This question is not related to "convergence".
Let $X$ be a random variable.
I wonder which of the two following conditions is stronger.
$ \mathbb{E}\{X\} = 0 $ (the expected value of $X$ is $0$)
$Pr\{ X=0 \} = 1$ (i.e. $X=0$ "almost everywhere")
This question is not related to "convergence".
Let $X$ be a random variable.
I wonder which of the two following conditions is stronger.
$ \mathbb{E}\{X\} = 0 $ (the expected value of $X$ is $0$)
$Pr\{ X=0 \} = 1$ (i.e. $X=0$ "almost everywhere")
The second condition is much stronger because it defines the whole distribution, i.e. $\mathbb{P}\{X=0\}=1$ implies that the density function of $X$ is $f(x)=\cases{1&x=0 \cr 0&otherwise}$ . This tells you everything about $X$.
On the other hand, $\mathbb{E}(X)=0$ only tells you that the density $f(x)$ of $X$ satisfies $\int_{-\infty}^\infty xf(x)dx=0$ (in the continuous case) or $\sum_{-\infty}^\infty xf(x)=0$ (in the discrete case); this could be the case for many different probability densities $f(x)$. For example, any random variable whose density is symmetric around $0$ will have $\mathbb{E}(X)=0$.