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.
 A: $\newcommand\om{\omega}$ $\newcommand\Om{\Omega}$ 
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. 
A: 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 $h_\infty \not\in L^1$.
