Union of two nullsets must be a nullset too.

Question

Hello i have trouble seeing my mistake in this contradiction relating to Lebesgue measure.

Following problem: Suppose I have an elementary box $B \subset \mathbb{R}^{d}, B = [a_1,b_1] \times.... \times [a_d,b_d], a_i < b_i$. Therefore Lebesgue measure is positive. However i can split it into two boxes $B_1 := [a_1,b_1] \times ... \times [a_{d-1}, b_{d-1}] \times \{0\}$ and then i take $B_2 := \{0\} \times .... \times \{0\} \times [a_d,b_d]$.

Both are nullsets since they are degenerative but the union is our box before. Where is my mistake?

Answer 1

...but the union is our box before is not correct. In the two dimensional case, let

$B=[0,1]\times[0,1]$

$B_1=[0,1]\times\{0\}$

$B_2=\{0\}\times[0,1]$

Then $B$ is a square, of which $B_1,B_2$ are two sides, and $B\neq B_1\cup B_2$. Higher dimensions are similar.

Answer 2

$B_1 \cup B_2 \neq B$. For example $(a_1, \dots, a_b) \notin B_1$ and $(a_1, \dots, a_b) \notin B_2$ while $(a_1, \dots, a_b) \in B$.