Showing $\{X_n\}$ is uniformly integrable so I want to prove that $\{X_n\}$ is uniformly integrable, given that E[$|X_n|^2$] &lt; C for every n
where C is a constant.
so using the markov inequality we get P($|X_n|^2$ $\geq$ N) $\leq$ $\frac{(E[|X_n|^2])}{N}$  $\leq$  $\frac{C}{N}$ , $\lim \limits_{N \to \infty}$$\frac{C}{N}$ = 0 therefore we can conclude
that:$\lim _{N\rightarrow \infty}\int_N^\infty P(|X_n|^2\geq N)dx \rightarrow 0$
but then I get that $\{X_n^2\}$ is uniformly integrable
how can I continiue from here or did I missed something?
Note that
$$1\{|X_n| \geq N\} N |X_n| \leq 1\{|X_n| \geq N\} |X_n|^2,$$
hence
$$\int_{|X_n| \geq N} |X_n| dP\leq \frac{1}{N} \int_{|X_n| \geq N} |X_n|^2 dP.$$
Can you conclude?
Note that
$$1\{|X_n| \geq N\} N |X_n| \leq 1\{|X_n| \geq N\} |X_n|^2,$$
hence
$$\int_{|X_n| \geq N} |X_n| dP\leq \frac{1}{N} \int_{|X_n| \geq N} |X_n|^2 dP.$$
Can you conclude?