Lebesgue outer measure intervals may all be assumed to be open On page 6-7 of Invitation to Ergodic Theory by C. E. Silva
Proposition 2.1.1 Lebesgue outer measure satisfies the following properties.
(1) The interval $I_j$ in the definition of outer measure may all be assumed to be open. 
With the proof for (1) being:
Let $\alpha(A)$ denote the outer measure of $A$ when computed using only open bounded intervals in the covering. Clearly, $\lambda^*(A)\leq\alpha(A).$ 
I don't understand why $\lambda^*(A)\leq\alpha(A)$ is true as a closed interval $I_k$ being $[a, b]$ contains $(a,b)$.
edit:
Following that snippet. The book says:
Now for $\epsilon>0$. For any covering ${I_j}$ of $A$ let $K_j$ be an open interval containing $I_j$ such that $\mid K_j\mid < \mid I_j\mid + \frac{\epsilon}{2^j}, j\geq1$. Then 
$$\sum_{j=1}^\infty\mid K_j\mid<\sum_{j=1}^\infty\mid I_j\mid+\epsilon$$
Taking the infimum of each side gives $\alpha(A)\leq\lambda^*(A)+\epsilon$, as this holds for all $\epsilon$, $\alpha(A)\leq\lambda^*(A)$.

The outer measure is defined as an infimum. If you take the infimum over less sets (only the open ones) then the infimum is possibly greater. 
