I just learned about Radon-Nikodym theorem. However, I do not seem to have any intuition on how to apply it...

For example :

Let $(X,\mathscr{M},\lambda)$ be a $\sigma-$finite measure.

Let $f$ be $\mathscr{M}$ measurable.

Let $\mathscr{N} \subset \mathscr{M}$ be a $\sigma-$algebra.

Prove that there exists an $\mathscr{N}$ measurable function $g$ such that $$ \int_B f d\lambda = \int_B g d\lambda $$ for every $B\in \mathscr{N}$

Clearly this question feels like an application for the Radon-Nikodym theorem. However how can I find all the ingredients ? I have to find some signed measure $\mu$ which is absolutely continuous with respect to $\lambda$. Maybe $\mu(B) = \int_B f d\lambda$ is simply the answer.


Consider the measure space $(X,\mathcal{N},\lambda)$

Let $\mu(B)=\int_Bfd\lambda$

It is not difficult to see that $|\mu|<< \lambda$

Thus from Radon-Nikodym exists a function $g: X \to [0,+\infty)$ such that $$\int_Bfd\lambda=\mu(B)=\int_Bgd\lambda,\forall B \in \mathcal{N}$$

  • $\begingroup$ Why is $\mu \geq 0$? The argument still works without that assumption, treating $\mu$ as a signed measure with $|\mu| \ll \lambda$. $\endgroup$ – anomaly Dec 17 '17 at 4:25
  • $\begingroup$ I put $\mu \geq 0$ for convenience...i will edit my answer.. $\endgroup$ – Marios Gretsas Dec 17 '17 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.