Proving That A Sequence Does Not Obey The Weak Law Of Large Numbers I have a sequence $f_n$ of random variables such that $\mu( \{ f_n = n \}) = \mu( \{f_n = -n \}) = \frac{1}{2}$, and need to prove that it doesn't satisfy WLLN.
Clearly, $E(f_n) = 0$, so I have to disprove that $$ \frac{1}{n} \sum_{j=1}^n f_j (\omega) \to 0$$ in measure. In other words, $$\lim_{n \to \infty} \mu \left( \omega: | \frac{1}{n} \sum_{j=1}^n f_j (\omega)| \ge \delta \right) \ne 0$$ $\forall \delta > 0.$
In my textbook, I haven't really learned much about how to do these types of questions. As a hint, I was told to show that if $\frac{1}{n} \sum_{j=1}^n f_j \to 0$ in measure $\mu$, then also $\frac{1}{n} \sum_{j=1}^{n-1} f_j \to 0$ in measure $\mu$.
However, I dont know how to show this and even if I were able to, I don't see why that means $f_n$ doesn't obey the Weak Law of Large Numbers.
I would gladly appreciate any help.

Take the difference, i.e.,
$$
\frac{1}{n}\sum_{i=1}^n f_i-\frac{1}{n}\sum_{i=1}^{n-1} f_i=\frac{f_n}{n}.
$$
The LHS converges to $0$ (in $\mu$-measure). Thus, $f_n/n$ must converge to $0$ as well. However,
$$
\mu(|f_n/n|>1/2)=1
$$
for each $n$.

The observation in the hint follows from the fact that if $\sum_{i=1}^n f_i/n\to 0$, then
$$
\frac{1}{n}\sum_{i=1}^{n-1}f_i=\frac{n-1}{n}\times \frac{1}{n-1}\sum_{i=1}^{n-1}f_i\to 0
$$
in $\mu$-measure.
