Why is ${|f_n-f|^p}$ uniformly integrable and tight iff {$|f_n|^p$} is uniformly integrable and tight ($f_n \rightarrow f$ pointwise)? Why is ${|f_n-f|^p}$ uniformly integrable and tight iff {$|f_n|^p$} is uniformly integrable and tight ($f_n \rightarrow f$ pointwise)?
This is from the last sentence in the proof in the following theorem. 

 A: Using the elementary inequality 
$$\left|a+b\right|^p\leqslant 2^{p-1}\left(\left|a\right|^p+\left|b\right|^p\right),\quad a,b\in\mathbb R$$
we have
$$\left|f_n\right|^p \leqslant 2^{p-1}\left|f_n-f\right|^p+2^{p-1}\left|f\right|^p.$$
Now, we use the following facts:


*

*if $\left(g_n\right)_{n\geqslant 1}$ and $\left(h_n\right)_{n\geqslant 1}$ are uniformly integrable (respectively tight over $E$) sequences, then so is $\left(g_n+h_n\right)_{n\geqslant 1}$;

*if $\left(G_n\right)_{n\geqslant 1}$ and $\left(H_n\right)_{n\geqslant 1}$ are such that $0\leqslant G_n \leqslant H_n$ and $\left(H_n\right)_{n\geqslant 1}$ is uniformly integrable (respectively tight over $E$), then so is $\left(G_n\right)_{n\geqslant 1}$. 


This proves that if $ \left(\left|f_n-f\right|^p \right)_{n\geqslant 1}$ is uniformly integrable and tight over $E$, then so is $ \left(\left|f_n\right|^p \right)_{n\geqslant 1}$. To see the converse, we use this time 
$$\left|f_n-f\right|^p \leqslant 2^{p-1}\left|f_n\right|^p+2^{p-1}\left|f\right|^p$$
and the previously mentioned facts.
