Suppose $1<p<\infty$ and q is conjugate exponent to p. Suppose $f_{n} \rightarrow f$ a.e where $sup_{n} ||f_{n}||_{p} < \infty$. If $g \in L^{q}$, then prove that $$Lim_{n\to \infty} \int f_{n}g =\int fg$$
Does this extend to the case when $f_{n} \rightarrow f$ a.e where, p=1 and $q=\infty$.
My attempt: I have the following $$|\int f_{n}g- \int{fg}\hspace{0.25cm} |=|\int (f_{n}g-fg)\hspace{0.25cm} | $$ $$=|\int (f_{n}-f)g\hspace{0.25cm} |\leq \int |(f_{n}-f)g| \leq ||f_{n}-f||_{p}||g||_{q} $$ By Holder's inequality, but now I suppose , that I can show that $||f_{n}-f||_{p} \to 0$, which will do the job. But can't show that.
I have also found the counter example.Consider $L^{1}[0,1]$. Let $f_{n}= n^{2}\chi_{[0,n^{-1}]}$, and $g=\chi_{[0,1]}$, then $g \in L^{\infty}([0,1])$ and $f_{n} \in L^{1}([0,1])$. Also $f_{n} \to 0$. This example I think perfectly works.
Thanks in advance for any help!!
You can proceed in this way (it's only a sketch of the proof).
Lemma 1: If $f_n\rightharpoonup h$ (meaning that $\int f_n g \to \int h g$ for every $g\in L^q$) and $f_n \to f$ a.e., then $h = f$ a.e.
Since $\|f_n\|_p \leq C$ for every $n$, then we can extract a subsequence $(f_{n_j})$ weakly convergent to some function $h \in L^p$.
From Lemma 1, we have $h=f$ a.e., hence by uniqueness of the limit the whole sequence $(f_n)$ weakly converges to $f$.