If $m^*(E)=0$, prove that $m^*(E_n)=0$ for each $n$

Question

Define $\displaystyle m^*(E)=\inf\left\{\sum_{n=1}^\infty\ell(I_n):E\subset\bigcup _{n=1}^\infty I_n\right\}$.

Let $E=\bigcup_{n=1}^\infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.

My attempt:

Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.

QED

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Is this correct?

Answer 1

$E_n \subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.