5
$\begingroup$

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{d\mu}{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), $\dfrac{d\mu}{dm}=F'$ 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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $\mu \ll m$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .