Question about the Riemann-Lebesgue Lemma proof

Question

Ok, so one of the formulations of the Riemann-Lebesgue Lemma says: $$ f\in L^1(\mathbb{R}) \implies \hat{f}(\omega)\to 0\;\mbox{ when } \;|\omega|\to\infty.$$

I get all the steps of the proof, except the one which says that if $f$ is an arbitrary integrable function, it may be approximated in the $L^1$ norm by a compactly supported smooth function $g\in L^1$. My question is: Why? How do you know that for all $\epsilon>0$, and for all arbitrary function $f\in L^1$ exists a compactly supported smooth function $g$ so that $||f-g||_{L^1}<\epsilon$? The "proof" can be found in many places, but this step can be found in Wikipedia https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma

Answer 1

Let $f \in L^1(\mathbb{R})$. Given an $\epsilon > 0$ you can find a simple compactly supported function $\phi$ such that $\|f - \phi\|_{L^1} < \epsilon/2$. Now you just have to find a smooth and compactly supported function near $\phi$. One way to do this is by doing a convolution against a mollifier.

There are other ways to prove this. Another proof can be found on Folland's book Real Analysis: Modern Techniques and Their Applications.

Answer 2

There are several ways to construct $L^1(\mathbb{R})$.

One way is to let $L^1(\mathbb{R})$ be the smallest Banach space containing $C_0^\infty(\mathbb{R})$ under the norm $\|\cdot\|=\|\cdot\|_{L^1}$.

For other definitions of $L^1$ this is also true (because it is the same space), however then this would need a proof based on the particular construction.

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To clarify myself, what I mean is that:

If we define $L^1(\mathbb{R})$ as the closure of the space of compactly supported smooth functions. Then, by definition, we can always approximate the $L^1$-functions by a compactly supported smooth function.

If we define $L^1(\mathbb{R})$ in some other way then we can still approximate the $L^1$-functions by a compactly supported smooth function - but then it needs a proof depending on the particular definition.