# Inequality Screwing with my Mind

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?

• 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. – saz Jan 17 '19 at 14:30

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 $$f_n=n\chi_{[0,1/n]}$$.