# Why the Komlós theorem is not valid for any sequence of measurable functions?

I read an article, and they use a certain theorem, called Komlós theorem, which says:

Theorem 1 (Komlós theorem)

Let $$(E,\mathcal {A}, \mu )$$ be a finite measure space and $$(f_n)_{n\geq 1} \subset \mathcal {L}_{\mathbb {R}}^1$$ is a sequence with : $$\sup_n \int_{E}{|f_n| d\mu} < \infty .$$ Then there exist $$h _{\infty} \in \mathcal {L}_{\mathbb {R}}^1$$ and a sub-sequence $$(g_k)_k$$ of $$(f_n)_n$$ such that for every sub-sequence $$(h_m)_m$$ of $$(g_k)_k$$ : $$\frac{1}{i}\sum_{j=1}^{i}{h_j}\to h _{\infty} \text{ a.s. }$$

The original Komlós theorem concerns $$\mathcal{L}^1_\mathbb{R}$$-bounded sequence of functions. The following theorem gives a similar result for nonnegative valued measurable functions.

Theorem 2

Let $$(f_n)_{n\geq 1}$$ be a sequence of nonnegative valued measurable functions.

Then there exist a sub-sequence $$(g_k)_k$$ of $$(f_n)_n$$ and a measurable function $$h _{\infty}$$ such that for every sub-sequence $$(h_m)_m$$ of $$(g_k)_k$$ : $$\frac{1}{i}\sum_{j=1}^{i}{h_j}\to h _{\infty} \text{ a.s. }$$

My problem: Why the theorem 2 is not valid for any sequence of measurable functions? I am looking for a counterexample and would appreciate any ideas.

• I made a small edit to the last sentence in order to slightly soften its tone. – Jochen Glueck May 7 at 21:15


Let $$(R_n)$$ be the sequence of independent Rademacher random variables (r.v.'s) defined on some probability space $$(\Om,\mathcal F,P)$$, so that $$P(R_n=\pm1)=1/2$$ for all $$n$$; such a probability space exists. Let $$(E,\mathcal A,\mu):=(\Om,\mathcal F,P)$$. Let $$f_n:=X_n:=n!R_n$$ for all natural $$n$$. Let $$(g_k):=(Y_k)$$ be any subsequence of the sequence $$(X_n)$$, so that $$Y_k=X_{n_k}$$ for some strictly increasing sequence $$(n_k)$$ of natural numbers and all natural $$k$$. Let $$Y_\infty$$ be any r.v. on the probability space $$(\Om,\mathcal F,P)$$ with values in $$[-\infty,\infty]$$. It suffices to show that $$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to Y_\infty\Big)\overset{\text{(?)}}=0. \tag{1}$$

Note that for some r.v. $$U_{1,K}$$ with values in $$[-1,1]$$ we have $$\sum_{k=1}^K Y_k=\sum_{k=1}^K (n_k)!R_{n_k} =(n_K)!R_{n_K}+U_{1,K}\sum_{j=1}^{n_K-1}j!.$$ Next, for natural $$n$$, $$\sum_{j=1}^{n-1}j!\le(n-2)(n-2)!+(n-1)!=o(n!).$$ So, $$\frac1K\,\sum_{k=1}^K Y_k\sim\frac{(n_K)!}K\,R_{n_K}.$$ Therefore and because $$n_K\ge K$$ and $$|R_n|=1$$, for each $$\om\in\Om$$, $$\frac1K\,\sum_{k=1}^K Y_k(\om)$$ may only converge to $$\infty$$ or $$-\infty$$; that is, $$\text{ on the event \Big\{\frac1K\,\sum_{k=1}^K Y_k\to Y_\infty\Big\} we must have Y_\infty\in\{\infty,-\infty\}.} \tag{2}$$

Moreover, $$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to\infty\Big)=P\Big(\bigcup_{K=1}^\infty A_K\Big),$$ where $$A_K:=\{R_{n_K}=1,R_{n_{K+1}}=1,\dots\}$$. Obviously, $$P(A_K)=0$$ for each natural $$K$$. Hence, $$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to\infty\Big)=0.$$ Similarly, $$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to-\infty\Big)=0.$$ Now, in view of (2), we see that (1) follows, as desired.

Adeed: Having now looked at Komlós's paper, I see that Theorem 2 there presents a stronger counterexample, as follows: For any sequence $$(a_n)$$ of positive real numbers such that $$a_n\to\infty$$ there exists a sequence $$(\eta_n)$$ of iid r.v.'s with $$E|\eta_1|=1$$ such that for the sequence $$(\xi_n)$$ with $$\xi_n:=a_n\eta_n$$ and for any of its subsequences the strong law of large numbers is not valid. Thus, the factor $$n!$$ in my example can be replaced by arbitrarily slowly growing $$a_n$$, with the Rademacher $$R_n$$'s replaced by iid r.v.'s $$\eta_n$$ with a (barely) finite expectation.

• Fixed a typo: Of course, $\sum_{j=1}^{n-1}j!$ is $o(n!)$, not $o(n)$. – Iosif Pinelis May 4 at 14:17
• @losif Theorem 2 is true for non-negative functions (see the proof of Lemma 5.1 on page 243 in the article "Komlós theorem for unbounded random sets G. KRUPA") – Karim KHAN May 7 at 16:15
• @StephanSturm : You are right, thank you for your comment. – Iosif Pinelis May 7 at 20:08

As a simple example let $$E$$ be a one point set and $$\mu$$ the one point measure. Let then $$f_n \equiv 2^n, ~ n \in \mathbb{N}$$. Then $$\frac{1}{i} \sum_{j=1}^i h_j$$ is dominated by the largest element in $$\{h_1,\ldots,h_i\}$$, in particular necessarily $$h_\infty = \infty$$. Thus the second assertion is not true with finite $$h_\infty$$ and for the first $$h_\infty \not\in L^1$$.

• you did not understand my question, I want a counterexample for which theorem 2 is not valid for a sequence of measurable functions, because the theorem 2 is valid for all sequences of measurable functions – Karim KHAN May 3 at 23:03
• Hmm? What is the meaning of your comment? It seems to me that you allow $h_\infty$ to be infinite. I really don't understand your problem. – Dieter Kadelka May 4 at 8:26
• @DieterKadelka : I too had trouble understanding the question. At this point, I am pretty sure the intended question was this: Is there an example of a sequence of measurable functions $f_n$ for which the main conclusion of the Komlós theorem, about the a.s. convergence of the averages, does not hold? By the Komlós theorem itself, that conclusion holds assuming the $L^1$ boundedness or the nonnegativity of the $f_n$'s. So, in such a counterexample, the $f_n$'s must take values of both signs and must be $L^1$-unbounded. – Iosif Pinelis May 4 at 14:12
• Previous comment continued: Also, in such a counterexample, the a.s. limit $h_\infty$ must be allowed to take infinite values -- because otherwise the "nonnegative" version of the Komlós theorem would obviously be false in general. – Iosif Pinelis May 4 at 14:14
• @DieterKadelka I want a counterexample for Theorem 2 – Karim KHAN May 5 at 16:47