$L^1(\mathbb{R})$ functions are continuous in the metric 
I try to prove the $L^1(\mathbb{R})$ functions are continuous in the metric i.e. if $f\in L^1(\mathbb{R})$, given $\epsilon>0$ there exists $\delta >0$ s.t. 
  $$\int_{\mathbb{R}}\vert f(x+y)-f(x)\vert\leq \epsilon.$$

It is enough to show that 
$$
\lim_{y\to 0}\int_{\mathbb{R}}\vert f(x+y)-f(x)\vert=0.$$
Here is my proof:
Firstly, consider the indicator function $f(x)=\chi_K(x)$ which is supported on the interval $K:=[a,b]\subset \mathbb{R}$. By the Dominated convergence theorem, We have
[\begin{aligned} \lim_{h\to 0} \int \mid f(x+h)-f(x)\mid d\lambda(x)&=\lim_{h\to 0} \int \mid \chi_K(x+h)-\chi_K(x)\mid d\lambda(x)\\&=\int_{K_h\setminus K}d\lambda(x)+\int_{K\setminus K_h}d\lambda(x)\\&=\lambda(K_h\setminus K)+\lambda(K\setminus K_h)\to 0 \text{ as  } h\to 0\end{aligned}]
where $K_h=K-h=\{x-h: x \in K\}$.
So for simple function $f=\sum_{i}^{n} c_i\chi_{K_i}(x)$ also satisfies $\lim_{h\to 0} \int \mid f(x+h)-f(x)\mid d\lambda(x)=0$.

Is it right? Or any more easy way to prove that?

 A: The functions $\chi_{K}$ is not continuous, much less to say about its behavior of the pointwise convergence of $\chi_{K}(x+h)$ as $h\rightarrow 0$.
But thing is not that bad, at least we know that integrable functions can be approximated by the continuous one, and we focus on the continuous functions and then use density to complete the proof:
Let $\varphi$ be a continuous function with compact support that approximate $f$ in $L^{1}$ sense, so that $\|f-\varphi\|_{L^{1}}$ is small. And we note that 
\begin{align*}
\|f(\cdot+h)-\varphi(\cdot+h)\|_{L^{1}}=\int|f(x+h)-\varphi(x+h)|dx=\int|f(x)-\varphi(x)|dx=\|f-\varphi\|,
\end{align*}
which is also small.
Now Lebesgue Dominated Convergence Theorem gives
\begin{align*}
\int|\varphi(x+h)-\varphi(x)|dx\rightarrow 0.
\end{align*}
Finally we invoke the inequality that 
\begin{align*}
&\int|f(x+h)-f(x)|dx\\
&\leq\int|f(x+h)-\varphi(x+h)|dx+\int|\varphi(x+h)-\varphi(x)|dx+\int|\varphi(x)-f(x)|dx\\
&=2\|f-\varphi\|_{L^{1}}+\int|\varphi(x+h)-\varphi(x)|dx
\end{align*}
which is small, we are done.
