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 $.

  • $\begingroup$ Thanks! Can you please explain how to move to integrals in part b, I understood your approach but did not manage to connect it to integrals. Is what I solved in c and d seems fine? $\endgroup$
    – user726608
    Nov 16 '20 at 16:14
  • 1
    $\begingroup$ @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 $\int \chi _{h^{-1}(E)}\mathop{}\!d \mu =\mu (h^{-1}(E))=\eta (E)$ $\endgroup$
    – Masacroso
    Nov 16 '20 at 17:33

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