What is the difference between sup P{|x(t)|>a}=0 and P{sup |x(t)|>a}=0

Question

Consider that $x(t)$ is a stochastic process, which is a solution of a stochastic differential equation issuing from $x_0$. I want to study the stochastic stability of the zero equilibrium. Two definitions are found in the literature

Weak stochastic stability

$\lim_{x_0 \to 0}\sup P\{|x(t)| \geq a \}=0$

and

Stochastic stability

$\lim_{x_0 \to 0} P\{\sup |x(t)| \geq a \}=0$.

I do not understand the difference between the two definitions. Why the first definition is weaker?

Answer 1

Note that

$$\begin{align} & & \sup_{t}|X(t)| &\geq X(t) \\ &\Rightarrow & \left\{\sup_{t}|X(t)| \geq a\right\} &\supseteq \left\{|X(t)| \geq a\right\} \\ &\Rightarrow & \Pr\left\{\sup_{t}|X(t)| \geq a\right\} &\geq \Pr\left\{|X(t)| \geq a\right\} \\ &\Rightarrow & \Pr\left\{\sup_{t}|X(t)| \geq a\right\} &\geq \sup_{t}\Pr\left\{|X(t)| \geq a\right\} \\ \end{align}$$ the last line follows from the fact that LHS is independent of $t$ since the first line. So by squeezing principle, if $$ \lim_{x_0 \to 0}\Pr\left\{\sup_{t}|X(t)| \geq a\right\} = 0$$ then it implies $$ \lim_{x_0 \to 0}\sup_{t}\Pr\left\{|X(t)| \geq a\right\} = 0 $$ but not the converse. So the first one is a weaker definition.