The continuous embedding of weighted $L^1$ space. Take $\omega_0$: $\mathbb R^N\to \mathbb R^+$ such that $\omega$ l.s.c. and  $\omega_0\geq 1$ and satisfies
$$
\frac{1}{|{B}|}\int_{B(x,r)} \omega_0(y)\,dy\leq C\omega_0(x) \tag 1
$$
for any ball $B\subset \mathbb R^+$.
We define weighted $L^1$ space with weight $\omega$ as
$$
\int_{\mathbb R^N}|u|\omega\,dx<\infty.
$$
Then, take $\omega_1$ which satisfies same condition as $\omega_0$, I want to show
$$
L_{\omega_0}^1(\mathbb R^N)\hookrightarrow L_{\omega_1}^1(\mathbb R^N), \tag 2
$$
i.e., $L^1_{\omega_0}(\mathbb R^N)$ is continuous embedded in $L^1_{\omega_1}(\mathbb R^N)$, if and only if
$$
\omega_1(x)\leq C\omega_0(x),\tag 3
$$
for some $C>0$.

My try: the if part is easy. Now for only if part. Assume $(3)$ does not hold, then I can have a sequence of points $x_n$ such that $\omega_1(x_n)>n\omega_0(x_n)\geq n$. Then I try to define $u_n:=\chi_{B(x_n,1)}$ and use condition $(1)$ and $(2)$ and hope to draw some contradiction. But I got stuck here...
Any help is really welcome!
 A: Suppose that $L_{\omega_0}^1(\mathbb{R}^N)\hookrightarrow L_{\omega_1}^1(\mathbb{R}^N)$.  This means there is a constant $A>0$ such that for any $u\in L_{\omega_0}^1(\mathbb{R}^N)$,
$$
\int |u|\omega_1\,dx \leq A\int |u|\omega_0\,dx.
$$
We will show that for any $x\in\mathbb{R}$, $\omega_1(x)\leq AC\omega_0(x)$.
Indeed, since $\omega_1$ is lower semi-continuous, 
\begin{align*}
\omega_1(x) &\leq \liminf_{r\to 0} \frac{1}{|B(x,r)|}\int_{B(x,r)} \omega_1(y)\,dy \\
&\leq \liminf_{r\to 0} \frac{A}{|B(x,r)|}\int_{B(x,r)} \omega_0(y)\,dy\\
&\leq AC\omega_0(x).
\end{align*}
In the second inequality, we used the embedding, and in the final step we used our first assumption on $\omega_0$.
EDIT:  Now I will show why, for lower semi-continuous $\omega_1$, we have
$$
\omega_1(x)\leq \liminf_{r\to 0} \frac{1}{|B_r|}\int_{B(x,r)} \omega_1(y)\,dy.
$$
Suppose on the contrary that there exists a $\varepsilon>0$ and a sequence $r_n\to 0$ such that 
$$
\frac{1}{|B_{r_n}|}\int_{B(x,r_n)} \omega_1(y)\,dy \leq \omega_1(x)-\varepsilon
$$
for all $n$.  Then there must be some $x_n\in B(x,r_n)$ with $\omega_1(x_n) \leq \omega_1(x)-\varepsilon$.  Notice that, since $r_n\to 0$ and $|x_n-x|\leq r_n$, it follows that $x_n\to x$.  But this is a contradiction to lower semicontinuity, since
$$
\omega_1(x) \leq \liminf_{n\to\infty} \omega_1(x_n) \leq \omega_1(x)-\varepsilon.
$$
