I think it is time to give an answer. Warning: I give a full answer, so you might want to spot reading halfway if you just want a good hint.
We can write:
$\{L > 1\} = \bigcup_{x > 1} \{L > x\},$
but since we have an uncountable number of event on the right, we won't be able to conclude that the left side has measure $0$ even if each of the $\{L > x\}$ has measure $0$. So we use the usual trick (discretization):
$\{L > 1\} = \bigcup_{k \in \mathbb{N}^*} \{L > 1+1/k\},$
and all we have to show is that, for all $k \in \mathbb{N}^*$, we have $\mathbb{P} (L > 1+1/k) = 0$. Let us choose any positive $k$. Then:
$\{L > 1+1/k\} = \{ \exists \varepsilon > 0: X_n \geq (1+1/k+\varepsilon) \ln (n) \ \text{i.o.} \} \subset \{ X_n \geq (1+1/k) \ln (n) \ \text{i.o.} \}.$
If you have the good answer for the first question, then you know that $\mathbb{P} (X_n \geq (1+1/k) \ln (n) \ \text{i.o.}) = 0$ because $1+1/k>1$. Hence, $\mathbb{P} (L > 1+1/k) = 0$, and this holds for all $k \geq 1$. By the argument I gave in the beginning, $\mathbb{P} (L > 1) = 0$.
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Just a quick remark: I did not do exactly what I said in my comment up there. Indeed, if we choose to use the events $\{L \geq 1+1/k\}$, we don't benefit from the inclusion:
$\{L \geq 1+1/k\} \subset \{ X_n \geq (1+1/k) \ln (n) \ \text{i.o.} \},$
which is false. Instead, we must use something like:
$\{L \geq 1+1/k\} \subset \{ X_n \geq (1+1/(2k) \ln (n) \ \text{i.o.} \},$
which is messy. So, for the sake of elegance, the strict inequality in $\{L > 1+1/k\}$ is quite useful.