# Radon-Nykodym approach to classic result: Left-cont. nondecreasing f on $[a,b] \Rightarrow$ exists pos. Borel measure s.t. $f(x)-f(a)=u([a,x))$

EDIT!!! The problem has been reduced to confirming that if for g strictly increasing left continuous and V open then g(V) is open. The proposition is to decompose g into a continuous and singular part, but then I'm unsure of the set manipulations to confirm that for an open interval, $(g_c+g_s)(I)$ is a potentially countable union of disjoint potentially degenerate intervals. See answer and comments, below.!!!

Struggling with exercise 7.12 part b in Rudin's Real & Complex Analysis.

$I = [a,b]$

f:I$\to\Bbb{R}$ is left continuous nondecreasing. Show that there is a positive Borel measure s.t. $f(x)-f(a)=\mu[a,x)$.

The hint is to imitate the proof of theorem 7.18:

$g:I\to\Bbb{R}$ is continuous nondecreasing. The following are equivalent.

1. $g$ is AC on I.

2. $g$ maps sets of Lebesgue measure zero to sets of Lebesgue measure zero.

3. $f$ is differentiable; $f'\in L^1; f(x) - f(a) = \int_a^xf'd\mu$.

$1 \rightarrow 2$ by approximating the Lebesgue measure zero set by an open set (hence a disjoint union of open segments) and applying abs. cont..

$2 \rightarrow 3$

• A) by defining $h(x) = x + g(x)$-- $h$ is injective; decompose any Lebesgue measurable set E to a Borel union of compact sets and a set of Lebesgue measure zero to show $m(g(E))$ is a measure;

• B) the Radon-Nikodym theorem gives the integral representation.

$3 \rightarrow 1$ is obvious.

I want to proceed analogously to the proof of $ii \rightarrow iii$ A). But, I am struggling to show that m((f+x)(E)) is a measure. It is clear that there are countably many points of discontinuity D by the monotonicity of f. I'm not sure how to show that f+x sends Borel sets to Lebesgue measureable sets.

• Please stop removing the probability-theory tag: problem 7.12 provides a nonstandard approach to a fundamental existence theorem from probability theory; search of similar questions on the exchange suggest this tag will likely attract the attention of responders most familiar with things of this sort. – entprise Apr 5 '16 at 14:19

I'm going to assume that $g$ is a continuous and strictly increasing function on $[a,b]$, and denote the set of Lebesgue measurable subsets of $[g(a), g(b)]$ by $\mathfrak{M}$. Put $$\Omega = \{E\subset [a,b]: g(E) \in \mathfrak{M}\}.$$ It is clear that $\Omega$ is a $\sigma$-algebra. For any open subset $V$ of $[a,b]$, $g(V)$ is also open. Hence $\Omega$ contains all open subsets of $[a,b]$. By definition, $\Omega$ contains all Borel subsets of $[a,b]$. That is, for any Borel set $E$, $g(E)$ is Lebesgue measurable.