TITLE: $\lim_{n\rightarrow \infty}\int_0^1f_nhdm=\int_0^1fhdm$, prove $f\in L^p(m)$ , where $1\le p<\infty$.
QUESTION [1 upvotes]: On $[0,1]$, suppose $\|f_n\|_p\le 1$,
$\lim_{n\rightarrow \infty}\int_0^1f_nh\, dm=\int_0^1fh\, dm$, for any  $h\in L^\infty(\mu)$,
I need to prove $f\in L^p(\mu)$ , where $1\le p<\infty$.
I only find out when setting $h(x)=\dfrac{\overline{f(x)}}{|f(x)|}$, it can get $f\in L^1(\mu)$, But I have no idea about other situation.

REPLY [1 votes]: Is anything known about the measure $\mu$? If it was finite, you could use weak-compactness of bounded sets in $L^p(\mu)$ together with the given limit condition to reach the conclusion.
{"set_name": "stack_exchange", "score": 1, "question_id": 241076}