# If $f_n\to f$ a.e. and are bounded in $L^p$ norm, then $\int f_n g\to \int fg$ for any $g\in L^q$

Suppose $p>1$ and $q$ is its conjugate exponent. Suppose $f_n \rightarrow f~a.e.$ and $\sup_n\|f_n\|_p < \infty$. prove that if $g \in L^q,$ then $\lim_{n \rightarrow \infty} \int f_ng=\int fg.$ Does this extend to the case where $p=1$ and $q=\infty$? If not, give a counter example.

### Progress

I know I need to prove $|\int (f_ng-fg)| < \epsilon$ and $|\int (f_ng-fg)| < |\int (f_n-f)g|$, if $f_n \rightarrow f$ in $L^p$. I can use Holder’s inequality to get the result. My question is how to get $f_n \rightarrow f$ in $L^p$ by the hypothesis $f_n \rightarrow f~a.e.$ and $\sup_n\|f_n\|_p < \infty$ or use other method to get the result. thanks

#### Solutions Collecting From Web of "If $f_n\to f$ a.e. and are bounded in $L^p$ norm, then $\int f_n g\to \int fg$ for any $g\in L^q$"

You have $\int |f|^p = \int \liminf_n |f_n|^p \le \liminf_n \int |f_n|^p < \infty$, so $f \in L^p$.

Counterexample for $p=1$:
Let $X=[0,1]$, $f_n=n1_{[0,{1 \over n}]}$, $\|f_n\|_1 = 1$, $f_n(x) \to 0$ ae.,
and let $g=1$. Then
$\int f_n g = 1$ for all $n$, but $\int 0 g = 0$.

For $p> 1$ it is true that $\int{f_ng}\rightarrow\int{fg}$. This follows from Theorem 3. Because the cited theorem has a different notation, I will write the body of it with your notation (in terms of $p$, $L_p$ ) and then apply it.

Theorem 3.

Let $M$ be a positive constant and $(X,\mu)$ be a complete measure space. Suppose that $0<p\leq \infty$ and that $f,\,f_1,\,f_2,….$ is a sequence of $\mu-$measurable functions, such that
$\|f_n\|_p\leq M,\,\,n=1,2,…$
and
$\lim\limits_{n\rightarrow\infty}{f_n(x)}=f(x)\quad$ $\mu-$a.e $x\in X$.
Then $\lim\limits_{n\rightarrow\infty}{\|f_ng-fg\|_s}=0$ for every $s\in(0,p)$ and every $g\in L_q(X)$, where $\frac{1}{q}+\frac{s}{p}=1$

Now you apply this theorem with $p>1$, $s=1\in(0,p)$ and $\frac{1}{q}+\frac{1}{p}=1$. You get that $\lim\limits_{n\rightarrow\infty}{\|f_ng-fg\|_1}=0\Leftrightarrow \int\limits_{X}{|f_ng-fg|d\mu}=0$. Consequently $|\int\limits_{X}{(f_ng-fg)d\mu}|\leq \int\limits_{X}{|f_ng-fg|d\mu}\rightarrow 0$