We list all the rational numbers in $\mathbb{R}\bigcap\lbrack0,3\rbrack$ here, $\{x_1,x_2,x_3,...\},$ with the order from small to large. Then we define the ball $B(x_i,r_i)$, $B(x_i,r_i)=\{x\arrowvert x_i-r_i\le x\le x_i+r_i\}$ We take $r_i=1/2^i(1\le i<\infty)$.
Now there exists one question here.
Do these balls $(\bigcup_{1\le i<\infty}B(x_i,r_i))$ cover the interval $[0,3]$?
If they can cover the interval,then $m(\bigcup_{1\le i<\infty}B(x_i,r_i))\le\sum_{1\le i<\infty} B(x_i,r_i)=2< m([0,3])$ it leads to a contradiction.
If they cannot cover the interval,then there must exist a positive integer $N$,such that $x_{N+1}-x_N>\frac{1}{2^N}+\frac{1}{2^{N+1}}>\frac{1}{2^N}$ I just cannot imagine there exist the two adjacent rational numbers that are so far apart (their distance is larger than one rational number $\frac{1}{2^N}$). That's impossible! We can insert another rational number $x_N+\frac{1}{2^N}$ between $x_N$ and $x_{N+1}$.This leads to a contradiction.
I just cannot find any mistakes from above words.Who can find an answer?