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 $$\int_Bfd\lambda=\mu(B)=\int_Bgd\lambda,\forall B \in \mathcal{N}$$
• Why is $\mu \geq 0$? The argument still works without that assumption, treating $\mu$ as a signed measure with $|\mu| \ll \lambda$. – anomaly Dec 17 '17 at 4:25
• I put $\mu \geq 0$ for convenience...i will edit my answer.. – Marios Gretsas Dec 17 '17 at 4:42