Krdytkyu

This is a question from Ph. D Qualifying Exam of real analysis.

Let $F$ be an increasing function on $[0,1]$ with $F(0)=0$ and $F(1)=1$. Let $mu$ be a Borel measure defined by $mu((a, b))=F(b-)-F(a+)$ and $mu({0})=mu({1})=0$. Suppose that the function $F$ satisfies a Lipschitz condition $$|F(x)-F(y)|le A|x-y|$$ for some $A>0$. Let $m$ be the Lebesgue measure on $[0,1]$.

(a) Prove that $mu ll m$.

(b) Prove that $dfrac{dmu}{dm} le A $ a.e.

My attempt: (a) Since $F$ is Lipschitz continuous, it is clear that $F$ is absolutely continuous and $F$ is differentiable a.e. and $$mu((a, b))=int_{a}^{b}F'dm$$ by absolute continuity. Since $mu$ is a Borel measure, it extends to $mu(E)=int_E F'dm$ for every Borel set $E$.(I'm not sure for this part. Is there any related theorem or counterexample for this one?)

Therefore, it suffices to show that $int_E F' =0$ whenever $E$ is a Borel set and $m(E)=0$, and this is obvious from the definition of Lebesgue integral.

(b) From (a), $dfrac{dmu}{dm}=F'$ and by Lipschitz continuity, $|F'|le A$ a.e. and the result is obvious.

Am I correct? Is there any errors or logical jumps in my attempt?

You don't need the full extension to Borel sets - open sets will suffice. Recall that any open set $G subset mathbb R$ may be written as a union of disjoint open intervals ${(a_k,b_k)}$.

Suppose that $N subset (0,1)$ is a Borel set with Lebesgue measure zero. For any $epsilon > 0$ there exists an open set $G subset (0,1)$ with the property that $N subset G$ and $m(G) < epsilon$. Writing $G = cup (a_k,b_k)$ as above you find
$$mu(G) = sum_k mu(a_k,b_k) = sum_k int_{(a_k,b_k)} F' , dm = int_G F' , dm.$$

The monotonicity of $mu$ and the fact that $|F'| le A$ almost everywhere give you
$$mu(N) le mu(G) = int_G F' , dm le A mu(G) < A epsilon.$$
Since $epsilon > 0$ is arbitrary you get $mu(N) = 0$.

In general, if $N subset [0,1]$ is a Borel set with Lebesgue measure zero then
$$mu(N) le mu({0}) + mu(N cap (0,1)) + mu({1}) = 0$$
so that $mu(N) = 0$ too. Thus $mu ll m$.

I would suggest a simpler approach: First note that even though $F$ was only assumed Borel-measurable, we also assume $F$ is Lipschitz and thus continuous. Since $F$ is also increasing, we see that in fact $mu((a,b))=F(b)-F(a)=|F(b)-F(a)| leq A |b-a| = A : m((a,b))$. It follows immediately that $mu leq A m$ (in the sense of on every measurable set). Absolute continuity is clear now, and using the Radon-Nikodym theorem on $mathbb{R}$, i.e.
begin{equation}
frac{d mu}{d m}(x)=lim_{epsilon rightarrow 0} frac{mu((x-epsilon,x+epsilon))}{m((x-epsilon,x+epsilon))}
end{equation}
For almost all $x in mathbb{R}$, we can immediatly conclude
begin{equation}
frac{d mu}{d m}(x) leq A
end{equation}

I would‘t see what‘s wrong with your approach though, maybe this is a just a bit more immediate and uses less machinery.