I have a bit trouble showing that $\lambda(D)=\infty$.
I have been giving the following problem:
Given $D=\{(x,y)\in \mathbb{R}^2 | 0<x=y<1\}$ and $\mathcal{N}=\{N \in \mathcal{B} | m_I(N)=0\}$ (where $m_I$ is the Lebesgue measure on $(\mathbb{R}, \mathcal{B})$ restricted on the interval $(0,1)$) - and the measure $\lambda:\mathcal{B}\otimes \mathcal{B} \to [0,\infty]$, where $$\lambda(G)= \begin{cases} 0 & \text{ if }\, \exists N_1,N_2 \in \mathcal{N}, B_1,B_2 \in \mathcal{B}, \text{ then } G \subseteq (B_1 \times N_1) \cup (N_2 \times B_2) \\ \infty & \text{ or else }\\ \end{cases} $$ Show that $\lambda(D)=\infty$.
I get that I might have to show it by contradiction, but I have only gotten so far:
Assume by contradiction, that $\lambda(D)\neq \infty \Longrightarrow \lambda(D)=0$, then $\exists N_1, N_2 \in \mathcal{N}$ and $B_1,B_2 \in \mathcal{B}$, where $D\subseteq (B_1 \times N_1) \cup (N_2 \times B_2)$.
For $D\subseteq (N_2\times B_2)$, I have that for $N_2\in \mathcal{N}$ it applies that $m_I(N_2)=0$.
Can somebody give me a hint, on how to get moving? I assume that for $D \subseteq (B_1 \times N_1)$, it follows analogously. Thank you.
Assume that $D \subseteq (B_1 \times N_1) \cup (N_2 \times B_2)$. We want to find a contradiction.
Note that you can write the elements of $D$ in this way $$D= \{ (t,t) : t \in (0,1) \}$$
Define $$I_1 = \{ t \in (0,1) : (t,t) \in (B_1 \times N_1) \}$$ and analogously $$I_2 = \{ t \in (0,1) : (t,t) \in (N_2 \times B_2) \}$$ By our assumption you have that $(0,1) \subseteq I_1 \cup I_2$.
On the other hand, if you look carefully at the definition of $I_1$ and $I_2$ you can see that $$I_1 \subseteq N_1 \qquad I_2 \subseteq N_2$$ To show this, let $t \in I_1$. Then $(t,t) \in B_1 \times N_1$. This means that $t \in B_1$ and $t \in N_1$. In particular this implies $t \in N_1$. By arbitrarity of $t \in I_1$, we have $I_1 \subseteq N_1$ (actually we can say more: $I_1 = B_1 \cap N_1$). The proof of $I_2 \subseteq N_2$ is similar.
Hence $$(0,1) \subseteq N_1 \cup N_2$$ This contradicts monotonicity of the Lebesgue measure, since $N_1 \cup N_2$ has measure zero.