided that one works, of course...)?

Answer 1

From $\nu(A)=\int_{A} w \, d \mu$ your equation hold for characteristic functions $g=\chi_A$ of measurable sets $A$, and hence, by linearity of integral, also for simple functions.

For an abritrary measurable function $g$, take sequence of simple functions $g_n \uparrow g$. We already proved that $$ \int_X g_n \, d \nu = \int_X g_n w \, d \mu$$ and since $g_n w \uparrow gw$, we get the result passing to limits by monotone convergence theorem. I don't see what could be simpler than that.