# Absolute continuity, implies bounded variation.

I am trying to directly prove from the definition of absolute continuity, that if a function $$f:[a,b]\rightarrow\mathbb{R}$$ is absolutely continuous, then it is of bounded variation, i.e $$Vf$$ is finite.

I have tried taking a uniform partition of [a,b] and played around with the definition. But it just does not seem intuitive to me. What is the main idea behind the proof?

image below is as described in the comment to the answer. https://i.sstatic.net/TzljL.png

• The total variation over $[a,b]$ is the sum of the total variations taken over the subintervals of a partition of $[a,b]$. Absolute continuity tells us the total variation can be made less than $1$ over an interval if the length of that interval is sufficiently small, say, less than $\delta>0$. You partition $[a,b]$ into a finite number of subintervals each of length $\delta$. Commented Apr 21, 2019 at 14:56

So then since $$f$$ is absolutely continous on $$[a,b]$$, for $$\epsilon=1 >0$$ there exists $$\delta > 0$$ such that for all finite disjoint open intervals on $$[a,b]$$, $$\{(x_1,y_1),...,(x_n,y_n)\}$$ with sum of their length less than $$\delta$$, we have $$\sum_{i=1}^n |f(y_i)-f(x_i)|<\epsilon$$.

let $$P*=\{a_0=a,...,a_n=b\}$$ be a partition of $$[a,b]$$ with tthe propertiy that $$a_i-a_{i-1}=\frac{\delta}{2}$$ for all $$i\in \{1,2...,n-1\}$$, and $$a_n-a_{n-1} \leq \frac{\delta}{2}$$.

Then $$n= \lfloor \frac{2(b-a)}{\delta} \rfloor +1$$. Now pick any partition $$P$$ and let $$P'=P\cup P^*$$ and $$P'=\{z_0=a,...,z_n=b\}$$. Let $$P_i'$$ be the set of points in $$P'$$ contained in $$[a_{i-1},a_i]$$ for $$i\in \{1,...,n\}$$, i.e.,

$$P'_i = \{z_{i_k}\in P'\,:\, z_{i_k}\in[a_{i - 1},a_{i}] \}$$

So variation $$V(P,f)=\sum_{i=1}^s|f(c_i)-f(c_{i-1})|\leq V(P',f)=\sum_{i=1}^n\sum|f(z_{i_k})-f(z_{i_{k-1}})|\leq n$$. Here $$\{c_0,...,c_s\}$$ is the partition $$P$$.

• What does the notation $V(P,f)$ mean?
– kam
Commented Apr 21, 2019 at 15:46
• I added the definition, it is the "variation" of the specific partition $P$ like it is one of the partition in the all possible partition $[a,b]$ in the definition of total variation! I got it from wiki: en.wikipedia.org/wiki/Bounded_variation I also refered: mathonline.wikidot.com/… Commented Apr 21, 2019 at 15:54
• In the second link, there is a diagram which shows that the gap between two endpoints in the partition $P*$ is $\delta$ not $\frac{\delta}{2}$. is there any reason why we choose $\frac{\delta}{2}$ in the restriction for $a_i-a_{I-1}$ and not just $\delta$? I'll try attach the picture in the main question.
– kam
Commented Apr 21, 2019 at 17:04
• I think that is because in $\{(x_1,y_1),...,(x_n,y_n)\}$ , we need sum of their length to be $\underline{strictly}$ less than $\delta$. I verified Wikipedia also it confirms that: en.wikipedia.org/wiki/Absolute_continuity Commented Apr 21, 2019 at 17:14
• I think this answer contains a little mistake (please confirm if so!). Suppose you have the point $z = a_i$ in the last sum. Because it is both in $[a_{i-1}, a_i]$ and $[a_i, a_{i+1}]$, it gets counted twice. To correct, you can work with half open intervals and the last sum can be made over the closed interval? Commented Apr 21, 2019 at 17:23

