I'm working on a problem from Royden's Real Analysis:

Show that if a function $f$ is monotone increasing on $[a,b]$, then $f$ can be represented as the sum of an absolutely continuous function and a singular function.

I understand the general idea of the proof (I think), but there are a few details I'm unclear on. Here's my proof so far:

Let $f$ be monotone increasing on $[a,b]$. Let $g(x) = \int_a^x f'(t) dt + g(a)$. Since $g$ is an indefinite integral, $g$ is absolutely continuous.

Let $h(x) = f(x) - g(x)$. Then $h'(x) = f'(x) - g'(x)$.

Since $f$ is monotone increasing, by Theorem 5.3 $\ f'$ is measurable. Then, by Lemma 5.9 $g'(x) = f'(x)$ almost everywhere, so $h'(x) = 0$ almost everywhere, and $h$ is thus singular.

$\textbf{Theorem 5.3:}$ If $f$ is monotone increasing on $[a,b]$, then $f$ is differentiable a.e. and $f'$ is measurable.

$\textbf{Theorem 5.9:}$ If $f$ is bounded and measurable on $[a,b]$, and $F(x) = \int_a^x f(t) dt + F(a),$ then $F'(x) = f(x)$ a.e.

Here are my questions:

What allows me to say $h= f - g \implies h' = f' - g'$?

I would think it's just the differentiability of $f, g,$ and $h$. I know $f$ is differentiable because it's monotone increasing, and $g$ is differentiable because it's absolutely continuous and thus of bounded variation. Is this correct?

Is f' bounded?

In order to apply Lemma 5.9, f' must be both bounded and measurable. I could also use Thm 5.10, which requires $f'$ to be integrable, but I'm not sure if I have integrability, either.

$\textbf{Theorem 5.10:}$ If $f$ is integrable on $[a,b]$, and $F(x) = \int_a^x f(t) dt + F(a),$ then $F'(x) = f(x)$ a.e.

Is it necessary to let $g(x) = \int_a^x f'(t) dt + g(a)$, or can I let $g(x) = \int_a^x f'(t) dt$?

I would think I need the former in order to use 5.9 or 5.10, but I saw a proof that used the latter. Is there any difference in the two approaches?

  • $\begingroup$ (1)Since$ h=f-g$ we have $h'(x)=f'(x)-g'(x)$ wherever $f'(x)$ and $g'(x)$ exist.....(2) Theorems 5.9 and 5.10 will apply if you replace $b$ with any $c\in (a,b)$. You can then let $c$ approach $b$. I think this may work. $\endgroup$ Commented Nov 18, 2015 at 5:53

1 Answer 1


I am 4 years late but I figure I will answer this for anyone else who search up this question.

1) Yes just from differentiability, but only on the intersection of the set on which $f'$ is defined and the set on which $g'$ is defined, but since each of them is the complement of a set of zero measure, $h'$ is also defined a.e.

2) $f'$ is integrable on [a,b] by theorem 3 in the same chapter, which says that if $f$ is an increasing real-value function on the interval [a,b]. Then $f'$ defined a.e. and measurable and we have that:

$\int_a^b f' \leq f(b)-f(a)$.

This allows us to use Theorem 10 which says that if $g(x) =: \int_a^x f' + f(a)$, $g' = f'$ a.e.

3) Once you have $g'(x) = f'$ by answer to your question (1), $(g-f(a))' = (\int_a^x f')' = g' - 0 = g'$.


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