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Let $g$ be Riemann integrable on $[a,b]$, $f(x)=\int_a^x g(t)dt $ for $x \in[a,b]$.

Can I show that the total variation of $f$ is equal to $\int_a^b |g(x)| dx $?

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Yes you can. $ $ – Did May 28 '12 at 16:27
@did, How to show ? – Leitingok May 29 '12 at 1:20
@Leitingok Didier is trying to make you see "can I" should be replaced by "could you", maybe. – Pedro Tamaroff May 30 '12 at 2:49
From a perspective of prudence, may I ask why a 500 rep bounty was necessary for such a question? Though I understand it's your decision, I have to be honest, I am curious. – ThisIsNotAnId May 30 '12 at 5:53
This bounty is a freebie. – Norbert Jun 5 '12 at 9:26
up vote 15 down vote accepted

If $g$ is non-negative, $f$ is non-decreasing and the total variation is $f(b)-f(a)$ which coincides with $\int_a^b{|g(x)|}dx$, so the theorem is true.

For arbitrary $g$, write $g=g^+-g^-$, with at least one of $g^+(x)$ and $g^-(x)$ equal to 0.

Fix $\varepsilon>0$. There is a mesh $\delta^+>0$ such that for all partitions $a=x_0<\dots<x_n=b$ of $[a,b]$ finer than $\delta^+$ (i.e. $\max_i x_{i+1}-x_i\le\delta^+$), $$\sum_{i=1}^n (x_i-x_{i-1})\inf g^+([x_{i-1},x_i]) \ge \int_a^b{g^+(x)}dx - \varepsilon/2$$ We define $\delta^-$ symmetrically, and let $\delta=\min (\delta^+,\delta^-)$.

Now let's compute the total variation for a partition $x_0<\dots<x_n$ of $[a,b]$ finer than $\delta$: $$V=\sum_{i=1}^n |f(x_i)-f(x_{i-1})|$$ For any interval $I=[x_i,x_{i-1}]$, if $\inf g^+(I)>0$, then $g^+$ is always non-zero on this interval and $g^-$ must be identically zero. Then $$\begin{aligned} |f(x_i)-f(x_{i-1})| =&\int_I{g^+(x)}dx\\ \ge&(x_i-x_{i-1})\inf g^+([x_{i-1},x_i])\\ =&(x_i-x_{i-1})\inf g^+([x_{i-1},x_i]) + (x_i-x_{i-1})\inf g^-([x_{i-1},x_i]) \end{aligned}$$ The bound holds similarly when $\inf g^-(I)>0$. Finally when $\inf g^+(I)=\inf g^-(I)=0$, $$\begin{aligned} |f(x_i)-f(x_{i-1})| \ge&0\\ =&(x_i-x_{i-1})\inf g^+([x_{i-1},x_i]) + (x_i-x_{i-1})\inf g^-([x_{i-1},x_i]) \end{aligned}$$

So we can write $$V\ge \sum_{i=1}^n (x_i-x_{i-1})\inf g^+([x_{i-1},x_i]) + (x_i-x_{i-1})\inf g^-([x_{i-1},x_i])$$ and because the partition is finer than $\delta$: $$V \ge \left(\int_a^b{g^+(x)}dx - \varepsilon/2\right)+\left(\int_a^b{g^-(x)}dx - \varepsilon/2\right)$$ that is $$V \ge \int_a^b{|g(x)|}dx - \varepsilon$$

We also have the obvious upper bound $$V=\sum_{i=1}^n \left|\int_{x_{i-1}}^{x_i} {g(x)} dx\right|\le \sum_{i=1}^n \int_{x_{i-1}}^{x_i} {|g(x)|} dx = \int_a^b{|g(x)|}dx$$

Since this holds for all $\varepsilon$, the total variation (the upper bound of the total variation over all partitions) is precisely $\int_a^b{|g(x)|}dx$.

