# Can anyone explain the connection between reverse fatou's lemma and Fatou's lemma?

Here is the version of reverse Fatou's lemma I am looking at.

$E_n$ is a sequence of events. $P(\limsup E_n) \geq \limsup P(E_n)$

Here is Fatou's lemma.

Let $f_1,f_2,\ldots$ be a sequence of non-negative measurable functions on a measure space $(S, \Sigma, \mu)$. Define the function $f:S \to [0,1]$ a.e. pointwise limit by :$$f(s) =\liminf_{n\to\infty} f_n(s),\qquad s\in S.$$ Then $f$ is measurable and $$\int_S f\,d\mu \le \liminf_{n\to\infty} \int_S f_n\,d\mu\,.$$

What is the connection of the two theorems?

• I believe you can define $f_n(s) = 1\{s \in E_n^c\}$, where $1\{\}$ denotes an indicator function that is either 1 or 0 depending on the event inside the braces. – Michael Jun 20 '16 at 23:14
• Can you write out what is in your mind? I do not see the reverse relationship in here. – user1559897 Jun 20 '16 at 23:16
• I mean apply Fatou to those functions and see what happens. For example, can you compute $\int f_n(s) d\mu$ ? – Michael Jun 20 '16 at 23:16
• By the same reasoning you can prove $P[\liminf E_n] \leq \liminf P[E_n]$, which is a more direct application of Fatou as it does not need complements. To build intuition, you might want to prove that one first. – Michael Jun 20 '16 at 23:27
• I know the proof of inverse fatou's lemma already. And ∫fn(s)dμ is 1 - P($E_n$) if I get you correctly. – user1559897 Jun 20 '16 at 23:31

To each event $E_n$ there is an "indicator" random variable $$X_n = \begin{cases} 1 & \text{if the event E_n occurs,} \\ 0 & \text{if the event E_n does not occur.} \end{cases}$$ Let $Y_n=1-X_n$. Then $\liminf\limits_{n\to\infty} Y_n \vphantom{\dfrac 1\int}$ is a random variable that has some probability of being equal to $1$ and is otherwise $0$. And $\operatorname{E}(Y_n) = 1 - \Pr(E_n) = \Pr(\text{not } E_n)$. Since a random variable is a function whose domain is a probability space and an expected value is its integral over that space, the version of Fatou's lemma for measurable functions says $$\operatorname{E} \left(\liminf_{n\to\infty} Y_n\right) \le \liminf_{n\to\infty} \operatorname{E} (Y_n).$$ Now show that
• the event $\limsup\limits_{n\to\infty} E_n$ occurs if and only if the random variable $\liminf\limits_{n\to\infty} Y_n$ is equal to $0$; and
• $\operatorname{E}\left( \liminf\limits_{n\to\infty} Y_n \right) = \Pr\left(\liminf\limits_{n\to\infty} (\text{not }E_n) \right) = \Pr\left( \text{not } \left(\limsup\limits_{n\to\infty} E_n\right) \right)$. Those last named events have equal probabilities because they are the same event: either one of them occurse if and only if the other one occurs.