$f \in \mathcal{L}^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R}),f' \in \mathcal{L}^1(\mathbb{R}) \Longrightarrow f \in \mathcal{C}_{0}(\mathbb{R})$ I want to show the following theorem:

Suppose $f \in \mathcal{L}^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R})$ and $f' \in \mathcal{L}^1(\mathbb{R})$. Then it holds that $f \in \mathcal{C}_{0}(\mathbb{R})$.

How can I show that $\lim_{|x| \to \infty} f(x)=0?$
 A: First it suffices to show that $f$ has a limit as $x\to +\infty$ and a limit as $x\to -\infty$, as theses limits will then necessarily be $0$, because one of these limits not being zero would violate the $L^1$ hypothesis made on $f$. Now, to show that $f$ has a limit as $x\to +\infty$, as $\mathbf{R}$ is complete (yes yes !), it suffices to show that $f$ satisfies the Cauchy criterion as $x\to +\infty$, that is, that : $$\forall \varepsilon >0, \exists M\in\mathbf{R}, y\geq x\geq M \Rightarrow |f(x)-f(y)| \leq \varepsilon.$$ Now, remember that, still by the Cauchy criterion, $g$ being $L^1$ (i.e. $|g|$ is integrable) is equivalent to say that $$\forall \varepsilon >0, \exists M\in\mathbf{R}, y\geq x\geq M \Rightarrow \int_x^y |g(u)|du \leq \varepsilon.$$ So let $\varepsilon >0$ be given. As $f'$ is $L^1$, we can therefore find an $M$ such that $y\geq x\geq M \Rightarrow \int_x^y |f'(u)|du \leq \varepsilon$, and the latter sum is bigger than $|\int_x^y f'(u)du|$ which is equal to $|f(y)-f(x)|$, showing thereby that $f$ satisfies the Cauchy criterion as $x\to +\infty$. An analogue argument would show that $f$ satisfies the Cauchy criterion as $x\to -\infty$. We have shown that $f$ tends to $0$ as $x$ tends to $+\infty$ and $-\infty$, that is, as $|x|$ tends to $+\infty$. To show that $f$ is in $C_0$, it remains to show that it is in $C^0$, but that's trivial, as $f$ is already in $C^1$.
