Weakening demands on Law of Large Numbers I'm reading about the Law of Large Numbers in the book Probability Theory: Independence, Interchangeability, Martingales, by Yuan Shih Chow and Henry Teicher:

Theorem 4 (Feller): If $\{X_n\}$ are i.i.d then $ (S_n-C_n)/n $ converges in probability to $0$ for some choice of $(C_n)$ iff $nP(|X_1|&gt;n) \rightarrow 0$. And then $\lim C_n/n=E(X_1)$.


  But what if $X_n$ has no expected value? 


Assume that $ nP(|X_1|&gt;n)\rightarrow 0$. Is $C_n/n$ still converging to some limit, and what is the limit?
My first though was "this must be wrong", so I constructed a simple, positive valued distribution
$$
X=\frac{1}{U\cdot\ln(1+1/U)}\text{ where }U\sim\text{Uniform}[0,1]
$$
which satisfies $n\cdot\Pr[X&gt;n]\rightarrow 0$ as $n\rightarrow\infty$, but has $\text{E}[X]=\infty$.

The following are observations, not proofs, but should be sufficient to understand what is going on.

If you sample values of $S_n/n=(X_1+\cdots+X_n)/n$, they tend to be clustered together around a value $C_n/n$, but with some extreme values in the upper tail.

As $n$ increases, the centre of the cluster $C_n/n$ will increase slowly without converging, while the distribution will become narrower (in the sense that for any $a&gt;0$ you have $\Pr[|S_n/n-C_n/n|&gt;a]\rightarrow 0$ as $n$ increases). In other words, $S_n/n-C_n/n$ will converge in probability to zero, but $C_n/n\rightarrow\infty$ as $n\rightarrow\infty$.

The criterion $n\cdot\Pr[X&gt;n]\rightarrow 0$ is important because it ensures (almost surely) that the jumps in the sequence $S_n/n$ will be smaller and smaller as $n$ increases. However, when the expected value is infinite (or undefined), these "jumps" caused by the extreme values of the distribution are sufficiently frequent to ensure $S_n/n$ will keep increasing rather than converge.
If $X_1$ does not have an expected value, then the law of large numbers does not apply and you will find  $nP(|X_1|&gt;n) \not \rightarrow 0$ in probability.  

For example, with a standard Cauchy random variable it converges to $2/\pi$.