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very nice! thank you! – Leitingok May 31 '12 at 12:51
looks good. (+1) – robjohn Jun 1 '12 at 0:40

$\smash{\rlap{\phantom{\Bigg\{}}}\newcommand{\Var}{\mathrm{Var}}$ Break up $g(t)=g_+(t)-g_-(t)$ where $$ g_+(t)=\left\{\begin{array}{}g(t)&\text{if }g(t)\ge0\\0&\text{if }g(t)<0\end{array}\right.\tag{1} $$ and $$ g_-(t)=\left\{\begin{array}{}0&\text{if }g(t)\ge0\\-g(t)&\text{if }g(t)<0\end{array}\right.\tag{2} $$ Define $$ f_+(x)=\int_a^xg_+(t)\,\mathrm{d}t\tag{3} $$ and $$ f_-(x)=\int_a^xg_-(t)\,\mathrm{d}t\tag{4} $$ Then, $f(x)=f_+(x)-f_-(x)$, where $f_+$ and $f_-$ are montonic increasing.

Note that $$ |g(t)|=g_+(t)+g_-(t)\tag{5} $$ and $$ \Var_a^b(f)=\Var_a^b(f_+)+\Var_a^b(f_-)\tag{6} $$ In light of $(5)$ and $(6)$, assume that $g(t)\ge0$ and $f$ is monotonic increasing.

For any partition $P=\{t_i:0\le i\le n\}$ where $a=t_0\le t_{i-1}< t_i\le t_n=b$, define $$ \Var_{\lower{3pt}P}(f)=\sum_{i=1}^n|f(t_i)-f(t_{i-1})|\tag{7} $$ Since $g(t)\ge0$ and $f$ is monotonic increasing, for any partition of $[a,b]$, $$ \begin{align} \Var_{\lower{3pt}P}(f) &=\sum_{i=1}^n|f(t_i)-f(t_{i-1})|\\ &=\sum_{i=1}^nf(t_i)-f(t_{i-1})\\ &=f(b)-f(a)\\ &=\int_a^bg(t)\,\mathrm{d}t\\ &=\int_a^b|g(t)|\,\mathrm{d}t\tag{8} \end{align} $$ Combining $(5)$, $(6)$, and $(8)$, we can remove the restriction on $g$: $$ \int_a^b|g(t)|\,\mathrm{d}t=\mathrm{Var}_a^b(f)\tag{9} $$ for all Riemann integrable $g$, as required.

Detailed Explanation of $\mathbf{(6)}$:

The idea is that the increases in $f_+$ and $f_-$ are disjoint because $g(t)$ cannot be both positive and negative.

