# How to show that $\mu(\lim_{n\rightarrow\infty}\sup A_n)\geq \lim_{n\rightarrow\infty}\sup\mu(A_n)$?

Let $$\mu$$ be a measure function. I want to show that and $$\mu(\lim_{n\rightarrow\infty}\sup A_n)\geq \lim_{n\rightarrow\infty}\sup\mu(A_n)$$ provided that $$\mu(\cup^\infty_{n=1}A_n)<\infty$$.

I have already proved that $$\mu(\lim_{n\rightarrow\infty}\inf A_n)\leq \lim_{n\rightarrow\infty}\inf\mu(A_n)$$ from answers given here. Can I use this to prove that $$\mu(\lim_{n\rightarrow\infty}\sup A_n)\geq \lim_{n\rightarrow\infty}\sup\mu(A_n)$$?

Could you please give me a hint on how to start proving this.

• This is reverse Fatou's lemma specialized to characteristic functions. Reverse Fatou's lemma says that if $(f_n)$ is a sequence of nonnegative measurable functions and $|f_n| \leq g \in L^1(X, \mu)$, then $\int \limsup f_n \geq \limsup \int f_n$. It is easy to prove using Fatou's lemma. Oct 30, 2021 at 4:21
• @Mason, why is the condition $\mu(\cup^\infty_{n=1}A_n)<\infty$ provided in the question? Oct 30, 2021 at 4:42
• That condition gives you your $g \in L^1(X, \mu)$. Oct 30, 2021 at 4:44

Let $$B=\bigcup_n A_n$$. Then $$B \setminus \lim \sup A_n=\lim\inf (B\setminus A_n)$$. So $$\mu( B \setminus \lim \sup A_n)$$ $$\leq \lim \inf \mu( B \setminus A_n)$$ $$=\lim \inf (\mu (B)-\mu (A_n))=\mu (B)-\lim \sup \mu(A_n).$$
Can you finish? [$$\mu (B) <\infty$$ is needed at this last step]]
• So now we have $\mu( B \setminus \lim \sup A_n) \leq \mu (B)-\lim \sup \mu(A_n)$ then $\mu( B) -\mu(\lim \sup A_n) \leq \mu (B)-\lim \sup \mu(A_n)$ hence $\mu(\lim \sup A_n) \geq \lim \sup \mu(A_n)$ is this right? Oct 30, 2021 at 4:54
• but where did we use $\mu(\cup A_n)<\infty$. Is it so we can subtract $\mu(\cup A_n)$ from both sides? Oct 30, 2021 at 4:57
• @user3911153 Yes, $\infty -1=\infty -2$ but $1 \neq 2$, right? You cannot subtract infinity from both sides of an equation. Oct 30, 2021 at 4:59
• Just one last question. If it is not too much trouble for you could you please explain how we get $B \setminus \lim \sup A_n=\lim\inf (B\setminus A_n)$ and $\mu( B \setminus \lim \sup A_n) \leq \lim \inf \mu( B \setminus A_n)$. Oct 30, 2021 at 5:05