Prove that if the Measure Space is Finite ($\mu(X)<\infty)$, then $\lim\sup \mu(A_n)\le \mu(\lim\sup A_n)$ Where $\lim\sup A_n=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}A_k$
I understand the solution, what I don't get is why do we have to demand the space to be finite? Yet we don't demand that in the similar equation for the infimum: $\mu(\lim\inf A_n)\le \lim\inf\mu(A_n)$ Where $\lim\inf A_n=\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}A_k$
For if $\mu(A_n)=\infty$ surely $\mu(\lim\sup A_n)=\infty$ right? so the equation will hold.
Thanks,