Given a partition of $[a,b]$, $P=\{t_i:0\le i\le n\}$ and its intervals $I_i=(t_{i-1},t_i)$, define the upper and lower Riemann sums as $$ {\sum_P}^+g=\sum_{i=1}^n\sup_{t\in I_i}g(t)\;|I_i|\quad\text{and}\quad{\sum_P}^-g=\sum_{i=1}^n\inf_{t\in I_i}g(t)\;|I_i|\tag{10} $$ Choose an $\epsilon>0$. Since $g$ is Riemann integrable, there is a partition of $[a,b]$, $P=\{t_i\}$ , so that $$ {\sum_P}^+g_+-{\sum_P}^-g_+<\epsilon\quad\text{and}\quad{\sum_P}^+g_--{\sum_P}^-g_-<\epsilon\tag{11} $$ Let $\mathcal{I}_\pm$ be the subcollection of $\{I_i\}$ where $\sup\limits_{I_i}g_+>0$ and $\sup\limits_{I_i}g_->0$. Since only one of $g_+(t)$ or $g_-(t)$ can be non-zero, we have that for $I_i\in\mathcal{I}_\pm$ $$ \inf\limits_{I_i}g_+=\inf\limits_{I_i}g_-=0\tag{12} $$ which implies that the lower Riemann sums of $g_+$ and $g_-$ restricted to $\mathcal{I}_\pm$ must be $0$: $$ {\sum_{\mathcal{I}_\pm}}^-g_+={\sum_{\mathcal{I}_\pm}}^-g_-=0\tag{13} $$ Thus, $(11)$ and $(13)$ imply that the upper Riemann sums of $g_+$ and $g_-$ restricted to $\mathcal{I}_\pm$ must be small: $$ {\sum_{\mathcal{I}_\pm}}^+g_+<\epsilon\quad\text{and}\quad{\sum_{\mathcal{I}_\pm}}^+g_-<\epsilon\tag{14} $$ Estimates $(14)$ show that $$ \Var_{\lower{3pt}\mathcal{I}_\pm}(f_+)<\epsilon\quad\text{and}\quad\Var_{\lower{3pt}\mathcal{I}_\pm}(f_-)<\epsilon\tag{15} $$ Let $\mathcal{I}_+$ be the subcollection of $I_i$ where $\sup\limits_{I_i}g_-=0$ and $\mathcal{I}_-$ be the subcollection where $\sup\limits_{I_i}g_+=0$. Then, $$ \Var_{\lower{3pt}\mathcal{I}_-}(f_+)=0\quad\text{and}\quad\Var_{\lower{3pt}\mathcal{I}_+}(f)=\Var_{\lower{3pt}\mathcal{I}_+}(f_+)>\Var_{\lower{3pt}P}(f_+)-\epsilon\tag{16} $$ $$ \Var_{\lower{3pt}\mathcal{I}_+}(f_-)=0\quad\text{and}\quad\Var_{\lower{3pt}\mathcal{I}_-}(f)=\Var_{\lower{3pt}\mathcal{I}_-}(f_-)>\Var_{\lower{3pt}P}(f_-)-\epsilon\tag{17} $$ Combining $(16)$ and $(17)$ yields $$ \begin{align} \Var_{\lower{3pt}P}(f) &\ge\Var_{\lower{3pt}\mathcal{I}_+}(f)+\Var_{\lower{3pt}\mathcal{I}_-}(f)\\ &>\Var_{\lower{3pt}P}(f_+)+\Var_{\lower{3pt}P}(f_-)-2\epsilon\tag{18} \end{align} $$ Since we can refine $P$ and make $\epsilon$ as small as we want, $(18)$ shows that $$ \Var_a^b(f)\ge\Var_a^b(f_+)+\Var_a^b(f_-)\tag{19} $$ The opposite inequality is immediate, so we get $(6)$.

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How do you get $(6)$? – leo May 31 '12 at 1:33
Let $P[a,b]$ the set of partitions of $[a,b]$. For a partition $P=\{a=x_0\lt\ldots\lt x_m=b\}\in P[a,b]$, define $S_P(f)=\sum_{k=1}^n|f(x_k)-f(x_{k-1})|$, so that $$V_f[a,b]=\sup_{P\in P[a,b]}S_P(f)$$ is the total variation of $f$ over $[a,b]$, when the sup exist. I think it is worth to point out that the $\lim_{||P||\to0}S_P(f)$ not always agree with $V_f[a,b]$. However, continuity of $f$ is enough to get $\lim_{||P||\to0}S_P(f)=V_f[a,b]$. That's the Theorem stated in my answer (to be continued...) – leo May 31 '12 at 1:48
(...continuation) In this case everything is fine because Riemann integrability of $g$ insures continuity of map $x\mapsto\int_a^x g(t)\ \mathbb{d}t$. So the limit step holds, but it is not a trivial step I think. – leo May 31 '12 at 1:50
@robjohn, why (6) holds? – Leitingok May 31 '12 at 2:54
(6) is definitely the difficult part of the proof. If $g$ is non-negative, there is an easy proof by seeing that $f$ is non-decreasing, thus the total variation of an arbitrary partition is $f(b)-f(a)$ (which coincides with $\int_a^b{|g(x)|}dx$) and the theorem is true. – Generic Human May 31 '12 at 11:54

$\newcommand{\P}{\mathcal{P}[a,b]}\newcommand{\R}{\mathbb{R}}$ Let's to establish the notation first.

