Show that $f(g(x))g′(x)$ is measurable where $f$ is bounded and measurable

Question

Let $g$ be absolutely continuous and increasing on $[a,b]$ with $g(a)=A$ and $g(b)=B$. Suppose $f$ is any measurable and bounded function on $[A,B]$.

Show that $f(g(x))g'(x)$ is measurable on $[a,b]$.

My ideas:

1. Show that if a function $G$ is increasing and $F$ is absolutely continous in $[a,b]$, then $G(F(x))$ is absolutely continous. (1)
2. If $f$ is bounded, then is $L^ 1$, then we can write a function $G = \int_a^xf(t) dt$ that is absolutely continous and then by a theorem $G'= f(x)$ a-e
3. Then $G(g(x))$ is absolutely continous(by (1)) and then $G'(x)$ exist.
4. $(G(g(x)))' = G'(g(x)) g'(x) = f(g(x)) g'(x)$
5. Using the fact that if a function $H$ is increasing and continuous, then $H'$ exists almost everywhere and $H'$ is measurable, I get that $H'(x) = f(g(x)) g'(x)$ is measurable.

I think that the first assumption is not true, can someone help me to prove it?, also, is it true that $G(g(x))$ is continous and increasing?

Answer 1

I don't think the monotonic assumption is required other than to establish that $g([a,b]) \subset [A,B]$. (Perhaps there is a simpler proof with this assumption, but it escapes me.)

You have pointed out that $G$ can be written as $G(x) = \int_a^x f(t)dt$. Since $f$ is bounded, say by $B$, it follows that $G$ is Lipschitz continuous.

The composition of a Lipschitz function with an ac. function is easily shown to be ac., hence $G \circ g$ is ac. and so $(G \circ g)'(x)$ exists for ae $x$.

To finish, we need to establish that $(G \circ g)'(x) = G'(g(x))g'(x)$ for ae. $x$. This does not follow immediately from the usual composition rule, since $g$ may map into a non differentiable point of $G$ for a non null set.

Note that if $g'(x_0) = 0$ then the expression $G'(g(x))g'(x)$ is taken to be zero, regardless of whether $G'$ is differentiable at $g(x)$.

The following proof is copied from here (see Theorem 3.1). Let $E$ be such that $g'$ and $(G \circ g)'$ exist on the complement of $E$.

If $g(x_0)' = 0$ then for any $\epsilon>0$ there is some $\delta$ such that $|g(x)-g(x_0)| \le \epsilon |x-x_0|$ for $x \in B(x_0,\epsilon)$. Since $G$ is Lipschitz with rank $B$ we have $|G(g(x))-G(g(x_0))| \le M \epsilon |x-x_0|$ and so $(G \circ g)'(x_0) = G'(g(x)) g'(x_0) = 0$.

Now suppose $g'(x_0) \neq 0$. Note that for $x$ close to $x_0$ we have $g(x) \neq g(x_0)$. It is straightforward to see that $(G \circ g)'(x_0) = \lim_{x \to x_0} { G(g(x))-G(g(x_0)) \over x-x_0 } = \lim_{x \to x_0} { G(g(x))-G(g(x_0)) \over g(x)-g(x_0) } { g(x)-g(x_0) \over x-x_0 } = d g'(x_0)$, where $d=\lim_{x \to x_0} { G(g(x))-G(g(x_0)) \over g(x)-g(x_0) }$.

It is easier to finish using sequences. Suppose $y_n \to g(x_0)$, then there are $x_n \to x_0$ such that $g(x_n) = y_n$. Then $\lim_{n} { G(g(x_n)))-G(g(x_0)) \over g(x_n)-g(x_0) } = \lim_{n} { G(y_n))-G(g(x_0)) \over y_n-g(x_0) } = d$ from which we see that $d=G'(g(x_0)$.