Showing that the Lebesgue-Stieltjes measure is absolutely continuous with respect to Lebesgue measure (and one more thing).

Question

I aim to show the following result:

Let $f:\mathbb{R}\to\mathbb{R}$ be a nondecreasing, continuously differentiable function and let $\lambda_f$ be the corresponding Lebesgue-Stieltjes measure generated by $f$. Prove:

(a) $\lambda_f <<\lambda$ (that is, $\lambda_f$ is absolutely continuous with respect to Lebesgue-measure $\lambda$);

(b) $\frac{d\lambda_f}{d\lambda}=f'$ (that is, $f'$ is the Radon-Nikodym derivative of $\lambda_f$).

-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-

For item (a), I tried consider the case when all the things happen in a compact interval $[a,b]$ (the general case easily follow from this one). So $f$ is uniformly continuous in $[a,b]$. Let $E\subset[a,b]$ with $\lambda(E)=0$ and let $\epsilon>0$ arbitrary. We want to find a cover $\{(a_k, b_k]\}_{k=1}^\infty$ of $E$ such that $$\sum_{k=1}^\infty [f(b_k)-f(a_k)]<\epsilon.\tag{1}$$ If we find $\{(a_k, b_k]\}_{k=1}^\infty$ sufficiently fine so that $f(b_k)-f(a_k)<\epsilon/2^k$ for all $k\in \mathbb{N}$, then (1) follows.

Here we have a problem: $\epsilon/2^k$ depends on $k$.

Since $\lambda(E)=0$, there exists a cover $\{(a_k, b_k]\}_{k=1}^\infty$ of $E$ such that $$\sum_{k=1}^\infty(b_k-a_k) < \frac{\epsilon}{\star},$$ where $\star\in\mathbb{R}^+$ I can control. But the number $\star$ cannot depends on $k$.

Now, what should I do?

Right now I will try to use the Mean Value Theorem in each interval and see what happen.

-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-#-

For item (b) (considering that item (a) is valid), the things become complicated to write. I just solve a half of this item. My attempt was to show that $\lambda_f(E)=\int_Ef'd\lambda$ for all $E$ measurable, because from Radon-Nikodym theorem, the derivative is $\lambda$-a.e. unique.

Since $f'\geq0$ ($f$ is nondecreasing), if we set

$$\mathcal{A} = \left\{\sum_{k=1}^\infty[f(b_k)-f(a_k)]:E\subset\bigcup_{k=1}^\infty(a_k, b_k]\right\}=\left\{\sum_{k=1}^\infty f'(c_k)(b_k-a_k):E\subset\bigcup_{k=1}^\infty(a_k, b_k]\right\}$$

(we used the Mean Value Theorem) and

$$\mathcal{B}=\left\{\int_E sd\lambda:0\leq s\leq f'~\text{is simple}\right\}=$$ $$=\sup\left\{\sum_{k=1}^n\alpha_k\lambda(A_k):E =\sum_{k=1}^n A_i~\text{and}~0\leq\alpha_k\leq f', \forall k\in A_k, k\in\{1,2,...,n\}\right\},$$

from the definition we get

$$\lambda_f(E) = \inf\mathcal{A}~~~~~~~~\text{and}~~~~~~~~\int_E f' d\lambda=\sup\mathcal{B}.$$

We need to prove that $\inf\mathcal{A}=\sup\mathcal{B}$.