Let $\P$ the set of all the partitions $\Gamma=\{a=x_0\lt x_1\lt\ldots\lt x_m=b\}$ of $[a,b]$. For such a $\Gamma$, define $$|\Gamma|=\max_{i\in \{1,\ldots,m\}} x_i - x_{i-1}$$ and $$S_\Gamma[a,b]=\sum_{i=1}^m |f(x_i)-f(x_{i-1})|$$

We say that a function $h$ is of bounded variation on $[a,b]$ if $\{S_\Gamma[a,b]:\Gamma\in\P\}$ is bounded. In that case, $$V_f[a,b]=\sup\{S_\Gamma[a,b]:\Gamma\in\P\}.$$

Now, consider the following theorem taken from Zygmund & Wheeden, Measure and Integral.



Provided that $g$ is Riemann integrable on $[a,b]$, the function $$f(x)=\int_a^x g(t)\ \mathbb{d}t$$ is continuous. For any partition $\Gamma=\{a=x_0\lt\ldots\lt x_m=b\}$ we have $$\begin{equation} \sum_{i=1}^m |f(x_i)-f(x_{i-1})|= \sum_{i=1}^m \left| \int_{x_{i-1}}^{x_i} g(t)\ \mathrm{d}t \right|\tag{1} \end{equation}$$ By the Theorem 2.9 cited above, it is enough to show that $$\lim_{|\Gamma|\to 0} \sum_{i=1}^m \left| \int_{x_{i-1}}^{x_i} g(t)\ \mathrm{d}t \right|=\int_a^b |g(t)|\ \mathrm{d}t$$ because in that case in view of (1) we get the desired result.

If we assume $g$ continuous on $[a,b]$ then $g$ is integrable on $[a,b]$ and therefore $$\begin{align*} \sum_{i=1}^m \left| \int_{x_{i-1}}^{x_i} g(t)\ \mathrm{d}t \right| &= \sum_{i=1}^m \left| g(\theta_i)(x_i-x_{i-1}) \right|\\ &= \sum_{i=1}^m |g(\theta_i)|(x_i-x_{i-1}) \end{align*}$$ for some $\theta_i\in (x_{i-1},x_i)$, so $$\lim_{|\Gamma|\to 0} \sum_{i=1}^m \left| \int_{x_{i-1}}^{x_i} g(t)\ \mathrm{d}t \right|=\lim_{|\Gamma|\to 0}\sum_{i=1}^m |g(\theta_i)|(x_i-x_{i-1})=\int_a^b |g(t)|\ \mathrm{d}t$$ as we want.

Another approach. Remember that for a Riemann integrable function on $[a,b]$ the Riemann and Lebesgue integrals over $[a,b]$ are the same.

Theorem 1. Let $f:\R\to\R$ be monotone increasing and finite on an interval $(a,b)$. Then $f$ has a measurable, nonnegative derivative $f'$ almost everywhere in $(a,b)$. Moreover, $$0\leq\int_a^b f'\leq f(b-)-f(a+).$$

Theorem 2. Let $f$ of bounded variation on $[a,b]$. Write $$V(x)=V_f[a,x].$$ Then $$V'(x)=|f'(x)|$$ for almost every $x\in [a,b]$.

Corollary 3. Let $f$ of bounded variation on $[a,b]$. Then $$\int_a^b |f'|\leq V_f[a,b].$$

Proof. Since $f$ is of bounded variation on $[a,b]$, by Theorem 2 we have $$\int_a^b |f'|=\int_a^b V',$$ since $V$ is increasing by Theorem 1. we get $$\int_a^b |f'|=\int_a^b V'\leq V(b)-V(a)=V(b)=V_f[a,b].$$

Theorem 4 (Lebesgue's Differentiation Theorem). Let $f\in L([a,b])$. Then $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\ \mathrm{d}t=f(x)$$ for almost every $x\in [a,b]$.

Now, since $g$ is Riemann integrable, $g\in L[a,b]$. By the above Theorem $$f'(t)=g(t)$$ for almost every $t\in [a,b]$.

The estimate (already dealt with in this answer) $$S_\Gamma\leq \int_a^b |g|\quad \forall\Gamma\in \P$$ shows that indeed $f$ is of bounded variation on $[a,b]$. Moreover, it says that $$V_f[a,b]\leq\int_a^b |g|\tag{2}$$ Therefore Corollary 3 is applicable and by the above observation ($f'(t)=g(t)$ a.e. in $[a,b]$) we get $$\int_a^b |g|=\int_a^b |f'|\leq V_f[a,b],$$ in view of $(2)$ we conclude $$\int_a^b |g|=V_f[a,b]$$ as we wanted.

