0

Let $g$ increasing functions absolutely continuous on $[a,b]$ with $g(a)=c$ and $g(b)=d.$

Show that for any open set $O\subset [c,d],\ m(O)=\int_{g^{-1}(O)}g'(x)dx$ (lebesgue integral).

I have the following formal calculations, but I need to justify some things.

To simplify the case, let $O=(c',d')\subset [c,d]$ open interval. Then, $m(O)=m((c',d'))=d'-c'=g(b')-g(a')=\int_{[a',b']}g'(x)dx=\int_{(a',b')}g'(x)dx=\int_{g^{-1}((c',d'))}g'(x)dx=\int_{g^{-1}(O)}g'(x)dx$.

How can I ensure that there exists an $a'$ such that $g(a')= c'$? (and also $g (b') =d'$?

How can I ensure that $g^{-1}((c',d')) = (a',b')$?

eraldcoil
  • 3,508

1 Answers1

1

Since $g$ is continuous, by definition the inverse image of every open set is open. Thus, $g^{-1}[(c',d')]$ is open.

Suppose $g^{-1}[(c',d')]$ was not an interval. Then there would exist points $x < y < z$ such that $x,z \in g^{-1}[(c',d')]$ and $y \notin g^{-1}[(c',d')]$. But since $g$ is increasing, $g(x) < g(y) < g(z)$, so if $g(x),g(z)\in (c',d')$, then so is $g(y)$, and thus $y \in g^{-1}[(c',d')]$, a contradiction. Thus, $g^{-1}[(c',d')] = (a',b')$ for some $a'$ and $b'$.

Since $g$ is increasing, we must have $g(a') \le c'$ and $g(b') \ge d'$. If $g(a') < c'$, there exists $x$ and $y$ with $g(a') < x< c' < y < d'$. From this, we have $a' < g^{-1}(x) < g^{-1}(y)$. However, $g^{-1}(y)\in (a',b')$ and $g^{-1}(x) \notin (a',b')$, so $g^{-1}(x) \le a' < g^{-1}(y)$. Combining these gives $a' < a'$, a contradiction. Thus, $g(a') = c'$. A similar argument shows $g(b') = d'$.

Note that to complete the proof for all open sets $O$, you need the theorem that all open sets of reals are countable unions of disjoint intervals.

eyeballfrog
  • 22,485