The definition of A is a Null set is that $A\subseteq \mathbb{R}$,forall $\varepsilon>0$, exists open cover $\{I_n\}$,so that $\displaystyle \sum_{n=1}^\infty |I_n|\leq \varepsilon.$
Could I change open intervals set to close intervals set?
I think the essence of the question is how to define the length of interval?
For example, we define the length of close interval $[a,b]$ is $b-a.$If also define the length of open interval $(a,b)$ is $b-a$,obviously,two definition is equivalent.
Besides,how to define the length of$(a,b],[a,b)$?Is there have well-defined?
Suppose $|\cdot|$ is a nonnegative measure that acts on a sigma algebra of $\mathbb{R}$ that includes all intervals of $\mathbb{R}$, and that satisfies for all real numbers $a,b$ with $a<b$: $$ |[a,b]| = b-a$$ Fix $x \in \mathbb{R}$. Then $\{x\} = \cap_{n=1}^{\infty}[x, x+1/n]$ (so $\{x\}$ is in the sigma algebra) and for any positive integer $n$ we have $$ \{x\} \subseteq [x, x+1/n] \implies |\{x\}|\leq \underbrace{|[x,x+1/n]|}_{1/n}$$ Taking $n\rightarrow\infty$ in the above gives $|\{x\}|=0$, which holds for all $x \in \mathbb{R}$. This means for all real numbers $a,b$ with $a<b$: $$b-a=|[a,b]| = \underbrace{|\{a\}|}_{0} + |(a,b)| + \underbrace{|\{b\}|}_{0} = |(a,b)|$$ Similarly $$|[a,b)| = |\{a\}| + |(a,b)| = 0 + (b-a)=b-a$$ $$|(a,b]| = |(a,b)| + |\{b\}| = b-a + 0 = b-a$$
Therefore, if you have a set $A\subseteq\mathbb{R}$ such that for all $\epsilon>0$ there is a sequence of open intervals $\{I_n\}_{n=1}^{\infty}$ (of the type $I_n=(a_n,b_n)$ for $a_n<b_n$) such that $A \subseteq \cup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n|\leq \epsilon$, then you can define closed intervals $\hat{I}_n=[a_n,b_n]$ and, since $|I_n|=|\hat{I}_n|$ for all $n$, we immediately have $A \subseteq \cup_{n=1}^{\infty} \hat{I}_n$ and $\sum_{n=1}^{\infty} |\hat{I}_n|=\sum_{n=1}^{\infty} |I_n|\leq \epsilon$.
Conversely, Suppose you have a set $A\subseteq\mathbb{R}$ such that for all $\epsilon>0$ there is a sequence of closed intervals $\{I_n\}_{n=1}^{\infty}$ (of the type $I_n=[a_n,b_n]$ for $a_n<b_n$) such that $A \subseteq \cup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n|\leq \epsilon$. Then for any $\delta>0$ you can define open intervals $\tilde{I}_n=(\tilde{a}_n, \tilde{b}_n)$ such that $A\subseteq \cup_{n=1}^{\infty} \tilde{I}_n$ and $$\sum_{n=1}^{\infty} |\tilde{I}_n|\leq \delta$$ I leave to you as an exercise to construct such $\tilde{I}_n$. (Hint: Define $\epsilon = \delta/2$.)
This means: Assuming your $I_n$ sets in your definition of "null set" are open intervals, then you can indeed change $I_n$ to be closed intervals and the corresponding new definition is equivalent to the old.