Strong Law of Large Numbers for i.i.d. with undefined (not infinite) mean Consider a i.i.d. $\{X_n\}$ whose mean $E[X_1]$ is undefined (not infinite). Then, is it possible for this i.i.d. to have a form of Strong Law of Large Numbers hold on it - that is, can there exist a constant $c$ (possibly infinite) such that
\begin{equation}
P\left(\lim_{n\to\infty} \frac{S_n}{n} \to c\right) = 1?
\end{equation}
Searching online just brings out results on the cases where $E[X_1]$ is defined (possibly infinite), and I haven't been able to figure out this on my own.
 A: If $S_n/n \to c\: $ a.s. where $c$ is finite, then $S_{n+1}/n \to c \:$ a.s., so
$X_{n+1}/n \to 0$ a.s.
This implies that $|X_{n+1}|/n \ge 1$ finitely often a.s. By the second Borel Cantelli Lemma,
$$\infty> \sum_{n \ge 0} P(|X_{n+1}| \ge n)=\sum_{n \ge 0} P(|X_{1}|\ge n) =E\Bigl(\sum_{n \ge 0} {\bf 1}_{|X_{1}|\ge n}\Bigr) \ge E|X_1|\,.$$

On the other hand,  if the almost sure  limit $c=\lim S_n/n$ exists, but is infinite, then  it is possible that $E(X_1)$ is undefined, i.e., we can have  $E(|X_1 \wedge 0|)=E(X_1 \vee 0)= \infty$.
Example:: Suppose that for all $x \ge 0$, we have $$P(X_1 \ge x)=\frac1{2\ln(e+x)}$$ and
$$P(X_1 \le -x)=\frac1{2(1+x)} \,.$$
Then $E(|X_1 \wedge 0|)=E(X_1 \vee 0)= \infty$.
Now suppose $X_k$ are i.i.d. with this law.  Then $P(X_k \le -k^2) \le 1/k^2$ is summable, so By Borel-Cantelli, a.s.  all but finitely many of the $X_k$ satisfy  $X_k >-k^2$. On the other hand,
$$P\bigl(\, \forall 1 \le k \le n, \quad X_k \le e^{\sqrt{n}}-e \bigr)=\Bigl(1-\frac1{2\sqrt{n}}\Bigr)^n \le e^{-\sqrt{n}/2} \,,
$$
which is summable. Thus by Borel Cantelli,
almost surely, for all $n$ large enough we have
$$S_n=X_1+\ldots +X_n \ge e^{\sqrt n}-e-n^3-C \,,$$
where $C$ is some random constant coming from the finitely many $X_k$ that satisfy $X_k \le -k^2$.
