What the (strong) law of large numbers says: $$ \require{cancel} \begin{align} \text{right: } & \Pr\left( \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n X_i = 1\right) = 1. \\ \\ \text{wrong: } & \xcancel{\lim_{n\to\infty} \Pr\left( \frac 1 n \sum_{i=1}^n X_i = 1\right) = 1.} \end{align} $$


If e.g. $X_1$ has continuous distribution then so has $\sum_{i=1}^nX_i$ for every $n$.

Consequence: $$\mathsf P(\frac1n\sum_{i=1}^nX_i=1)=0\text{ for every }n$$ so that also: $$\lim_{n\to\infty}\mathsf P(\frac1n\sum_{i=1}^nX_i=1)=0\neq1$$