Given a probability density
$$f_Y= \frac{1}{(y+1)^2}*1_{[0,\infty)}$$ Where 1 is indicator function
Is the Random variable $Y$ integrable? In other word, is $E[Y]<\infty$.
I am confused by, if I perform the integral from 0 to infinity, the integral becomes infinite. However, the probability is 0 for $Y->\infty$. Therefore the $Y$ cannot attain infinity as value, and if density cannot attain infinity and $Y$ also not, the expectation should be bounded. How can I calculate $E[Y]$
You are totally right, $$E[Y] = \int_0^\infty \frac{y}{(y+1)^2} dy = +\infty$$ And yes, $Y$ is finite almost surely. But there is no reason why the expectation should be bounded!
It's true that if the expectation is bounded then the random variable is finite almost surely. But it's not true the other way around… Y above is an example for it.