Show $\int_X f d\nu = \int_X fgd\mu$ if $\nu(E)=\int_E g d\mu$ .

Question

$f$ and $g$ are both non-negative functions where the integral of non-negative function is defined as the supremum over all simple functions dominated by the non-negative function.

Would going through the definition of supremum work?

This is an exercise in Royden. He gives this hint: Establish it for simple functions, use Simple approximation Lemma and Monotone Convergence theorem.

Answer 1

If $f=\chi_E$, then conjunction is obvious. Using linearity of the Lebesgue integral we conclude conjunction for simple positive function. If $f$ is positive measurable, then we can find sequence $\{s_n\}_{n=1}^{\infty}$ positive simple functions such that $s_n \uparrow f$, then $s_n g \uparrow fg$, and then we can apply Monotone Convergence theorem to find $$\int_X f \,d\nu=\int_X \lim_{n\to \infty} s_n\, d\nu =\lim_{n\to \infty} \int_X s_n \,d\nu =\lim_{n\to \infty} \int_X s_n g\, d\mu= \int_X f g \,d\mu$$

Also, I think we need assumption that $g \in L^1 (\mu),$ but I suppose we say that when we say $\nu (E)=\int_E g d\mu$, or $d\nu = g d\mu$

Answer 2

Hint. Use the uniquiness of Radon-Nikodym derivative in Radon-Nikodym theorem. Define $m(\,E\,)=\int_{E}f\, d\nu$. Then $f$ is the Radon-Nikodym derivative (unique in $L^1(X)$) of $m$ with respect to $\nu$.