For a given $ε>0$ we have the assumption that
$$ Z(ε):=\begin{cases} Z, & \text{if }\Vert Z \Vert < \dfrac{1}{ε}\\ 0, & \text{otherwise} \end{cases} $$
We want to show that $$Z(ε) \xrightarrow{P} Z.$$
Idea of the solution:
we have to show that For $δ>0$
$$P \left\lbrace \vert Z(ε) -Z \vert > δ \right\rbrace = 0 \quad \text{as } ε \rightarrow 0.$$
Or equivalently,$$ P \left\lbrace \Vert Z \Vert > \frac{1}{\epsilon} \right\rbrace =0\quad \text{as } ε \rightarrow 0. $$
I could not get the full idea of the solution. I would appreciate any kind of help.
Best regards,
Mohammad
By dominated convergence theorem,$$ \lim_{ε \to 0^+} P\left( \|Z\| > \frac{1}{ε} \right) = \lim_{ε \to 0^+} E\left( I_{\{\|z\| > \frac{1}{ε}\}}(Z) \right) = E\left( \lim_{ε \to 0^+} I_{\{\|z\| > \frac{1}{ε}\}}(Z) \right) = 0. $$