Sigma Algebra Generated by Subset and Radon-Nikodym derivative

Question

So this is a question to study for qualifying exams, not a homework question!

Let (X,M,μ) be a measure space with μ(X) < ∞. Suppose A ∈ M and 0 < μ(A) < μ(X).

Let N be the σ-algebra generated by {A}, and denote the the restriction of μ to N by μ|N .

(a) Describe N and N-measurable functions from X to R.

(b) Let f be an M-measurable, μ-integrable function from X to R. Define ν to be the signed measure on (X,N) given by ν : E →$\int$ fdμ for every E ∈ N.

Determine, on (X, N ), the Radon-Nikodym derivative of ν with respect to μ|N .

a) The sigma algebra is N = {A, $A^c$, $\mathbb{R}$, $\emptyset$}. If B is a Borel set, and f is a measurable function, then: $f(A) \subseteq B$

$f(A^c) \subseteq B^c$

b) I'm really not sure how to approach this question. Also, if my understanding for part a is incorrect then I will be on the wrong track here too. Any suggestions for how to approach this? Also, is my understanding of part A correct?

Thank you for any and all help. Again, studying for quals, not doing homework!

Answer 1

a) It is correct that $N=\{\varnothing, A,A^{\complement},X\}$.

So a function $g:X\to\mathbb R$ is $N$-measurable if $g^{-1}(B)\in\{\varnothing, A,A^{\complement},X\}$ for every Borel set $B\subseteq\mathbb R$.

This implies that $g$ can take at most $2$ values (do you see why?).

We find that $g$ is $N$-measurable that if it can be written as $a\mathbf1_A+b\mathbf1_{A^{\complement}}$ or - equivalently - as $c+d\mathbf1_A$.

b) The Radon-Nikodym derivative of $\nu$ with respect to $\mu|N$ must be an $N$-measurable function $g$ so must have the form $a\mathsf1_A+b\mathsf1_{A^{\complement}}$ as proved above.

Next to that it must satisfy the condition:$$\int_Eg\;d(\mu|N)=\nu(E)=\int_E f\;d\mu\text{ for every }E\in N=\{\varnothing,A,A^{\complement},X\}$$

Substituting for $g=a\mathsf1_A+b\mathsf1_{A^{\complement}}$ we find $4$ equalities that must satisfied:

• $0=0$ for $E=\varnothing$
• $a\mu(A)=\int_A f\;d\mu$ for $E=A$
• $b\mu(A^{\complement})=\int_{A^{\complement}} f\;d\mu$ for $E=A^{\complement}$
• $a\mu(A)+b\mu(A^{\complement})=\int_Xf\;d\mu$ for $E=X$

The first is irrelevant and the fourth is a consequence of the second and third.

Since $0<\mu(A)<\mu(X)<\infty$ we know that $\mu(A)$ and $\mu(A^{\complement})$ are both positive and finite.

So the second and third enable us to find $a,b$ hence also $g=a\mathsf1_A+b\mathsf1_{A^{\complement}}$.