Let $f: (0,1] \times [0,1] \rightarrow \mathbb{R}$ , $ (x,y) \mapsto \begin{cases} \frac{1}{\sqrt{x^2+y^2}} \ \text{if} \ \ x^2 + y^2 < 1\\ 1 \ \text{otherwise}\\ \end{cases}$
$f$ is a continous and integrable function. I want to calculate its integral over $x$ if $y=0$. $f$ is also positive so Fubinis theorem should be valid. ( or is it?)
With $y=0$ you get $ g:(x,0) \mapsto \begin{cases} \frac{1}{x} \ \text{if} \ \ x^2 < 1\\ 1 \ \text{otherwise}\\ \end{cases}$ and $x=1$ is the only case where $f=1$. $\{1\}$ is a null set so it should not change the integral
If I want to calculate the integral if $y=0$ I assumed that I could calculate it with the one dimensional $\int_{0}^{1}{\frac{1}{x}dx}$. But it doesn't converge. This would be a cotradiction to Fubinis theorem.
Now I'm confused and have problems applying $y=0$ for Fubinis theorem. If I want to use Fubinis theorem for this, is
$\int_{(0,1] \times [0,1]}f(x,0) = \int_{[0,1)}\int_{[0,1]}g(x)dydx = \int_{(0,1]}g(x)dx$
this the right way to use it? $g(x)$ is independent from $y$ so $\int_{[0,1]}g(x)dy = g(x)$?
Or is its integral just zero because if $y=0$, $\int_{[0,1]}g(x)dy$ gets $0$ because the intervall gets zero or $dy$ with $y=0$ gets zero?