QUESTION
Let $(X,\alpha,\mu)$ be a measure space.
Let $(E_n)$ be a sequence in $X$.
Prove $\limsup\mu(E_n)\leq\mu(\limsup E_n)$ when $\mu(\bigcup E_n)<\infty$.
Show that this inequality may fail if $\mu(\bigcup E_n)=\infty$.
IDEAS
I was thinking of showing the reverse – $\mu(\limsup E_n)\geq\limsup\mu(E_n)$ because I know $\limsup E_n=\bigcap _{n=1}^\infty\bigcup_{m=n}^\infty E_n$. I know the measure of a union is equal to the union of the measures if the sequence is disjoint (which I am not guaranteed to have here). Besides I would have to deal with the intersection first.
I also have to prove a similar inequality for $\liminf$, but I figure I should be able to figure out the other if I can figure out one.
Please let me know what ideas you have. Thank you!
Best Answer
Let $A = \cup E_n$. By Fatou's lemma,
$$\int_A \liminf 1_{A\setminus E_n} \, d\mu \le \liminf \int_X 1_{A\setminus E_n}\, d\mu.$$
Thus
$$\int_A \liminf (1 - 1_{E_n})\, d\mu \le \liminf \int_A(1 - 1_{E_n})\, d\mu$$
or
$$\mu(A) - \int_A \limsup 1_{E_n}\, d\mu \le \mu(A) - \limsup \int_A 1_{E_n}\, d\mu.$$
Since $\limsup 1_{E_n} = 1_{\limsup E_n}$, $\int_A 1_{E_n}\, d\mu = \mu(E_n)$, and $\int_A 1_{\limsup E_n}\, d\mu = \mu(\limsup E_n)$, we have
$$\mu(A) - \mu(\limsup E_n) \le \mu(A) - \limsup \mu(E_n).$$
Since $\mu(A) < \infty$, we deduce
$$\limsup \mu(E_n) \le \mu(\limsup E_n).$$