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Prove Fatou's lemma: for a sequence of function $f_n : [0,1] \to [0,\infty)$, $\int_{[0,1]} \limsup f_n \le \limsup \int_{[0,1]} f_n$.

From what I understand, this problem has a mistake in it; the inequality should be $\limsup \int f_n \le \int \limsup f_n$. But consider the following example. Let $f_n : [0,1] \to [0,\infty)$, $f_n(x) = n 1_{[0,\frac{1}{n}]}(x)$. Then $\int f_n = n \int 1_{[0,\frac{1}{n}]} = n \frac{1}{n} = 1$, so $\limsup \int f_n = 1$. It isn't hard to show that $\lim_{n \to \infty}f_n(x) = 0$ for $x \in (0,1]$, so $\int \limsup f_n = 0$. So $\int \limsup f_n = 0 < 1 = \limsup \int f_n$, which verifies the allegedly incorrect inequality and shows the other is false...What did I do wrong?

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  • $\begingroup$ If you look up Fatou's lemma (e.g. on wikipedia) you will find that it states that $$\int \liminf f_n \leq \liminf_n \int f_n$$ for any sequence of non-negative functions $f_n$. Under additional assumptions (which you can also find on wikipedia) it is possible to obtain an inequality for $\limsup$ (the "corrected" inequality from your question). Searching for "Fatou's lemma" will provide you with plenty of related questions, e.g. this one. $\endgroup$ – saz Jan 17 at 14:30
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Yes, the statement in the quote is incorrect, as you noticed.

However, yours is not a counterexample because the $\limsup$ version for the Fatou's lemma (see e.g. here) also requires that there exists an integrable function $g$ such that $f_n\leq g$ for all $n$ (just like in the dominated convergence theorem). This is clearly not the case for $f_n=n\chi_{[0,1/n]}$.

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