Q:

Riemann-Lebesgue Lemma proof

$\textbf{Statement}: \text{ If } f \in L^1(\mathbb{R^n})$, then the Fourier transform of $f(x)$, that is $\hat{f}(w) \rightarrow 0$ as $|w| \rightarrow \infty$.
Proof: $|\hat{f}(w)|:= \int_{\mathbb{R^n}} |f(x) e^{-2\pi iwx}dx| 
                    \\=|\frac{1}{-2\pi i}|\int_{\mathbb{R^n}} |f'(x) e^{-2\pi iwx}dx|\\ \leq \frac{1}{2\pi |w|}||f'||_{L^1(\mathbb{R^n})} \rightarrow 0 $.
$\textbf{Question:}$ Does this implies that $f(x) \rightarrow 0 \text{ as }|x| \rightarrow \infty$
It is clear that $|\hat{f}(w)| \leq ||f(x)||_{L^1(\mathbb{R^n})}$ and $|w| \rightarrow \infty \implies f(x)=0$.

A:

Your argument only works when $f$ is differentiable and has compact support. But you have to prove it for all $f\in L^1$. This works because differentiable functions with compact support are dense. 
Also, I have no idea what you say is "clear", but the answer to your question is no. The Riemann-Lebesgue Lemma applies to any $f\in L^1$, and it states a property of the Fourier Transform. But there are lots of $f\in L^1$ such that $f(x)$ does not go to zero as $|x|\to\infty$. For instance take 
$$
f=\sum_n n\,1_{[n,n+\tfrac1{n^3}]}\in L^1(\mathbb R). 
$$

