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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.

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  • $\begingroup$ I made a small edit to the last sentence in order to slightly soften its tone. $\endgroup$ – Jochen Glueck May 7 at 21:15
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$\newcommand\om{\omega}$ $\newcommand\Om{\Omega}$ We need to construct an example of a finite measure space $(E,\mathcal A,\mu)$ and a sequence $(f_n)$ of real-valued measurable functions on $E$ such that for any subsequence $(g_k)$ of the sequence $(f_n)$ and any measurable function $g_\infty$ with values in $[-\infty,\infty]$ we have $\mu(\{x\in E\colon \frac1K\,\sum_{k=1}^K g_k(x)\not\to g_\infty(x)\})>0$.

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.

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  • $\begingroup$ Fixed a typo: Of course, $\sum_{j=1}^{n-1}j!$ is $o(n!)$, not $o(n)$. $\endgroup$ – Iosif Pinelis May 4 at 14:17
  • $\begingroup$ @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") $\endgroup$ – Karim KHAN May 7 at 16:15
  • $\begingroup$ @StephanSturm : You are right, thank you for your comment. $\endgroup$ – Iosif Pinelis May 7 at 20:08
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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 $h_\infty \not\in L^1$.

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  • $\begingroup$ 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 $\endgroup$ – Karim KHAN May 3 at 23:03
  • $\begingroup$ 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. $\endgroup$ – Dieter Kadelka May 4 at 8:26
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    $\begingroup$ @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. $\endgroup$ – Iosif Pinelis May 4 at 14:12
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    $\begingroup$ 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. $\endgroup$ – Iosif Pinelis May 4 at 14:14
  • $\begingroup$ @DieterKadelka I want a counterexample for Theorem 2 $\endgroup$ – Karim KHAN May 5 at 16:47

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