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Exercise 11, chapter 6 from second edition of Baby Rudin:

Suppose $g$ is Riemann integrable on $[a,b]$, put $$\int_a^x g(t)dt,$$ and define $g^+(t)=\max(g(t),0)$, $g^-(t)=-\min(g(t),0)$. Prove that $f$ is of bounded variation on $[a,b]$ and that its variation functions are given by $v(x)=\int_a^x \left| g(t)\right|dt$, $p(x)=\int_a^x g^+(t)dt$ and $q(x)=\int_a^x g^-(t)dt$ where $p,q$ are positive and negative variation.

I'm aware of this question but I still have doubts, because he says "I also know that if F is BV, then F′ exists a.e.(almost everywhere)" I don't have the concept of almost everywhere yet. Also, I can't follow the proof because I don't know if $f$ is differentiable, that depends on $g$ being continuous!

I did this:

$$\left| f(x_i)-f(x_{i-1}) \right| = \left| \int_{x_{i-1}}^{x_i} g(t)dt \right| $$

My intention was to take sum at both sides and then limit as the diameter of the partition goes to zero but...If $g$ is continuous it's super easy using mean value theorem for integrals! Otherwise...I don't know which road to take here...

Also, if I could prove the last two equalities it'd be done, since: $|g|=g^-+g^+$, but doing that first was way harder...

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Lotte
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    Look at http://math.stackexchange.com/a/151501/27978. – copper.hat May 07 '15 at 04:19
  • Oh, haven't seen that one...Should I delete mine? – Lotte May 07 '15 at 04:25
  • I'll mark it as a duplicate... (Wow, didn't realise it just took one vote!) – copper.hat May 07 '15 at 04:27
  • I still have a little question...how do I prove the stated result for $p$ and $q$... – Lotte May 07 '15 at 05:13
  • In robjohn's answer, we have $p(x)=\operatorname{Var}a^x(f+)$, similarly for $q$. Is this what you are looking for? – copper.hat May 07 '15 at 05:16
  • In his answer I'd have to prove that $p(x)=f_-(x)$. I can easily prove that $v_f(x)=\int_a^xg_-+\int_a^x g_+$ but it does not implies that each of the terms equal $p,q$ – Lotte May 07 '15 at 14:47
  • I'm not sure what you are asking; his (robjohns's answer, (3)) $f_+$ is exactly the $p$ above? – copper.hat May 07 '15 at 14:57
  • Yes, $f_+$ it's my $p(x)$, but I have to prove that they are actually equal, is what my problem asks...I know that by definition $p(x)=\frac{1}{2}(v(x)+f(x)-f(a))$, so I guess I have to prove that $f_+=\frac{1}{2}(v(x)+f(x)-f(a))$ – Lotte May 07 '15 at 15:18
  • I'm still not sure what you are asking. If you show the formula for $v$ above then you have $v = p+q$, and you also have $f(x)-f(a) = p(x)-q(x)$. This gives $p(x) = {1 \over 2} (v(x)+f(x)-f(a))$? – copper.hat May 07 '15 at 15:44
  • I have to show 3 different things: $v(x)=\int_a^x \left| g(t)\right|dt$ (done), $p(x)=\int_a^x g^+(t)dt$ (not done), $q(x)=\int_a^x g^-(t)dt$ (not done but similar to the last).

    My definition of $p$ is $p(x)=\frac{1}{2}(v(x)+f(x)-f(a))$, his answers shows that $v(x)=\int_a^x \left| g(t)\right|dt$ which implies that $v(x)=\int_a^xg_-+\int_a^x g_+$ (because $|g|=g^-+g^+$).

    I have to show that indeed $f_+=p$

     – Lotte May 07 '15 at 16:17
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    Well, since $v=p+q = f_++f_-$ and $f(x)-f(a) = p(x)-q(x) = f_+(x)-f_-(x)$ then you get the desired answer? (I could easily be missing your point.) – copper.hat May 07 '15 at 16:23
  • Oh god, yes, such a simple answer! Sorry! :/ – Lotte May 07 '15 at 16:29
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    Glad to be able to help! – copper.hat May 07 '15 at 16:31

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