# Show that B is measurable

Let $\mathbf{g}:\Delta \subset \mathbb{R}^n \rightarrow D\subset \mathbb{R}^n$ be univalent, $C^{(1)}$ and with the Jacobian different from zero, $\forall t \in \Delta$. Let $B=\mathbf{g}^{-1}(A)$, where $A$ is a measurable (Lebesgue) subset of $D$. Show that B is measurable.

There is hint which is to first consider the case when $A\subset L\subset D$, where $L$ is compact. And also, from another exercise, we proved that for this g, and for a compact $L$, and a number $C$ (proved in yet another exercise), if $B\subset int(L)$, then $V(A)=V[\mathbf{g}(B)]\leq C^n V(B)$. And the last part of the exercise is to find compact sets $L_1 \subset L_2 \subset \dots$ with union $D$.

We know that $\mathbf{g}^{-1}$ will also have the Jacobian different from zero. So, I was thinking of using this last theorem this time to $\mathbf{g}^{-1}$, resulting in a different upper bound number $Q$, we would get $V(B)=V[\mathbf{g}^{-1}(A)]\leq Q^n V(A)$... But this doesn't seem to help at all. I have no idea how to prove that it's measurable. The hints seem only help me prove what would be the measure of B (bounds of the measure) instead of trying to prove that B has a measure. I'm probably not understanding something...

Any help would be appreciated. Thanks in advance.

P.S.: This is an exercise taken from Wendell Fleming's Functions of several variables, page 216, exercise 8. (It uses exercise 6 and 7) http://en.bookfi.org/book/580162

• What is $\Delta$? – tomasz Jan 3 '14 at 23:59
• It's the domain of $g$. – An old man in the sea. Jan 4 '14 at 9:26
• I know that much, but you should probably also say that it's open or something like that. – tomasz Jan 4 '14 at 15:36

This is pretty general : if $f: X \rightarrow Y$ is continuous, then f is measurable, i.e $\forall A \subset Y$, measurable, $f^{-1}(A)$ is measurable in X.
• What do you mean by measurable (function)? Also are you sure you wanted to write $\forall A\in Y$? – tomasz Jan 3 '14 at 23:46