Let's define Then by hypothesis $$ \Pr \left\{ \Omega \backslash F\right\} =1 $$

and $$ \Pr \left\{ F\right\} =0. $$

We have: \begin{eqnarray*} E[\left\vert X\right\vert ^{k}] &=&\int_{\Omega }\left\vert X\right\vert ^{k}dP \\ &=&\int_{\Omega \backslash F}\left\vert X\right\vert ^{k}dP+\int_{F}\left\vert X\right\vert ^{k}dP \\ &=&\int_{\Omega \backslash F}\left\vert X\right\vert ^{k}dP \\ &\leq &\int_{\Omega \backslash F}M^{k}dP \\ &=&M^{k}\Pr \left\{ \Omega \backslash F\right\} =M^{k}. \end{eqnarray*} 2) $\Longrightarrow $ 1)

Let's define \begin{eqnarray*} F &=&\left\{ \omega \in \Omega :\left\vert X\left( \omega \right) \right\vert >M\right\} \\ F_{n} &=&\left\{ \omega \in \Omega :\left\vert X\left( \omega \right) \right\vert >M+\frac{1}{n}\right\} \text{ \ }\forall n\geq 1. \end{eqnarray*} Then we have obviously $F_{n}\subseteq F_{n+1}$ (in fact, if $\omega \in F_{n}\Longrightarrow \left\vert X\left( \omega \right) \right\vert >M+\frac{1% }{n}>M+\frac{1}{n+1}\Longrightarrow \omega \in F_{n+1}$ ) and $% F=\bigcup_{n=1}^{\infty }F_{n}$ (in fact, let $\omega \in F$ ; let $% N=\left\lceil \frac{1}{\left\vert X\left( \omega \right) \right\vert -M}% \right\rceil $ ; then $\left\vert X\left( \omega \right) \right\vert >M+\frac{% 1}{N}$ , that is $\omega \in F_{N}$ ); this means that in the meaning of sets sequences convergence.

So the continuity from below property of probability can be applied: $$ \Pr \left\{ F\right\} =\Pr \left\{ \lim_{n\rightarrow \infty }F_{n}\right\} =\lim_{n\rightarrow \infty }\Pr \left\{ F_{n}\right\} . $$

Now, for any $k\geq 1$ , \begin{eqnarray*} M^{k} &\geq &E\left[ \left\vert X\right\vert ^{k}\right] \\ &=&\int_{\Omega }\left\vert X\left( \omega \right) \right\vert ^{k}dP \\ &=&\int_{\Omega \backslash F_{n}}\left\vert X\left( \omega \right) \right\vert ^{k}dP+\int_{F_{n}}\left\vert X\left( \omega \right) \right\vert ^{k}dP \\ &\geq &\int_{F_{n}}\left\vert X\left( \omega \right) \right\vert ^{k}dP \\ &\geq &\int_{F_{n}}\left( M+\frac{1}{n}\right) ^{k}dP \\ &=&\left( M+\frac{1}{n}\right) ^{k}\Pr \left\{ F_{n}\right\} . \end{eqnarray*} that is so that the only acceptable value for $\Pr \left\{ F_{n}\right\} $ is $$ \Pr \left\{ F_{n}\right\} =0 $$

whence the thesis.