Working on a larger problem relating to harmonic analysis I have come upon this measure theory issue. Suppose $\varphi$ is a measurable function on $U$ and $f \in L^p(U)$. $\varphi$ is not necessarily bounded, but consider the set $$\{ f \in L^p(U) \ : \ \varphi f \in L^p(U) \}.$$
How do I show this space is dense in $L^p$?
If we consider that $f \cdot \chi_{\{ \left| \varphi \right| \leq N\} } \to f$ in $L^p$ can this help us?
Your Idea of using $f\cdot \chi_{\{|\varphi|\leq N\}} \to f$ in $L^p$ is a good starting point.
We denote the set in question with $$ A= \{f\in L^p(U) : \varphi f \in L^p(U) \}. $$
Then $g\cdot \chi_{\{|\varphi|\leq N\}}$ is in $A$ for all $g\in L^p(U)$. Indeed: $$ \int |\varphi g\chi_{\{|\varphi|\leq N\}}|^p \mathrm dx \leq \int | N g\chi_{\{|\varphi|\leq N\}}|^p \mathrm dx \leq N^p \| g\|_{L^p(U)}. $$
Now let $f\in L^p(U)$ be given arbitrarily. Then $f\cdot \chi_{\{|\varphi|\leq N\}} \to f$ in the $L^p(U)$-norm. (This is because the measure of $\{\varphi>N\}$ goes to $0$ as $N\to \infty$).
Hence we can approximate $f$ with elements in $A$, so $A$ is dense in $L^p(U)$.