Showing that $f'$ is the Radon-Nikodym derivative of $\lambda_f$

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 $f$. Consider valid $\lambda_f << \lambda$ (that is, $\lambda_f$ is absolutely continous with respect to $\lambda$). Prove that $$\lambda_f(E)=\int_Ef'd\lambda,~~~~\text{for all}~E~\text{measurable}.$$ (that is, $f'$ is the Radon-Nikodym derivative of $\lambda_f$)

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. I'd already proved that $\lambda_f(E)\geq\int_Ef'd\lambda$ (just a half of the problem), but I found my proof was too long.

Any hint for the problem in general or at least for the reverse inequality?

Answer 1

Another approach is possible: since $f$ is continuous and nondecreasing we have: $$ \lambda_{f}((x,y])=f(y)-f(x)$$ Since $\lambda_{f} << \lambda$, from the differentiation theorem for monotone functions (they are a.e differentiable) and the theorem about measure derivation (see below), we have: $$ \frac{d\lambda_{f}}{d \lambda}=\lim_{r \rightarrow 0} \frac{\lambda_{f}(B(x,r))}{\lambda(B(x,r))}=\lim_{y \rightarrow x} \frac{\lambda_{f}((x,y])}{y-x}=f'(x) \ \ a.e $$ The claim follows by the Radon-Nikodym theorem.

For completeness, I post the theorems mentioned above:

Derivation of measures and Radon derivative

We define the derivative of the measure $\mu$ with respect to another measure $\lambda$ as: $$D_{\lambda}(\mu)(x)=\lim_{r \rightarrow 0} \frac{\mu(B(x,r))}{\lambda(B(x,r))}$$ (to be more precise we should define the upper and lower derivatives first, then the derivative would be defined only if lower and upper ones agree). Let now $\lambda$ be the lebesgue measure and $\mu$ a $\sigma$-finite measure absolutely continuous with respect to $\lambda$. Then the Radon-Nikodym derivative of $\mu$ wrt $\lambda$ and $D_{\lambda}(\mu)$ agree almost everywhere, that is: $$D_{\lambda}(\mu)(x)=\lim_{r \rightarrow 0} \frac{\mu(B(x,r))}{\lambda(B(x,r))}=\frac{d \mu}{d \lambda} \qquad a.e $$

This theorem can be proved (at least if $\lambda$ is the Lebesgue measure) with:

Lebesgue differentiation theorem

If $f \in L^1_{loc}(\mathbb{R})$, then $$ \lim_{r \rightarrow 0} \frac{1}{\lambda(B(x,r))} \int_{B(x,r)} f(t) \, d\lambda(t) = f(x) \qquad a.e $$

Note that this result is almost trivial is $f$ is continuous!

Answer 2

Since $f$ is nondecreasing, $f' \ge 0$ so $\int_E f'$ is well-defined for measurable $E$. Also, it's enough to prove the result on some interval $[a,b]$, since, if the result holds here, we can write $E = \bigcup_k [k-1,k] \cap E = \bigcup_k E_k$ where the $E_k$'s are almost disjoint. Hence $\lambda_f(E) = \lambda_f(\bigcup_k E_k) = \sum_k \lambda_f(E_k) = \sum_k \int_{E_k} f' = \int_E f'$ by the monotone convergence theorem. So wlog we'll assume that we're working over some finite interval $[a,b]$ and hence $\lambda_f(E) < \infty$ for all measurable $E$.

First note that the collection of all finite unions of intervals $\mathcal{A}$ is an algebra and that $\int_E f' = \lambda_f(E)$ for all $E \in \mathcal{A}$ by the fundamental theorem of calculus.

Next let $\mathcal{M}$ be the collection of all sets $E$ for which the theorem holds. So $\mathcal{A} \subset \mathcal{M}$.

Claim: $\mathcal{M}$ is a monotone class

To see this, consider any increasing sequence $E_1 \subset E_2 \subset ...$ in $\mathcal{M}$. Then

$\int_{\cup_n E_n} f' = \int \lim_{N \to \infty} f' 1_{\bigcup_{n=1}^N E_n} = \int \lim_{N \to \infty} f' 1_{E_N} = \lim_{N \to \infty} \int f' 1_{E_N} $

by the monotone convergence theorem. The last integral is just $\lambda_f (E_N)$ because $E_N \in \mathcal{M}$ and $\lambda_f(E_N) \to \lambda_f (\bigcup_n E_n)$ by the continuity of measures.

So $\bigcup_n E_n \in \mathcal{M}$.

Next if $E_1 \supset E_2 \supset...$ is in$\mathcal{M}$ then $\lambda_f(E_1) < \infty$. By the continuity of measures,

$\lambda(\bigcap_n E_n) = \lim_{n \to \infty} \lambda_f(E_n) = \lim_{n \to \infty} \int_{E_n} f' = \lim_{n \to \infty} \int f' 1_{E_n}$

Since $1_{E_n} \to 1_{\bigcap_n E_n}$ as $n \to \infty$, the dominated convergence theorem shows that the last limit is just $\int_{\bigcap E_n} f'$.

So we have that $\mathcal{M}$ is closed under monotone countable unions and monotone countable intersections. Hence $\mathcal{M}$ is a monotone class.

By the monotone class theorem, the smallest monotone class containing $\mathcal{A}$ is the Borel $\sigma$-algebra. Hence the Borel sigma algebra is in $\mathcal{M}$ and so we're done.