I've come to the following theorem and proof in Klenke's Probability Theory book, and I've hit a roadblock on the very last step of the proof. We know $\limsup_{n\to\infty}|k_n^{-1}S_{k_n}-\mathbf{E}[X_1]|=0$, but I can't figure out why $\limsup_{n\to\infty}k_n^{-1}S_{k_n}\leq \mathbf{E}[X_1].$ I'm sure it's something simple that I'm missing but my brain cannot figure it out for some reason, any help would be much appreciated.
For a non-negative sequence $a_n$, we have $\liminf a_n \geq 0$. Thus, showing that $\limsup a_n = 0$ is enough to conclude that $\lim_{n\to \infty} a_n =0 $. With this in mind, the statement:
$$ \limsup_{n\to \infty} \big|k_n^{-1}S_{k_n}-E[X_1]\big| = 0 \ \text{almost surely} \tag{1}$$
Lets us conclude that:
$$ \lim_{n \to \infty} \big|k_n^{-1}S_{k_n}-E[X_1]\big|= 0 \ \text{almost surely}$$
Which further implies that: $$ \lim_{n\to\infty} k_n^{-1}S_{k_n} = E[X_1] \ \text{almost surely} \tag{2}$$
When a limit exists, it is equal to the $\limsup$, so we have:
$$\limsup_{n\to\infty} k_n^{-1}S_{k_n} =\lim_{n\to\infty} k_n^{-1}S_{k_n} = E[X_1] \ \text{almost surely} $$
Thus, the author could have actually used an equality sign.