I'm sorry if this is a rather elementary question, but I'm a bit confused on the computation of a particular iterated integral. Let $\mu_1$ be the counting measure on $\mathbb{R}$, and let $\mu_2$ simply be the Lebesgue measure on $\mathbb{R}$. If $E = \{(x,y) \in \mathbb{R}^2: 0 \leq x = y \leq 1\}$, let $f(x,y) = 1_E(x,y)$. I am then asked to compute the following iterated integral:
$$\int_\mathbb{R} d\mu_2(x) \int_\mathbb{R} f(x,y) d\mu_1(y)$$
As far as I was aware, this was just another way of writing $$\int_\mathbb{R} \left( \int_\mathbb{R} f(x,y) d\mu_1(y) \right) d\mu_2(x)$$
Observing that $f(x,y) = 1_{S(E,x)}(y)$ where $S(E,x) = \{y \in \mathbb{R}: (x,y) \in E\}$, I compute the inner integral as follows: $$\int_\mathbb{R} f(x,y) d\mu_1(y) = \int_\mathbb{R} 1_{S(E,x)}(y) d\mu_1(y) = \int_{S(E,x)} d\mu_1(y) = \mu_1(S(E,x)) \underbrace{=}_{\text{$S(E,x)$ is a singleton}} 1$$
Since the inner integral is $1$, then for the iterated integral we have $$\int_\mathbb{R} \left( \int_\mathbb{R} f(x,y) d\mu_1(y) \right) d\mu_2(x) = \int_\mathbb{R} d\mu_2(x) = \mu_2(\mathbb{R}) = \infty$$
However, I have been informed that this iterated integral was supposed to evaluate to just $1$. Could anyone shed light on this? What have I failed to account for, or what have I computed incorrectly? Any help would be appreciated!