# A question about measurable and simple functions- measure theory

Let $$(X,A), (Y,B)$$ be measurable spaces. $$\mu$$ is a measure on $$A$$, and $$h:X\to Y$$ a measurable function. Define $$\eta: B\to [0,\infty]$$ such as $$\forall E\in B$$ , $$\eta(E)= \mu(h^{-1}(E))$$.

A. Show that $$\eta$$ is well defined and is a measure on $$B$$.

B.Prove: $$\forall E\in B \int_{Y} \chi_E d\eta= \int_X (\chi_E\circ h) d\mu$$.

C.Let $$\phi : Y\to [0,\infty)$$ be a simple measurable function. Prove $$\int_{Y} \phi d\eta= \int_X (\phi\circ h) d\mu$$.

D.Let $$f:Y\to\ [0,\infty]$$ be a measurable function. Prove $$\int_{Y}f d\eta= int_X (f\circ h) d\mu$$.

My solution:

A.$$\eta(\emptyset)=\mu(h^{-1}(\emptyset))$$ then I did not manage to show that it is exactly $$\mu(\emptyset)=0$$. Let $$E_1,E_2,\ldots \in B$$ nonoverlapping sets so $$\eta(\cup_{i=1}^{\infty}(E_i))=\mu(h^-1(\cup_{i=1}^{\infty}(E_i)))=\mu(\cup_{i=1}^{\infty} h^-1(E_i))= \sum_i \mu(h^-1(E_i))=\sum_i \eta(E_i)$$. What does it mean that $$\eta$$ is well defined?

B.$$\forall E\in B, \mu(E)=\int_{E} 1 d\mu=\int_Y \chi_E d\mu$$ I did not succeed to finish it!.

C. We can write $$\phi=\sum_{j=1}^{n} a_{j} \chi_{E_j}$$ where $$E_j$$ are measurable. Then $$\int_Y \phi d\eta= \int_Y \sum_{j=1}^{n} a_{j} \chi_{E_j} d\eta$$ = ($$\phi_n$$ is a sequence of measurable positive functions)= $$\sum_{j=1}^{n} a_{j} \int_Y \chi_{E_j} d\eta$$ =(by part b) = $$\sum_{j=1}^{n} (a_{j} \int_X (\chi_{E_j}\circ h)) d\mu = \int_X (\phi\circ h) d\mu$$.

D. f is measurable therefore by a theorem, there is an increasing monotone sequence of simple measurable sets $$\phi_n$$ such that: $$lim_{n\to \infty}\phi_n=f$$ and by using the monotone theorem of Lebesgue we get

$$\int_Y f d\eta= lim_{n\to \infty} \int _Y \phi_n d\eta$$= (by part c)= $$lim_{n\to \infty} \int_X (\phi_n\circ h) d\eta$$ = ($$\phi_n\circ h$$ is measurable as a composition of two measurable functions)= $$\int_X lim_{n\to \infty} (\phi_n\circ h) d\eta= \int_X (f\circ h ) d\eta$$.

I will be glad if you can help in the points which I did not succeed, and tell me if what i did manage was okay.

For part A, that $$\eta$$ is well defined means that the expression $$\eta (E)$$ makes sense, and it does because $$h^{-1}(E)$$ is $$\mu$$-measurable. By the other side the preimage of the empty set is empty for every function, just note that $$h^{-1}(A):=\{x:h(x)\in A\}$$ So, when $$A=\emptyset$$ then $$h^{-1}(\emptyset )=\emptyset$$.
For the part B note that $$(\chi _E\circ h)(x)=1\iff h(x)\in E\iff x\in h^{-1}(E)\iff \chi _{h^{-1}(E)}(x)=1$$
so $$\chi _E\circ h=\chi _{h^{-1}(E)}$$, and so the relation between the integrals is trivially true in view of the definition of $$\eta$$.
• @user726608 you already solved part C and D correctly. For the integral of part B this is immediate from my answer above, just note that, by definition $\int \chi _{h^{-1}(E)}\mathop{}\!d \mu =\mu (h^{-1}(E))=\eta (E)$ Nov 16 '20 at 17:33