This answer is spot on, but let me give something more verbose since you asked about intuition and ideas. First, let's recall what we are assuming in the hypothesis of the theorem: absolute continuity of $$f$$ which is defined on $$[a, b]$$. Let's explicitly spell out what that means first. It means that for any $$\epsilon$$, there exists some $$\delta$$ such that for any finite set of $$n$$ disjoint intervals $$I_i = (c_i, d_i)$$ contained within $$[a, b]$$, whenever the sum of their lengths is less than $$\delta$$, then the sum of the increments of $$f$$ over those intervals is less than $$\epsilon$$. More mathematically, forall $$\epsilon > 0$$ there exists $$\delta > 0$$ such that:

$$\Sigma_{i = 1}^n |I_i| \leq \delta \implies \Sigma_{i = 1}^n |f(d_i) - f(c_i)| < \epsilon$$

At this point, let's stop for a second and think - does the theorem we're about to prove make sense? To me it does. The above condition already links the fate of the increments of a function to the fate of the interval lengths across which those increments are calculated. We can shrink those intervals to make the increments as small as we like. On the other hand, a partition is nothing but a set of intervals. The only thing we're missing is the shrinking part - we have no control over the length of the intervals in the partition. Hmmm, so this is the work we need to do in our proof.

Suddenly, an idea comes to mind. This, I believe, is the main idea behind the proof: what happens to a partition when we add points to it? Well, the variation across the partition does not decrease. It can only increase by adding more points.

Now, we lamented earlier about having no control over the lengths of the intervals in an arbitrary partition. But, by the above observation, we can construct any other partition we want, making its intervals as small as we want, and then add it into the original partition to form a new partition whose variance is at least as large as that of the original partition. Furthermore, by construction we can then invoke the condition of absolute continuity to arrive at bounded total variation.

To prove these ideas more formally, first, let's recall what total variation across an interval $$I = [a, b]$$ is: it's the supremum of variations of all paritions $$P$$ across $$I$$. For brevity, let's denote the variation across a partition $$P$$ as $$V(P)$$.

$$V(P) = \sum_{p_i, p_{i+1} \in P} |f(p_{i+1}) - f(p_i)|$$

Note that the function $$f$$ is implicit in the notation above, and that the sum is only over consecutive pairs $$p_i, p_{i+1}$$, not all pairs. So to show that the supremum of all such variations across some interval $$I = [a, b]$$ is bounded, we must find some constant $$C$$ such that for any partition $$P$$ over $$I$$, $$V(P) \leq C$$.

To start the proof, let's call upon our hypothesis of absolute continuity of $$f$$ over $$[a, b]$$. Choosing $$\epsilon = 1$$, by our hypothesis there exists a $$\delta$$ such that for any set of $$n$$ disjoint intervals $$I_i = (c_i, d_i)$$:

$$\Sigma_{i = 1}^n |I_i| \leq \delta \implies \Sigma_{i = 1}^n |f(d_i) - f(c_i)| < 1$$

Note: the choice of $$\epsilon = 1$$ here is arbitrary. Any value whatsoever would do just fine.

Now, taking an arbitrary partition $$P$$ we want to show its variance is bounded, and the bound doesn't depend on the choice of $$P$$. First thing we do is construct a partition $$Q$$ whose intervals all have length $$\delta$$, except for the last one which has length at most $$\delta$$: $$a, a + \delta, a + 2\delta, \dots, b$$. Then form a new partition by unioning this with the original so that $$P' = P \cup Q$$. At this point we can apply the idea we had: since $$P \subseteq P'$$, then by Lemma 1 below $$V(P) \leq V(P')$$ and so it suffices to show that $$V(P') < C$$ where $$C$$ is some constant that does not depend on $$P$$.

