Question about Measure Theory "integral on $\mathbb{R}^n$" I have this : 
$u_n\rightarrow u~\text{in}~L^{\Phi}(\mathbb{R}^N)$ i.e., $\int_{\mathbb{R}^N}\Phi(|u_n-u|)dx\rightarrow0$
by measure theory, given $\varepsilon>0$ there exists $R>0$ such that $$\int_{B^c_R(0)}\Phi(|u_n|) dx\leq \varepsilon, ~\text{and}~ \int_{B^c_R(0)}\Phi(|u|) dx\leq \varepsilon$$
how to find this?
Edit: Where $\Phi: \mathbb{R}\rightarrow[0,+\infty)$ continuous,
convexe, even, increasing  and satisfy $\Phi(2t)\leq K \Phi(t), \forall t\geq 0$ 
Thank you 
 A: We start from the inequalities 
\begin{align}  \Phi\left(\left|u_n\right|\right) &\leqslant 
\Phi\left(\left|u_n-u\right| +\left|u\right|\right) \\
&=\Phi\left(2\frac{\left|u_n-u\right| +\left|u\right|}2 \right) \\
&\leqslant  K\Phi\left(\frac{\left|u_n-u\right| +\left|u\right|}2 \right)\\
&\leqslant \frac K2\left(\Phi\left(\left|u_n-u\right|  \right)+\Phi\left(\left|u\right|  \right)\right).
\end{align}
Now, fix $\varepsilon$ and consider $n_0$ such that $\int_{\mathbb R^n}  \Phi\left(\left|u_n-u\right|  \right)\mathrm dx\leqslant \varepsilon /K$ if $n\geqslant n_0$. Then for any positive $R$ and $n$ such that $n\geqslant n_0$, 
$$\int_{B_R(0)^c}\Phi\left(\left|u_n\right|\right)\mathrm  dx\leqslant  \frac K2\left(\int_{B_R(0)^c}\Phi\left(\left|u_n-u\right|  \right)\mathrm  dx+\int_{B_R(0)^c}\Phi\left(\left|u\right|  \right)\mathrm  dx\right)\\
\leqslant  \frac K2\left(\int_{\mathbb R^N}\Phi\left(\left|u_n-u\right|  \right)\mathrm  dx+\int_{B_R(0)^c}\Phi\left(\left|u\right|  \right)\mathrm  dx\right)\\
\leqslant \varepsilon/(2K)+\frac K2 \int_{B_R(0)^c}\Phi\left(\left|u\right|  \right)\mathrm  dx.$$
To conclude, use the fact that for any integrable function $g$, $$\tag{*} \lim_{R\to  +\infty} \int_{B_R(0)^c}\left|g(x)\right|\mathrm dx=0 $$
to get what we want. For the second inequality, we only need (*).
