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

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 $\lambda_f$).

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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.

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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.

• Since $f$ is continuously differentiable all you really need to prove is that $\int df=\int f'dx$ and this is straightforward. Feb 11, 2017 at 4:11
• @chilangoincomprendido I don't understand where in the problem I have to do this. Can you explain your idea a little more? Feb 11, 2017 at 4:16
• @chilangoincomprendido How can I prove this equality when integrating over all measurable sets? Feb 11, 2017 at 4:27
• Do you know the monotone class or $\pi$-$\lambda$ theorems? They make this almost trivial. You know the identity $\lambda_f(E) = \int_E f'\,d\lambda$ holds when $E$ is an interval, and these theorems give you just what you need to extend this to all Borel sets $E$. Feb 11, 2017 at 6:32
• @NateEldredge I can't use it, Nate. But I will take a look at them. Feb 11, 2017 at 13:13

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.

• $\mu_f$ is the Lebesgue-Stieltjes measure generated by $f$, right?You used $\mu$ too, what it is? I've never seen proofs like this. First of all, you used only intervals, and the statement is supposed to work for all Lebesgue Measurable sets (but for borelians it would be a hand). Furthermore, why $\lim_{x\to x_0}\frac{\mu((x,x_0])}{\lambda((x_0,x])}$ would be the Radon-Nikodym derivative? Feb 11, 2017 at 13:10
• I dropped the $f$ for convenience. I then used a standard result about the R-N derivative. See Cohn's Measure Theory text, for example. Feb 11, 2017 at 15:17
• see my edits for different approaches. Feb 11, 2017 at 17:34
• please explain you last assertion. $\int_A f'd\lambda$ is a supremum over partitions of $A$. How can I find these disjoint intervals such that $\int_Af'd\lambda >\int _{\bigcup _i(c_i,d_i)}f'd\lambda -\epsilon$? Feb 14, 2017 at 3:52