Unifrom Convergence in $L^1(\mathbb{R})$ Let $\mathcal{A}\subset L^1(\mathbb {R})$, relatively compact. And I want to prove that:
$$\lim_{N\rightarrow \infty}\sup_{f\in \mathcal{A}}||1_{[-N,N]^c}f||=0.$$
My idea:
If we define for each function $f$ in $\mathcal{A}$ a sequence $f_n=1_{[-n,n]}f$, we have that $||f-f_n||_1\rightarrow0.$
In other words, for $\epsilon>0$ there is $N$ such that: $||f-f_n||=||1_{[-n,n]}f||<\epsilon.$ for all $n\geq N$.
now, we notice that $$\mathcal{A}\subset \cup_{f\in \mathcal{A}}B(f_N,\epsilon)$$ and since $\mathcal{A}$ is relatively compact. then: there is $K\in \mathbb{N^*}$ such that: 
$$\mathcal{A}\subset \cup_{j=1}^KB(f^j_N,\epsilon)$$ 
But I could not deduce the uniform convergence from this.
 A: Fix $\varepsilon > 0$. We want to show that for all $N$ sufficiently large, $\sup_{f \in \mathcal{A}} \|f 1_{[-N,N]^c} \|_{L^1} \leq \varepsilon$.
Since $\mathcal{A}$ is relatively compact, we can find an $\frac{\varepsilon}{2}$-net for $\mathcal{A}$. That is, there are $f_1, \dots, f_k \in \mathcal{A}$ such that for every $f \in \mathcal{A}$ there is $j$ such that $\|f-f_j\|_{L^1} \leq \frac{\varepsilon}{2}$.
In particular, we have that 
\begin{align}
\|f1_{[-N,N]^c}\|_{L^1} \leq& \|f_j 1_{[-N,N]^c}\|_{L^1} + \|(f-f_j)1_{[-N,N]^c}\|_{L^1}
\\ \leq& \max_l \|f_l 1_{[-N,N]^c}\|_{L^1} + \|f-f_j\|_{L^1}
\\ \leq& \max_l \|f_l 1_{[-N,N]^c}\|_{L^1} + \frac{\varepsilon}{2}
\end{align}
Now, using the DCT, one can check that $\max_l \|f_l 1_{[-N,N]^c}\|_{L^1} \to 0$ and hence there is a $K \in \mathbb{N}$ such that $N \geq K$ implies that $\max_l \|f_l 1_{[-N,N]^c}\|_{L^1} \leq \frac{\varepsilon}{2}$.
This allows us to conclude that for $N \geq K$, we have that
$$\sup_\mathcal{A} \|f 1_{[-N,N]^c}\|_{L^1} \leq \varepsilon$$ as desired.
