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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.: $$ \sup \left\{ \int h \, d\mu : h \text{ is } \nu \text{-integrable, simple, } h \leq gf \; \mu \text{-a.e.} \right\} $$

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    $\begingroup$ Doesn’t an appeal to monotone convergence work? $\endgroup$ – Theoretical Economist Jul 30 '18 at 22:37
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    $\begingroup$ 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. $\endgroup$ – Daniel Schepler Jul 30 '18 at 22:37
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I think you are too focused on trying to use the definition of the integral directly. There are lots of other methods to compute integrals that are much more useful (we just don't use them as the definition itself since it is less clear that they are well-defined).

In particular, any nonnegative measurable function can be integrated by considering it as an increasing limit of simple functions. Think about how you could use that here. A full solution using this idea is hidden below.

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 $$\int g \, d\nu = \int g \cdot f \, d\mu.$$

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