While trying to answer this question I initially tried to argue via contradiction and was led to the following result:

Unproved Theorem: For each positive integer $n$ let $J_n$ be a union of finite number of non-overlapping (closed) sub-intervals of $[a, b] $. And further let the combined length of sub-intervals in $J_n$ be greater than or equal to $\delta>0$ for all $n$. Then there is at least one point $c\in[a, b] $ which lies in infinitely many $J_n$.

Two intervals are non-overlapping if their interiors are disjoint. Unfortunately I was unable to prove the theorem and I need some help here. Also approaches without the use of measure theory are desired as I was able to answer the linked question using measure theory.

The relation between the above theorem and the linked question is based on the fact that if a Riemann integral is positive then the function being integrated is positive on some union like $J_n$. More formally Apostol proves the following theorem in an exercise (see Exercise 7.35, page 180, Mathematical Analysis):

Theorem: Let $f:[a, b] \to\mathbb {R} $ be a non-negative Riemann integrable function such that $I=\int_{a} ^{b} f(x) \, dx>0$. Let $M$ be a positive bound for $f$ and $\delta=\dfrac{I} {2(M+b-a)}$. Then the set $A=\{x\mid f(x) \geq \delta\} $ contains a finite number of non-overlapping intervals whose combined length is at least $\delta$.

Here is one approach which sounds promising but there are certain doubts.

Let's assume on the contrary that no such point $c$ exists. Then for each $x\in[a, b] $ there is a positive integer $n_x$ such that $x\notin J_n$ for all $n\geq n_x$. Since $J_n$ is closed it follows that there is a neighborhood $I_x$ such that $I_x\cap J_n=\emptyset $ for all $n\geq n_x$. By Heine Borel we can choose a finite number of such neighborhoods say $I_{x_1},\dots,I_{x_p}$ which cover $[a, b] $. Let $N$ be the maximum of $n_{x_1},\dots,n_{x_p}$. Then $[a, b] \cap J_n=\emptyset $ for $n\geq N$ and this is a contradiction.

The problem with above proof is that the value $\delta$ is used nowhere. That part of hypotheses seems crucial.

The problem with the above proof is found and the wrong inference is striked out. I will let the above wrong proof be there so that readers can avoid it.

Update: On further investigation as well as looking at this answer of the linked question, it appears that the unproved theorem mentioned above is equivalent to the theorem of Arzelà:

Arzelà's Theorem: Let $\{f_n\} $ be a sequence of functions $f_n:[a, b] \to\mathbb {R} $ such that each $f_n$ is Riemann integrable on $[a, b] $ and let $|f_n(x)|\leq M$ for all positive integers $n$ and all $x\in[a, b] $. Further let $f_n$ converge point wise to a Riemann integrable function $f$ on $[a, b] $. Then $\int_{a} ^{b} f_n(x) \, dx\to\int_{a} ^{b} f(x) \, dx$.

By considering the functions $f_n-f$ the above theorem is reduced to the special case when $f(x) =0$ for all $x\in[a, b] $. Thus let $f_n(x) \to 0$ for all $x\in[a, b] $ and $|f_n(x)| \leq M$ for all $x\in[a, b] $ and all $n\in\mathbb {N} $. Also since $|\int_{a} ^{b} f_n(x) \, dx|\leq \int_{a} ^{b} |f_n(x) |\, dx$ the theorem is reduced to the case when each $f_n$ is non-negative. Thus Arzelà's theorem is reduced to the equivalent simpler formulation:

Arzelà's Theorem (Simplified): Let $\{f_n\} $ be a sequence of non-negative Riemann integrable functions $f_n:[a, b] \to\mathbb{R} $ which converges to $0$ point wise on $[a, b] $ and let $f_n(x) \leq M$ for all $x\in[a, b] $ and all $n\in\mathbb {N} $. Then $\int_{a} ^{b} f_n(x) \, dx$ converges to $0$.

Arguing by contradiction let us suppose that $\int_{a} ^{b} f_n(x) \, dx$ does not converge to $0$. Then there is an $\epsilon>0$ and a subsequence $f_{n_k} $ such that $\int_{a} ^{b} f_{n_k} (x) \, dx\geq \epsilon$. Without any loss of generality we can assume the subsequence $f_{n_k} $ to be the entire sequence $f_n$. Thus we have $\int_{a} ^{b} f_n(x) \, dx\geq \epsilon $. And by the theorem from Apostol's exercise this means that if $\delta=\dfrac{\epsilon} {2(M+b-a)}$ then the set $A_n=\{x\mid f_n(x) \geq\delta\} $ contains a union $J_n$ of non-overlapping subintervals of $[a, b] $ whose combined length is not less than $\delta$. By the theorem stated at the beginning of this post there is a point $c\in[a, b] $ which lies in infinitely many $J_n$ and hence $f_n(c) \geq \delta$ for infinitely many $n$. This contradicts the hypotheses that $f_n$ converges to $0$ point wise on $[a, b] $. This completes the proof of Arzelà's theorem.

Assuming the truth of Arzelà's theorem one can prove the theorem mentioned in the beginning of this post. Let's just take $f_n$ to be the indicator function of $J_n$. If every $x\in[a, b] $ lies only in a finite number of $J_n$ then $f_n$ converges point wise to $0$ on $[a, b] $. It can be checked that all hypotheses of Arzelà's theorem are satisfied and hence the integrals $\int_{a} ^{b} f_n(x) \, dx$ converge to $0$. But these integrals represent the length of $J_n$ which is at least $\delta$ and thus we reach a contradiction. In this manner we see that the theorem mentioned in beginning of the post is equivalent to the theorem of Arzelà.


