Lebesgue-counting product measure of the diagonal

$\newcommand{\l}{\mathcal{L}} \newcommand{\n}{\mathcal{C}}$ I assume here that $\mathcal{I} = [0,1]$. Using the definition of the product measure: $$(\l \times \n)(D) = \inf \left\lbrace \sum_{n=1}^\infty \l (A_n) \n (B_n) \, : \, (A_n\times B_n)_{n=1}^\infty \text{ are rectangles covering }D \right\rbrace$$

Note that if $(A_n \times B_n)_{n=1}^\infty$ covers $D$, then $(A_n \cap B_n)_{n=1}^\infty$ covers $[0,1]$, so that $\l (A_n \cap B_n) >0$ for some $n\in \mathbb{Z}^+$. But this particular set cannot be finite (otherwise it would have Lebesgue measure zero), so we have $\l (A_n) > 0$ and $\n (B_n) = \infty$. Thus, $$\sum_{n=1}^\infty \l (A_n) \n (B_n) = \infty$$ for any cover of $D$ by rectangles, giving you $(\l \times \n )(D) = \infty$.

First note that Fubini's theorem holds even without $\sigma$-finiteness hypothesis: Let $(X, \mu), (Y, \nu)$ be measure spaces. I claim that if $f \in L^1(X \times Y, \mu \times \nu)$, then $$\int_{X \times Y}f(x, y)\,d(x, y) = \int_{X}\int_{Y}f(x, y)\,dy\,dx.$$ I'll give a proof here. First, by writing $f = f^+ - f^-$, we may assume $f \geq 0$. Since $f \in L^1(X \times Y, \mu \times \nu)$, it follows that $E = \{(x, y) \in X \times Y : f(x, y) \neq 0\}$ is a $\sigma$-finite set. Thus $E = \bigcup_{j}E_j$, with $E_j$ disjoint and $(\mu \times \nu)(E_j) <\infty$. Since $f = \sum_{j}f\chi_{E_j}$, by the monotone convergence theorem, it suffices to prove the theorem for $f\chi_{E_j}$. Hence it suffices to prove the theorem under the additional assumption that $E$ has finite measure. A simple consequence of the definition of $\mu \times \nu$ is that there are measurable $A_j \subset X, B_j \subset Y$ with $E \subset \bigcup_{j}A_j \times B_j$, $\{A_j \times B_j\}$ disjoint, and $\sum_{j}\mu(A_j)\nu(B_j) < (\mu \times \nu)(E) + 1 < \infty$. We have $f = \sum_{j}f\chi_{A_j \times B_j}$. Hence it suffices to prove the theorem for $f\chi_{A_j \times B_j}$. But this is easy since $\mu(A_j) < \infty$ and $\nu(B_j) < \infty$ and Tonelli's theorem applies: \begin{align} \int_{X \times Y}f(x, y)\chi_{A \times B}(x, y)\,d(x, y) &= \int_{A \times B}f(x, y)\chi_{A \times B}(x, y)\,d(x, y) \\ &= \int_{A}\int_{B}f(x, y)\chi_{A \times B}(x, y)\,dy\,dx \\ &= \int_{A}\int_{Y}f(x, y)\chi_{A \times B}(x, y)\,dy\,dx \\ &= \int_{X}\int_{Y}f(x, y)\chi_{A \times B}(x, y)\,dy\,dx. \end{align}

Now if you know that the first two are equal, then it follows that the third must be infinite since otherwise Fubini's theorem would imply they are equal.