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what's the defination of $S_{\Tau}$ and $V$ in your picture. – Leitingok May 30 '12 at 5:44
@Leitingok Given a partition $\Gamma=\{a=x_0\lt\ldots\lt x_m=b\}$ of $[a,b]$, $$S_\Gamma=\sum_{i=1}^m |f(x_i)-f(x_{i-1})|$$ and $V$ is the total variation of $f$ over $[a,b]$. – leo May 30 '12 at 6:00
@leo , what happans, if $g$ is not continuous ? – Leitingok May 30 '12 at 7:29
@Leitingok, I'm on it. I'll post the answer in some hours. The answer I'll give is via Lebesgue integration, since the Lebesgue and Riemann integrals agree for Riemann integrable functions, the result holds for Riemann integrable functions. – leo May 30 '12 at 10:06
@Leitingok answer updated... – leo May 31 '12 at 23:46

Just reading the definition in Wikipedia, I can give you some ideas:

$$V_a^b\left( f \right) = \mathop {\sup }\limits_P \sum\limits_{i = 0}^{{n_P} - 1} {\left| {f\left( {{x_{i + 1}}} \right) - f\left( {{x_i}} \right)} \right|} $$

where $\sup$ runs over the set of all partitions $P$ of $[a,b]$,

$$\mathcal P=\{P=\{x_0,\dots, x_{n_P}\}:P\text{ is a partition of }[a,b]\}$$

I assume $g$ is continuous in $[a,b]$. This means $f$ differentiable in $[a,b]$ and continuous, and we can apply the MVT, which means


where $y_i \in (x_i,x_{i+1})$. Thus the sum becomes:

$$V_a^b\left( f \right) = \mathop {\sup }\limits_P \sum\limits_{i = 0}^{{n_P} - 1} {\left| f'(y_i) \right||x_{i+1}-x_i|} $$

If $g$ is continuous then it is bounded, and therefore it is Riemann integrable.

Note that if $f(x)=\int_a^x g(t) dt$ then $f'(x)=g(x)$. Then the sum is

$$V_a^b\left( f \right) = \mathop {\sup }\left\{ \sum\limits_{i = 0}^{{n_P} - 1} {\left| g(y_i) \right||x_{i+1}-x_i|}:P=\{x_0,\dots,x_{n_P}\} \text{ is a partition of }[a,b] \right\}$$

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One direction's pretty easy, for a partition $a:= x_0 < x_1 < \ldots < b := x_n$, we have $$\sum |f(x_{i+1}) - f(x_i)| = \sum |\int_{x_i}^{x_{i+1}} g(t)dt| \leqslant \sum \int_{x_i}^{x_{i+1}} |g(t)|dt = \int_a^b |g(t)|dt$$So total variation is bounded above by this.

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The only hypothesis given on $g$ is that $g$ is Riemann integrable over $[a,b]$. Riemann integrability of $g$ over $[a,b]$ is not enough to conclude $f(x_{i+1}) - f(x_i) = g(x'_i)(x_{i+1} - x_i)$, for some $x_i < x'_i < x_{i+1}$. – leo May 30 '12 at 4:46
@leo (i) $f$ is differentiable with derivative $g$, (ii) divide that equality you wrote by $(x_{i+1} - x_i)$ and you have the mean value theorem. (i) (ii) – uncookedfalcon May 30 '12 at 4:50
If $g$ is continuous on $[a,b]$ then $f$ is differentiable with derivative $g$. I mean the identity $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^x g(t)\mathrm{d}t=g(x)$$ holds at the continuity points of $g$ as already stated here – leo May 30 '12 at 5:01
My goodness, you are absolutely right. Take any continuous function and change its value at a point, what I claimed can't be true ($\int$ is unchanged)! Thanks mate, back to the drawing board. – uncookedfalcon May 30 '12 at 5:12
No problem ${}$ – leo May 30 '12 at 6:02

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