Using a modicum of measure theory, let $F_n = \bigcup_{k=n}^\infty J_k$.

We have a decreasing sequence of sets where $F_{n+1} \subset F_n$ and, thus,

$$m \left( \bigcap_{n=1}^\infty F_n \right) = \lim_{n \to \infty} m(F_n) \neq 0$$

Thus, there is a point common to all of the sets $F_n$ and hence infinitely many sets $J_n$.

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  • $\begingroup$ Well I want something similar: a nested sequence of closed and bounded sets. The $F_n$ here may not be closed. $\endgroup$ – Paramanand Singh Dec 14 '18 at 7:47
  • $\begingroup$ @ParamanandSingh: ... meaning you want to apply nested cells or Cantor intersection theorem, correct? $\endgroup$ – RRL Dec 14 '18 at 7:58
  • $\begingroup$ Yes, but I don't think it is easily possible. $\endgroup$ – Paramanand Singh Dec 14 '18 at 8:16
  • $\begingroup$ I found that the theorem in my question is equivalent to the theorem of Arzela and any simple proof will lead to a simple of Arzela. See updated question. $\endgroup$ – Paramanand Singh Dec 14 '18 at 10:55
  • $\begingroup$ @ParamanandSingh: Are you now satisfied with this proof of your Unproved Theorem as a consequence of Arzela's theorem? Would you still like to find an elementary proof so that it stands alone as a lemma that can be used to prove Arzela's theorem? $\endgroup$ – RRL Dec 15 '18 at 2:02

I have finally found a proof of the theorem in my question. The essential idea is same as that given in the answer by user RRL but it does not use measure theory (instead uses something similar).

Thus we consider the sets $F_n=\bigcup\limits_{k=n} ^{\infty} J_k$ and show that the intersection of all $F_n$ is non-empty. What follows now is an exposition of a proof by Jonathan Lewin given in his paper A Truly Elementary Approach to The Bounded Convergence Theorem which appeared in The American Mathematical Monthly, Vol 93, No 5, May 1986, pp 395-397.

Note that each $J_n$ consists of a finite number of non-overlapping closed intervals. But the structure of each $F_n$ is bit different. To describe the structure of $F_n$ Lewin introduces what he calls elementary sets. An elementary set is a union of a finite number of bounded intervals (closed, open, whatever). The length of an elementary set $E$ denoted by $l(E) $ is defined to be the integral $\int_{a} ^{b} \chi_{E} (x) \, dx$ where $E\subseteq [a, b] $ and $\chi_E$ is the indicator function of $E$ (note that it is a step function). Thus $E$ is also a finite union of non-overlapping bounded intervals and its length is the sum of lengths of its constituent non-overlapping intervals.

Now we note that $F_n\supseteq F_{n+1}$ and each $F_n$ is bounded. And then Lewin considers a sequence $\{a_n\} $ defined by $$a_n=\sup\, \{l (E) \mid E\text{ is an elementary set and }E\subseteq F_n \} $$ By replacing $\delta$ with a smaller number we can assume that $a_n> \delta$ for all $n\in\mathbb {N} $. For each $n$ we can now choose an elementary set $E_n$ such that $E_n \subseteq F_n$, $E_n$ is closed and $l(E_n) > a_n-\dfrac{\delta} {2^n}$.

Next consider the sets $H_n=\bigcap\limits_{i=1}^{n}E_i$. Then each $H_n$ is closed and bounded and these form a nested sequence since $H_{n} \supseteq H_{n+1}$. If we show that each $H_n$ is non-empty then by Cantor intersection theorem there is at least one point which lies in all $H_n$ and hence in all $E_n$ and hence in all $F_n$.

The proof that $H_n$ is non-empty is given by analyzing the lengths of elementary sets contained in $F_n$ and the lengths of those contained in $F_n\setminus H_n$. First we examine the elementary sets contained in $F_n\setminus E_n$. If $E$ is an elementary subset of $F_n\setminus E_n$ then $$l(E) +l(E_n) =l(E\cup E_n) \leq a_n$$ and since $l(E_n) >a_n-\delta/2^n$ it follows that $l(E) <\delta/2^n$. Next if $E$ is an elementary subset of $F_n\setminus H_n$ then $$E=\bigcup\limits_{i=1}^{n}E\setminus E_i$$ Each set $E\setminus E_i$ is an elementary subset of $F_i\setminus E_i$ and hence its length is less than $\delta/2^i$ and therefore $l(E) <\delta$.

What we have proved so far is that if $E$ is any elementary subset of $F_n\setminus H_n$ then length of $E$ is less than $\delta$. On the other hand one can find an elementary subset of $F_n$ which is of length greater than $\delta$. It follows that each $H_n$ is non-empty and our job is done.

Let's now try to rephrase the above argument with a little less symbolism. The starting sequence of sets $J_n$ has a problem that it is not nested and hence it is difficult to find intersection of these sets. This obstacle is easily removed by constructing sets $F_n$ which form a nested sequence. But these $F_n$ are not necessarily closed. So we need to find subsets $E_n$ of $F_n$ which are closed and not too small in length compared to $F_n$. In that process we may lose the nested sequence property so we create sets $H_n$ via intersection of $E_n$'s. Finally we apply Cantor intersection theorem to the sets $H_n$. The fact that $H_n$ is non-empty is established by proving that $F_n$ contains larger elementary sets than those contained in $F_n\setminus H_n$.

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  • 1
    $\begingroup$ Very nice (+1) So we avoid "measure" by staying in the framework of an algebra rather than sigma-algebra of subsets. I think the analogy in probability is the events generated by a finite number of coin tosses. $\endgroup$ – RRL Dec 16 '18 at 2:32

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