Let $\lambda$ be the Lebesgue measure. Suppose $\{f_n\}$ is a sequence of functions on $[0,1]$ such that $\lambda(E)<\frac{8}{9} \implies \int_E f_n d \lambda<1$. Prove if $f_n \rightarrow f$, we have $\lambda(E)<\frac{8}{9} \implies \int_E f d \lambda \leq 1$.
Attempt: This problem seems too easy, and I do not know if I am missing something here.
Since $\{\int_E f_n d\lambda\}$ is a sequence of functions bounded above by $1$, we have $\liminf \int_E f_n d\lambda \leq 1$ for any measurable $E$ such that $\lambda(E)<\frac{8}{9}$. Then by Fatou's Lemma, for any measurable $E$ such that $\lambda(E)<\frac{8}{9}$, we have $$\int_E \liminf f_n d\lambda=\int_E f d\lambda \leq \liminf \int_E f_n d\lambda \leq 1$$
I would appreciate if someone just lets me know if this is correct or not. If it is wrong, I will try another approach
Best Answer
The approach taken won't work without the assumption that the functions $f_n$ satisfy the hypothesis of Fatou's lemma, namely that there is an integrable function $g$ such that $f_n\ge -g$ almost everywhere. Take the sequence $f_n = \chi_{[0,\frac{n-1}{n}]} - n^2\chi_{(\frac{n-1}{n},1]}$. It's clear that $f_n\to 1$ pointwise as $n\to\infty$. For any $E$ of measure less than $\frac89$ and any $n$, $\int_Ef_n<1$, and in fact $\int_{(\frac12,1]}f_n\approx -n$ while $\int_{(\frac12,1]}f = \frac12$.
For this example, the conclusion of the proposition still holds, just not the method of proof.