# How to show $\int g \, d\nu = \int g \cdot f \, d\mu$, where $\nu(K) = \int_K f \, d\mu$?

Let $\mu$ be a measure on $X$, and let $f:X \rightarrow [0,+\infty]$ be $\mu$-measurable. Define the measure: $$\nu(K) = \int_K f \, d\mu$$ I know that any $\mu$-measurable function $g:X \rightarrow [0,+\infty]$ is also $\nu$-measurable. I want to show that for any such function, we have: $$\int g \, d\nu = \int g \cdot f \, d\mu$$

I have already proven the case where $g$ is simple. I have also proven that if $h$ is simple, then $h \leq g \;$ $\nu$-a.e. if and only if $hf \leq gf \; \mu$-a.e. Using these facts and the definition of the lower integral, we can write: \begin{align*} \int g \, d\nu &= \sup \left\{ \int h \, d\nu : h \text{ is } \nu \text{-integrable, simple, } h \leq g \; \nu \text{-a.e.} \right\} \\ &= \sup \left\{ \int h \cdot f \, d\mu : h \text{ is } \nu \text{-integrable, simple, } hf \leq gf \; \mu \text{-a.e.} \right\} \end{align*} But at this point I'm lost as to how to transform this into the lower integral of $\int gf \, d\mu$, i.e.: $$\sup \left\{ \int h \, d\mu : h \text{ is } \nu \text{-integrable, simple, } h \leq gf \; \mu \text{-a.e.} \right\}$$

• Doesn’t an appeal to monotone convergence work? – Theoretical Economist Jul 30 '18 at 22:37
• I think once I know it's true for $g$ simple, I'd prove it for $g$ nonnegative by writing it as an increasing limit of simple functions, then applying MCT on both sides. – Daniel Schepler Jul 30 '18 at 22:37

Given any $g$, let it be the limit of an increasing sequence of simple functions $(g_n)$. You know that $$\int g_n \, d\nu = \int g_n \cdot f \, d\mu$$ for each $n$. Now just take the limit of both sides as $n\to\infty$. By the monotone convergence theorem (for integrals with respect to $\nu$ on the left and for integrals with respect to $\mu$ on the right), you get $$\int g \, d\nu = \int g \cdot f \, d\mu.$$