I am working on the question stated as below:
Let $(X_{k})$ be i.i.d random variable taking values in $\overline{\mathbb{R}}$ and let $M_{n}:=\max_{k=1}^{n}X_{k}$. Show that $M_{n}/n\longrightarrow 0$ a.s. if and only if $EX_{1}^{+}<\infty$ and $P(X_{1}>-\infty)>0$.
Similar questions are here:For i.i.d. $\{X_n\}$ proving $\frac1n\max\limits_{1\leqslant k\leqslant n}|X_k|\xrightarrow{\mathrm{a.s.}}0\Leftrightarrow E(|X_1|)<+∞$ and here: Show $n^{-1} \max_{k\leq n} X_k \rightarrow 0 \quad a.s.$ if and only if $\mathbb{E}(X_1)_+<\infty$ and $\mathbb{P}(X_1>-\infty)>0$, while the first link does not include the condition $P(X_{1}>-\infty)>0$, and the second post does not prove it completely.
I followed the second link and have got some progress:
$(\Leftarrow)$. Define $M_{n}^{+}:=\max_{m=1}^{n}X_{m}^{+}$, then I claim we have the following: $$E(X_{1}^{+})<\infty\iff X_{n}^{+}/n\longrightarrow 0\ \text{a.s.}\iff M_{n}^{+}/n\longrightarrow 0\ \text{a.s.}$$
Indeed, the first $\iff$ is due to the fact that $X_{n}/n\longrightarrow 0$ a.s. if and only if $E|X_{1}|<\infty$ and $E|X_{1}^{+}|=E(X_{1}^{+}).$ The second is due to an analytic fact that as $n\longrightarrow\infty$, we have $$a_{n}/n\longrightarrow 0\iff \dfrac{1}{n}\max_{m=1}^{n}|a_{m}|\longrightarrow 0.$$
Now, by hypothesis $P(X_{1}>-\infty)>0$, then if $P(X_{1}\geq 0)=k$, for any $0<k\leq 1$, then $P(M_{n}^{+}=M_{n})=1$, and thus $M_{n}/n\longrightarrow 0$ a.s.
However, I don't know how to show the case where $P(-\infty<X_{1}<0)=1$. The link said that I can show it directly, but I could not see how.
Also, how can I show $(\Rightarrow)$? Again it seems that $M_{n}/n\longrightarrow 0$ a.s. happens if and only if $EX_{1}^{+}<\infty$, but I don't see how $M_{n}/n\longrightarrow 0$ implies $P(X_{1}>-\infty)>0$.
Some detailed answers would be really appreciated :)
Edit 1:
I noticed something. Most of the time we consider $X_{k}$ takes regular real value, but this question assumes that it can also take $\infty$ and $-\infty$. I think that's why we need $P(X_{1}>-\infty)>0$, but this restriction is just saying that $X_{1}\neq-\infty$ a.s., how could I use this condition?