$L^2$ function with compact support is in $L^1$?

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I'm reading Real analysis(Stein) and confused by a statement: If $f\in L^2$, then $f\chi_R$ is in both $L^1$ and $L^2$, where $\chi_R$ is the characteristic function of $[-R, R]$.

But $f$ in $L^2$ does not mean that $f\in L^1$. Such an example given by Almost everywhere continuous functions. How could this function on $[-R, R]$ be in $L^1$?

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Due to the Cauchy-Schwarz inequality, we have $$ \|f\chi_R\|_1 \leq \|f\|_2\|\chi_R\|_2 = (2R)^{1/2}\|f\|_2 < \infty. $$