I really don't know how to solve the following problem.
Suppose $Y$ is a non-empty set and $\mu$ a measure defined by $\mu(\varnothing)=0$ and $\mu(A)=+\infty$ for all non-empty subset $A$ of $Y$. How can we determine $L^p(Y,\mu)$?
Any help is very much appreciated.
On
Integrals of arbitrary measurable functions are calculated in terms of simple functions, and this is obtained adding measures of sets. So any nonzero function will have an integral of at least the measure of a non-empty set. And any such set has infinite measure.
Conclusion: any nonzero function has an infinite integral and is not in $L^p$. So it has only the zero function
Summarizing the comments:
Suppose that $f$ is not the zero function, so there is some $y \in Y$ such that $f(y) \neq 0$. Since $f$ is constant on the set $\{y\}$, so is $|f|^p$, and so the integral of $|f|^p$ on that set is simply equal to the constant function value $|f(y)|^p$ times the measure of $\{y\}$: $$\int_{\{y\}} |f|^p\ d\mu = |f(y)|^p \mu(\{y\})$$ As $|f(y)|^p > 0$ and $\mu(\{y\}) = \infty$, we see that $|f(y)|^p \mu(\{y\}) = \infty$. (In the extended reals, a positive real number times $\infty$ is $\infty$.)
Now, since $Y$ is the disjoint union of $\{y\}$ and $Y \setminus \{y\}$, we have $$\int_Y |f|^p\ d\mu = \int_{\{y\}} |f|^p\ d\mu + \int_{Y \setminus \{y\}} |f|^p\ d\mu$$ We just showed that the first term on the RHS is $\infty$. Since the second term is nonnegative, this means that the LHS is also $\infty$. Conclusion: $$\int_Y |f|^p\ d\mu = \infty$$ and therefore $f \not\in L^p$. This is true for any function $f$ which is not the zero function. On the other hand, the integral of the zero function is zero (this is true for any measure), so $L^p$ contains the zero function and no other functions.