Let $(\Omega, \mathcal{A}, \mu)$ be a measure space and $f: \Omega \to \mathbb{R}$ a measure nonnegative function that is summable against $\mu$.

It is known that $$\nu(A) = \int_{A} f \mu$$ is also a measure on $\mathcal{A}$. I have to prove that $$\int_{\Omega} g \nu = \int_{\Omega} gf \mu$$ for all functions $g: \Omega \to \mathbb{R}$ that are summable against $\nu$.

I don't quite understand the notation, though. Specifically, on the proof statement, what does $gf \mu$ mean? Am I taking the composition of $g$ and $f$? How does one even evaluate $\int_{\Omega} g \nu$?

Thanks in advance for any help.

  • 2
    $\begingroup$ The more common notation is $\nu(A)=\int_Af\,d\mu$; the integral of $f$ with respect to the measure $\mu$. You're then supposed to show $\int_{\Omega}g\,d\mu=\int_{\Omega}g\cdot f\,d\mu$; the RHS has a product of $g$ and $f$. In words: show Lebesgue integral of $g$ with respect to $\nu$ equals Lebesgue integral of $gf$ (a product) with respect to $\mu$. These are all Lebesgue integrals with respect to the measures involved, so the definition goes like non-negative simple functions $\to$ non-negative measurable functions $\to$ real integrable $\to$ complex-valued integrable functions. $\endgroup$
    – peek-a-boo
    Mar 24, 2022 at 1:37
  • $\begingroup$ So to start off with simple functions, if I just consider indicator function, $g = a * X_A,$, $$\int g \nu = a \cdot \nu (A) = a \cdot \int_{A} f \mu = a \cdot \int X_{A} f \mu = \int a X_{A} f \mu = \int gf \mu.$$ Does that work? $\endgroup$
    – mathz2003
    Mar 24, 2022 at 1:45
  • $\begingroup$ Yes that's exactly right. Now use linearity to get all simple functions, and monotone convergence to get all non-negative measurable functions. $\endgroup$
    – peek-a-boo
    Mar 24, 2022 at 1:57
  • $\begingroup$ @peek-a-boo does "all functions summable against $\nu$" mean I have to go further than non-negative measurable functions? Do I need Dominated Convergence Theorem like how Mahdi said? Or is this much enough? $\endgroup$
    – mathz2003
    Mar 24, 2022 at 1:59
  • $\begingroup$ if you know the equality for all non-negative functions then for integrable functions the statement becomes clear by considering positive and negative parts. $\endgroup$
    – peek-a-boo
    Mar 24, 2022 at 2:31

1 Answer 1


Regarding your questions on the notation $\nu(A)=\int f d\mu$ and $\int g d \nu=\int fg d \mu$, and in particular $fg$ means multiplication of $f$ and $g$.

The question you are asking can be answered using Radon Nikodym theorem. However, you can also directly approach this question by assuming first that $g$ is a simple function. Then using the fact that you can approximate any measurable function with simple functions and dominated convergence theorem you can generalize it to any function.


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