TITLE: 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))$
QUESTION [5 upvotes]: 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.

$ g$ is AC on I.
$g $ maps sets of Lebesgue measure zero to sets of Lebesgue measure zero.
$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.
Thanks for your help!

REPLY [2 votes]: 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.
I believe you know how to proceed from here.