# 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.

Thanks for your help!

• 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

## 1 Answer

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.

• Thanks! I totally missed that it was an open map! I'll try it tomorrow and upvote if it works out! – entprise Apr 6 '16 at 4:28
• Sorry! Late in getting back to this. But, by assuming g is continuous it kind of makes the problem trivial. The problem is that the discontinuities might not be isolated?? – entprise Apr 14 '16 at 18:32
• @entprise You can write an increasing function as the sum of a continuous one and a series of jump functions. See Stein's real analysis for the treatment (or figure it out yourself). That's probably the only part I like about Stein's treatment of differentiability. – Qiyu Wen Apr 14 '16 at 18:43
• Ah. So. It's not the case that knowing it's left continuous allows you to directly show it sends intervals to elementary sets?? (And. You're absolutely right that separating the jumps makes it easy. I'm just apprehensive because the rest of the question doesn't head in "that direction") – entprise Apr 14 '16 at 18:55
• @entprise You may find a cleaner treatment. I haven't attempted much. – Qiyu Wen Apr 14 '16 at 19:02