Let $x\in \mathcal{A}$ and $y\in\mathcal{B}$ arbitraries. Then there exists a cover $\{(a_k, b_k]\}_{k=1}^\infty$ and a partition $\{A_r\}_{r=1}^n$ for $E$ and numbers $\alpha_1, ..., \alpha_n \geq 0$ with $\alpha_r\leq f'$ in $A_r$, $r=1, ..., n$, such that $$x = \sum_{k=1}^\infty f'(c_k)(b_k-a_k)~~~~~~~~\text{and}~~~~~~~~y=\sum_{r=1}^n\alpha_r\lambda(A_r).$$ We can assume that the family $\{(a_k, b_k]\}_{k=1}^\infty$ is disjoint (in fact, we can split it into more sets and split the expression of $x$ into more parcels). So each set $A_r$ in contained in some of the sets $\{(a_k, b_k]\}_{k=1}^\infty$. Call $I_r\subset\mathbb{N}$ the minimal set of indices such that $A_r\subset \sum_{k\in I_r}(a_k,b_k].$ Since $\{(a_k, b_k]\}_{k=1}^\infty$ is disjoint, $\{I_r\}_{r=1}^n$ is a partition of $\mathbb{N}$. Then $\alpha_r\leq f'$ in $A_r$ and $\lambda(A_r)\leq\sum_{k\in I_r}(b_k-a_k)$, what implies $\alpha_r\lambda(A_r)\leq\sum_{k\in I_r} f'(c_k)(b_k-a_k)$. From this we concludes $$y=\sum_{r=1}^n\alpha_r\lambda(A_r)\leq\sum_{r=1}^n\sum_{k\in I_r} f'(c_k)(b_k-a_k) = \sum_{k=1}^\infty f'(c_k)(b_k-a_k) = x.$$ Since $x, y$ are arbitrary, we have $\inf\mathcal{A}\geq\sup\mathcal{B}$.

How can I proceed to show that $\inf\mathcal{A}\leq\sup\mathcal{B}$?

Any hint will be really appreciated.

Answer 1

Define the premeasure $\mu_f$ on the algebra of half-open intervals $(a,b]$ by $\mu_f((a,b])=f(b)-f(a).$

Now, $\mu_f$ extends $uniquely$ to a measure $\mu$ on $\mathscr B(\mathbb R).$ Next, define, for each $A\in \mathscr B(\mathbb R)$, $\nu (A)=\int_A f'd\lambda. \ $ Since $\nu$ also extends $\mu_f,\ $ we have by uniqueness that $\nu=\mu$ and absolute continuity follows immediately, and clearly $\frac{d\mu}{d\lambda}=f'$.

edit: I think I have a proof from scratch:

The Lebesgue-Stieljes measure $\mu$ is the one that extends $\mu_f$. You want to prove from scratch (without the Monotone Class Theorem or similar) that

$\mu (A)=\int _Af'd\lambda.\ $

Now, clearly $\nu $ defined by $\nu(A)=\int _Af'd\lambda\ $ is a positive measure since $f$ is increasing. Therefore

$A\subseteq B\Rightarrow \int _Af'd\lambda)\le \int _Af'd\lambda.$

Note that $\mu_f(I)=f(b)-f(a)\ $ for $any$ interval with endpoints $a,b$ and that the result is clearly true if $A=(a,b]\ $ or if $\mu(A)=\infty.$

If $A$ is Borel such that $\mu(A)<\infty,\ $ there is a sequence of disjoint intervals $\left \{ (a_i,b_i) \right \}_{i\in \mathbb N}$ such that $A\subseteq \bigcup_i(a_i,b_i)\ $ and

$\mu(A)>\mu \left ( \bigcup_i (a_i,b_i) \right )-\epsilon=\sum_i (f(b_i)-f(a_i))-\epsilon=\int _{\cup_i (a_i,b_i)}f'd\lambda-\epsilon>\int_Af'd\lambda-\epsilon,\ $ so

$\mu (A)\ge \int_Af'd\lambda.$

Similarly, there is a sequence of disjoint intervals $\left \{ (c_i,d_i) \right \}_{i\in \mathbb N}$ such that $A\subseteq \bigcup_i(c_i,d_i)\ $ and

$\nu(A)=\int_Af'd\lambda >\int _{\bigcup _i(c_i,d_i)}f'd\lambda -\epsilon=\sum_i(f(d_i)-f(c_i))-\epsilon =\mu (\bigcup _i(c_i,d_i))-\epsilon>\mu(A)-\epsilon,$ so

$\mu (A)\le \int_Af'd\lambda.$

The result follows.