In short, this was proved in Lieb and Loss 2.16 p65.
Let $j \in L^1(\mathbb{R^n}), \int_{\mathbb{R}^n} j =1$. Let $f \in L^P (\mathbb{R^n})$.
If $j$ does not have compact support, we can find (by DCT) $0<R<\infty$ and $C>1$ such that $j^R = C \chi_{|x|< R } (x) j(x)$, $\int_{\mathbb{R^n}}j^R=1$ and $||f||_p ||j-j^R||_1 < \delta $.
What I don't understand is how one finds the $j^R by DCT - in particular also the condition of it being scaled to 1.
We first let $k^{R}=\chi_{|x|<R}j$, and we see that $k^{R}(x)\rightarrow j(x)$ pointwise since for each $x$, there is a large $R_{0}>0$ such that $|x|<R_{0}$, then $k^{R}(x)=j(x)$ for all $R\geq R_{0}$ and hence the pointwise convergence. Now $|k^{R}|\leq|j|$ where $j\in L^{1}({\bf{R}}^{n})$, so Lebesgue Dominated Convergence Theorem implies that $\|j-k^{R}\|_{L^{1}({\bf{R}}^{n})}\rightarrow 0$ as $R\rightarrow\infty$. As a consequence, $\displaystyle\int_{|x|<R}j(x)dx\rightarrow\int_{{\bf{R}}^{n}}j(x)dx=1$, so for a large $R_{0}$, $\displaystyle\int_{|x|<R_{0}}j(x)dx>\dfrac{1}{2}$ for all $R\geq R_{0}$.
Let $C_{R}=\left(\displaystyle\int_{|x|<R}j(x)dx\right)^{-1}$, then $\displaystyle\int_{{\bf{R}}^{n}}j^{R}(x)dx=C_{R}\int_{|x|<R}j(x)dx=1$.
Since $j^{R}(x)\rightarrow j(x)$ pointwise and $|j^{R}(x)|\leq 2|j(x)|$ for all $R\geq R_{0}$, and $j\in L^{1}({\bf{R}}^{n})$, then Lebesgue Dominated Convergence Theorem implies that $\|j-j^{R}\|_{L^{1}({\bf{R}}^{n})}\rightarrow 0$ as $R\rightarrow\infty$.