Consider the interval $[0, 1]$ with the Borel-$\sigma$-algebra $\mathcal{B}([0, 1])$. Let $\lambda$ be the Lebesgue-measure and $\mu$ the counting measure on $\mathcal{B}([0, 1])$. For the characteristic function $1_D: [0, 1] \to \mathbb{R}$ of the diagonal $D = \{(x, y) \in [0, 1]^2: x = y\}$ (meaning that $1_D(x, y) = 1$ if $(x, y) \in D$, and $0$ otherwise), I want to calculate the two iterated integrals:
$$\int_{[0, 1]}\left( \int_{[0, 1]} 1_D (x, y) d \lambda(x)\right) d \mu(y) \text{ and } \int_{[0, 1]} \left( \int_{[0, 1]} 1_D (x, y) d \mu(y) \right) d \lambda(x)$$
Finally, I'm asked on why this result doesn't contradict Tonelli's theorem.
To answer the question at the end (I'm assuming here that the two integrals are different from each other): I think the problem is that the counting measure isn't a $\sigma$-finite measure, as required for applying Tonelli's theorem, because $[0, 1]$ is an uncountable set, hence, can't be written as a countable union of sets which all have finite (counting) measure.
I'm having trouble though evaluating these two iterated integrals. For the first integral: if we "fix" an $y \in [0, 1]$, then $1_D(y, y) = 1$, but $1_D(x, y) = 0$ for $x ≠ y$. Since the inner integral is $≠ 0$ at only one point, the Lebesgue integral is $0$, I believe? But how does the integration over the counting measure work, then? Since the result of the inner integral is always $0$, does that mean the outer integral is $= 1$ since $0$ is only one point? I'm having trouble understanding how the integration over the counting measure works, and also, how to evaluate these iterated integrals in general.
Thanks in advance.
In the expression on the left hand side $$\int_{[0, 1]}\left( \int_{[0, 1]} 1_D (x, y) d \lambda(x)\right) d \mu(y),$$ you are right about the inner integral being $0$, so we are left with $$\int_{[0,1]} 0\ d \mu(y).$$ This integral is $0$. Maybe it helps your understanding to compare it to the integral $$\int_{[0,1]} 1_{\{y=0\}}\ d \mu(y),$$ whose value is $1$. The integrand is $1$ at exactly one point, which the counting measure tallies up. Let me give you a few other examples of integrating with respect to the counting measure. $$\int_{[0,1]} 1_{\{1/y\ \in\mathbb{N}\}}\ d \mu(y) = 1+1+1+1+\cdots = \infty,$$ because the contributions from the integral come from the values of $y$ in the set $\{1,1/2,1/3,1/4,1/5,\ldots\}$. Another example is $$\int_{[0,1]} y^21_{\{1/y\ \in\mathbb{N}\}}\ d \mu(y)=1+\frac{1}{4}+\frac{1}{9}+\cdots = \frac{\pi}{6},$$ where I use the well-known sum $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi}{6}$.
In general, the counting measure looks for points where the integrand is non-zero, and then sums up the values of the integrand at these non-zero points.