We know that the convergence almost surely/everywhere implies the convergence in probability/measure.
But is this also true when the measure space is not sigma finite since this can lead to different limits?
I'll explain myself with an example:
Consider $(\mathcal{R}, \mathcal{B(\mathcal{R})}, \lambda)$ the real line with the Lebesgue measure $\lambda$ and set with n $\in$ N:
$$ f_n(x) := \textbf{1}_{n,n+1} (x) $$
Convergence almost everywhere happens since, if we pick an x into the interval and push n to infinity, there's the pointwise convergence and so also the almost everywhere.
Convergence in measure in this case: $$ lim \lambda [f_n > \epsilon] = lim \lambda([n,n+1]) = lim 1 \neq 0 $$ Conclusion, there's no convergence in measure.
I suppose that this is happening since the lebesgue measure on R is not finite and so it's not possible to do this implication. Is that correct or am I missing anything?