Interval in $R$ does not have zero measure.

Consider $$Q$$ a interval in $$R$$. Number all the rationals in this interval. Then for rational $$q_i$$ consider interval $$[q_j-\epsilon /2^{j+2} ,q_j+\epsilon/2^{j+2}]$$. Now the whole interval lies inside the union of these intervals, sum of whose volumes is $$< \epsilon$$.
Why I am getting this contradiction? I know (with proof) that an interval does not have measure 0. What is wrong with the above argument?
I only know the concept of zero measure from analysis.

• You've given a proof by contradiction that the union of the balls contains the interval. Also maybe consider the set [1,2] and then consider the union of open balls around rationals $q\in[1,2]$ with radius $|q-\sqrt 2|$. This set won't have $\sqrt 2$ – Calvin Khor Feb 2 at 9:40

The gap in the proof is that you do not know that the interval $$[a,b]$$ lies in the union of these balls. In fact, if $$\epsilon < b-a,$$ you have shown that there are points in $$[a,b]$$ which are not in any of the balls.
You've given a proof by contradiction that the union of the balls contains the interval. As a simpler example of this sort of phenomenon, consider the set $$Q=[1,2]$$ and the union $$U$$ of open balls around rationals $$q∈[1,2]$$with radius $$|q-\sqrt 2|$$. $$U$$ doesn't have the point $$\sqrt 2 \in Q$$.
If the above does not convince you, one can directly prove that this set has other points: First consider the variant with open balls around each $$q_i$$, $$B_i := B_i(\epsilon) := (q_i - \epsilon2^{-i-2}, q_i +\epsilon2^{-j-2})$$ As each $$B_i$$ is open, the complement $$C_i=Q\setminus B_i$$ is closed. For $$\epsilon \ll 1$$, they are not empty. Moreover, $$D_i := Q \setminus \left (\bigcup_{j=1}^i B_j\right) = \bigcap_{j=1}^i C_j$$ is a sequence of nested, non-empty, closed, and bounded sets. By Cantor's Intersection Theorem, the set $$\bigcap_{j=1}^\infty C_i$$ is not empty, and thus $$\bigcup_{j=1}^\infty B_i$$ does not contain all of $$Q$$.
If you insist on using closed balls, you can use some $$\epsilon'<\epsilon$$; then notice that $$\overline{B_i(\epsilon')} \subset B_i(\epsilon)$$, so by the above, their union does not contain all of $$Q$$.
Further, the above works for every subset of $$Q$$ of measure more than $$\epsilon$$.