Can a function be square integrable without being integrable?
Solution 1:
As long as the measure space is of infinite measure, this ca happen: Consider $\frac{1}{x}$ on $(1,\infty)$.
If the measure space is finite, this cant happen by Cauchy Schwarz (with the constant 1 function as the second factor).
Solution 2:
The same way the square of $\frac{1}{n}$ converges, but itself does not converge.
So for functions $\frac{1}{x}$ is an example over $[1,\infty)$.