# $\lim_{n\rightarrow \infty}\int_0^1f_nhdm=\int_0^1fhdm$, prove $f\in L^p(m)$ , where $1\le p<\infty$.

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.

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You might be interested in the answer to this question. – commenter Nov 21 '12 at 5:39

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.