# Monotone increasing function can be expressed as sum of absolutely continuous function and singular function

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?

• (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. Commented Nov 18, 2015 at 5:53

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), $$(g-f(a))' = (\int_a^x f')' = g' - 0 = g'$$.