There exists $(z_n)_{n\geq 1}$ monotone $z_n\to x_0$ such that $\sum |\phi(z_i)-\phi(z_{i-1})|=\infty$

Question

I'm reading "Measure and integral" Wheeden & Zygmund. One of its exercises on bounded variation, asked me to prove :

If $V[\phi; a,b]=\infty$. Show there exists a point $z\in[a,b]$ and a monotone sequence $(z_n)_{n\geq 1}$ such that $z_n\to z$ and such that $\sum_{i=1}^{\infty} |\phi(z_i)-\phi(z_{i-1})|=\infty$

The variation of a function $\phi$ on $[a,b]$ is defined as: $$V[\phi;a,b]=\sup_{\Gamma}S_\Gamma=\sup_{\Gamma}\sum_{i=1}^m|\phi(x_i)-\phi(x_{i-1})|,\ \mathrm{where}\ \Gamma=\{x_0,\cdots,x_m\}\ \mathrm{it's\ a \ partition \ of\ } [a,b] $$

When $V[\phi;a,b]<\infty$ it is say that $\phi$ is of bounded variation on $[a,b]$. In order to prove this result, I already have demostrated the following two previous exercises:

If $V[\phi;a,b]=\infty$ there is a point $z\in [a,b]$ such that either $V[\phi;I]=\infty$ for every subinterval $I$ of $[a,b]$ having $z$ as left-hand endpoint or $V[\phi;I]=\infty$ if the interval has $z$ as right-hand endpoint.

Let $\phi$ is finite on $[a,b]$ and of bounded variation on every interval $[a+\varepsilon,b]$, $\varepsilon>0$, with $V[\phi;a+\varepsilon,b]\leq M<\infty$. Then $V[\phi;a,b]<\infty.$

Attempt: The idea is to use the previous two results. Let $z$ the point found, such that $V[\phi;z,z+\varepsilon]=\infty$ for every $\varepsilon>0$. Let $M>0$ as large as you want. Since variation in $[z,z+\varepsilon]$ is infinite, i.e., $\sup_{\Gamma} S_\Gamma=\infty$ there should be $\Gamma_{\varepsilon}=\{\tilde{z}_0,\cdots,\tilde{z}_m\}$ partition of $[z,z+\varepsilon]$ such that $S_{\Gamma_\varepsilon}=\sum_{i=1}^m|\phi(\tilde{z_i})-\phi(\tilde{z}_{i-1})|>M$. Since we want $(z_n)$ to be monotone, I was thinking in $z_0=\tilde{z}_m,z_1=\tilde{z}_{m-1},\cdots,z_{m-1}=\tilde{z}_1$, now since we want $z_n\to z=z_0$, I thought $z_k=({z_{k-1}+z})/2$, for all $k\geq m$ which works. I did all that, because I wanted to take advantage of $S_{\Gamma_\varepsilon}>M$, because $$\sum_{i=1}^\infty |\phi(z_i)-\phi(z_{i-1})|\geq S_{\Gamma_\varepsilon}>M $$

At this point I have two serious questions:

1. Can I conclude that since $$\sum_{i=1}^\infty |\phi(z_i)-\phi(z_{i-1})|>M $$ where $M$ was arbitrarily chosen, that it is infinite? I asked this because If I change $M$ then the partition $\Gamma_\varepsilon$ chance as well, and therefore the sequence, in order to have the "$>M$".
2. The reason why $$\sum_{i=1}^\infty |\phi(z_i)-\phi(z_{i-1})|\geq S_{\Gamma_\varepsilon} \ \ (**) $$ is based on a question that I have asked here: Triangle inequality infinite terms $|a-b|\leq \sum_{i=1}^\infty|x_{i}-x_{i-1}|$ but the main difference is that, as $i\to \infty$, we have $z_i\to z$ but there is no reason for $\phi(z_i)\to \phi(z)$. If $\phi(z_i)\to \phi(z)$ were true, then ( ** ) is clearly true.

If someone can bring solutions is ok, but I'm also interested on understand if I made mistakes, so I can correct my thinking. This is an independent learning.

Answer 1

I will post the answer, not occured to me before, for if someone will need it in future. The above approach was not fine.

Suppose W.L.O.G that $V[\phi,I]=\infty$ for any subinterval having $z$ as left-hand endpoint (see results above for the existence). There is an $n_0$ such that $z+\frac{1}{n_0}<b$ (to have the subinterval inside $[a,b]$) so for all $n\geq n_0$ there is $\Gamma_{n}$ partition of $[z,z+\frac{1}{n}]$ such that $S_{\Gamma_n}>n$, it follows that $S_{\Gamma_n}\to\infty$ as $n\to\infty$. Consider $$\Gamma = \bigcup_{n\geq n_0}\Gamma_n $$ now let's define $b\geq z_0>z_1>\cdots>z_{n_0-1}>z+\frac{1}{n_0} $ and for $n\geq n_0$ define $$z_n=\max\left\lbrace \Gamma\ \ \ \backslash \ \ \bigcup_{0\leq k\leq n-1} \{z_k\} \right\rbrace$$ for definition, we have for example $z_{n_0}=z+\frac{1}{n_0}$ and $z_n$ monotone decreasing. Indeed, for each $\varepsilon>0$ there is, $M$ and $\tilde{M}$ with $0<M\leq \tilde{M}$ such that $\frac{1}{M}<\varepsilon$ and $z_{\tilde{M}}=z+\frac{1}{M}$ (this is true because for a fix $k$ we have $z_k\leq z+\frac{1}{k}$ most of the time '<' ) so if $n\geq \tilde{M}$ we have $$|z_n-z|=z_n-z\leq z+\frac{1}{M}-z<\varepsilon $$ which proves that $z_n\to z$. This is our candidate for sequence. It remains only to show that $\sum_i|\phi(z_i)-\phi(z_{i-1})|=\infty$. For this notice that, $$ \sum_{i=0}^\infty |\phi(z_i)-\phi(z_{i-1})|\geq S_{\Gamma_n}>n,\ \ \forall n $$ taking limits as $n\to\infty$ we get the desired result. My answer comes after this idea: Is $\max$ a valid order for $\cup_{n\in\mathbb{N}}\Gamma_n$, each $\Gamma_n$ is finite.

Answer 2

Sorry for reopening the old post. I just came up with a different construction that gets round of the union of an infinite family of partitions.

Again, let us assume that $z\in[a,b)$ and $V_z^{z+\delta}=+\infty$ for every $0<\delta<b-z$.

Let $M=1$. By the second result mentioned above, there exists $u_1\in(z,b)$ such that $V_{u_1}^b(f)>1$. We let $y_1=\min\{u_1,(z+b)/2\}$. Then we also have $V_{y_1}^b(f)>1$. Consequently, we can find a partition $y_1=v_0<\cdots<v_k=b$ such that $\sum{|f(v_i)-f(v_{i-1})|}>1$. We simply let $x_1:=b,\ldots,x_{k+1}:=y_1$.

Then we apply similar idea to $[z,y_1]$ to find suitable $y_2$, since $f$ also has infinite variation on it.

Repeating such procedure for $n$ times, one can see that

1. $0<y_n-z<(b-z)\cdot 2^{-n}\to 0$.

2. Each time only finitely many points are introduced and so far their contribution to the total sum is $>n$.

The desired sequence thus follows.