While preparing for my exam, I stumbled upon the following two integrals:
$\int{\mu(D_{x}) m(dx) }$ and $\int{m(D_{y}) \mu(dy) }$
Where $D_{x,y}$ are the sections with respect to x and y for $D:= [(x,y) \in [0,1] \\ $x$[0,1] : x=y ]$ (So $D_x = [y \in [0,1] : (x,y) \in D] $ )
$\mu$ denotes the counting measure (thus infinite for [0,1]), $m$ denotes the Lebesgue measure.
The integral runs from 0 to 1.
My reader suggests that these integrals are not equal, but to me it seems that both integrals will have an 'infinite term' because of the counting measure and a 'finite term' because of the Lebesgue measure on the interval of finite length.
Can someone please explain why these integrals are not equal? Thank you very much in advance.
$D_{y}$ is by definition the set of $x$ such that $(x,y) \in D$. Note that $y$ is fixed. So, for each fixed single $y$, $D_{y}$ is by definition $\{y \}$ since only $(y,y) \in D$ by definition of $D$.
Now, $m$ is Lebesgue measure, so $m(D_{y}) = m( \{y\}) = 0$, right?
So $\int_{[0,1]} m(D_{y})\,d\mu = \int_{[0,1]} 0 \,d\mu = 0$.
Similarly, $D_{x}$ is by definition the set of $y$ such that $(x,y) \in D$. Note that $x$ is fixed. So, for each fixed $x$, $D_{x} = \{x \}$ since only $(x,x) \in D$ by definition of $D$.
Now, $\mu$ is counting measure, so $\mu( D_{x}) = \mu(\{x\}) = 1$, right?
So $\int_{[0,1]} \mu(D_{x})\,dm = \int_{[0,1]} 1 \,dm = 1$.
So, the iterated integrals are not equal. One equals $0$ while the other equals $1$.