Violation of Fatou's theorem?

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I have an assignment where the answer I have come up with violates Fatou's lemma. Clearly, I have either reached the wrong answer or misunderstood Fatou's lemma. Please help me find out which one it is!

We have a measure space (X, A, u) and a sequence of pair-wise disjoint subsets $(A_{n} )_{n \in \mathbb{N}}$. u($A_{n})=1$ for all $n \in \mathbb{N}$.

We also have a sequence of (indicator)functions $f_{n}=1_{A_{n}}$.

We are now asked to calculate $\underset{n\rightarrow \infty}{\text{limsup}}\:f_{n}$ and determine which of the values $\int \underset{n\rightarrow \infty}{\text{limsup}}\:f_{n}\: du$ and $\underset{n\rightarrow \infty}{\text{limsup}} \int f_{n} \: du$ are largest.

Fatou's lemma is that $\underset{n\rightarrow \infty}{\text{limsup}}\:f_{n}\leq \int \underset{n\rightarrow \infty}{\text{limsup}}\:f_{n}\: du$, as long as $f_{n}$ are all bounded.

The answer I've come up with is $\underset{n\rightarrow \infty}{\text{limsup}}\:f_{n}=1_{\emptyset}$, $\int \underset{n\rightarrow \infty}{\text{limsup}}\:f_{n}\: du=0$ and $\underset{n\rightarrow \infty}{\text{limsup}} \int f_{n} \: du=1$. Where am I going wrong?

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The hypothesis for this version of Fatou's lemma is that there exists a non-negative, integrable function $g$ such that $f_n \leq g$ for all $n$; i.e., $\int \sup_n f_n <+\infty$. This fails for your sequence, since $\sup_n f_n$ is the indicator function of a set of infinite measure.

Also, it looks like just a typo, but the conclusion should be $\limsup \int f_n \leq \int \limsup f_n$.