The next main idea in the proof is that the variation of $$P'$$ which is defined over $$I = [a, b]$$ can also be seen as the sum of variations of "sub partitions" across $$I_1 = [a, q_1], I_2 = [q_1, q_2], \dots, I_m = [q_{m-1}, q_m]$$. That is, let $$R_i = P' \cup I_i$$. Then:

$$V(P') = \sum_{i = 1}^m V(R_i)$$

Once the sum is rearranged this way, we notice a couple of nice things:

• The number of terms in the sum, $$m$$, does not depend on the partition $$P$$. This means any term depending on $$m$$ can be safely regarded as a "constant" w.r.t. the choice of $$P$$. Note: the value $$m$$ does depend on the partition $$Q$$, but that is OK- it only needs to be constant w.r.t. $$P$$.
• By construction of $$Q$$, the sum of all intervals in each sub-partition $$R_i$$ is less than or equal to the magic constant $$\delta$$, which sets us up nicely for applying our hypothesis of absolute continuity now.

Following on from that last observation, we are ready to apply the hypothesis of absolute continuity to each sub-partition $$R_i$$. Suppose $$R_i = \{x_0, x_1, x_2, \dots, x_r\}$$ with $$x_0 < x_1 < \dots < x_r$$. Then we can construct disjoint intervals $$J_1, J_2, \dots, J_r$$ where $$J_k = (x_{k-1}, x_k)$$. Clearly, the intervals $$J_k$$ are all disjoint. Also, since all intervals $$J_k \subset I_i$$, then we have that:

$$\Sigma_{k = 1}^r |J_k| \leq |I_i| \leq \delta$$

And so, applying our hypothesis of absolute continuity for $$\epsilon = 1$$, we have:

$$V(R_i) = \Sigma_{k = 1}^r |f(x_{k+1}) - f(x_k)| < \epsilon = 1$$

Therefore, putting it altogether:

$$V(P) \leq V(P') = \sum_{i = 1}^m V(R_i) < m$$

This shows that for any interval $$I$$, there is some constant $$m$$ bounding the size of the variation across any partition of $$I$$, which in turn proves that the supremum of all such variations i.e. the total variation of $$f$$ across $$I$$ is bounded.

Lemma 1: The variation across a partition does not decrease as we add more points. For this, it suffices to show that adding a single point does not decrease variation, because we can always just rinse, repeat as many times as we like. So, we are given a partition $$P = (p_0 = a, p_1, \dots, p_n = b)$$. Now suppose we add in some point $$x$$ with $$x \neq p_i$$. Then there is some open interval $$(p_k, p_{k+1})$$ containing $$x$$. Now, the variation across the new partition $$P' = P \cup \{x\}$$ is given by:

$$V(P') = \Sigma_{i = 0}^k |f(p_{i+1}) - f(p_i)| + |f(x) - f(p_k)| + |f(p_{k+1}) - f(x)| + \Sigma_{i = k+1}^n |f(p_{i+1}) - f(p_i)|$$

But since our original parition's variation is given by:

$$V(P) = \Sigma_{i = 0}^n |f(p_{i+1}) - f(p_i)| = \Sigma_{i = 0}^k |f(p_{i+1}) - f(p_i)| + |f(p_{k+1}) - f(p_k)| + \Sigma_{i = k+1}^n |f(p_{i+1}) - f(p_i)|$$

Then

$$V(P') = V(P) - |f(p_{k+1}) - f(k)| + |f(p_{k+1}) - f(x)| + |f(x) - f(p_k)|$$

So to show $$V(P') \geq V(P)$$ it remains to show:

$$|f(p_{k+1}) - f(k)| \leq |f(p_{k+1}) - f(x)| + |f(x) - f(p_k)|$$

Which is just the triangle inequality.

• Hey @Colm Bhandal, thanks for the awnser ! Very intuitive. I was just wondering, why isn't it working with a uniform continuous function on [a,b], by taking $\delta$ coming from the definition with $\epsilon = 1$ ? Because, it is not supposed to be working since $x\sin(1/x)$ on $[0,1]$ would be a counterexample, I guess. Could you help me clarify this point ? Commented May 28, 2021 at 11:14
• @Akurishen I cannot answer your query with complete confidence. But note that absolute continuity is a stronger condition than just continuity. I have a feeling you are thinking of continuous, but not absolutely continuous, functions. I think the sin function you mentioned is continuous at any given point but isn't absolutely continuous Commented May 30, 2021 at 13:24