Volum of the covering of $\bar{S} \geq S$?

Question

The proposition on GP Page 203 says:

Let $S$ be a rectangular solid and $S_1, S_2, \ldots$ a covering of its closure of $\bar{S}$ by other solids. Then $\sum$vol$(S_j) \geq$ vol($S$).

This does not quite make sense to be - why can we assert that Volum of the covering of $\bar{S} \geq S$? I think I must missed something. One thing I am not sure about is thatis defined to be the whole space?

Thank you very much for helping me clear my confusion.

Answer 1

Thanks to @Timkinsella. This is trivially true, since the closure of a set certainly is greater than or equal to the set.