Showing $\int_A f_jg\rightarrow 0$ Let $f_j\in L^P\cap L^1_{loc}$ and $g\in L^q$ where $p,q$ are conjugates and $||f_j||_p \lt M$ for all $j$ . I have that $\int_S f_j\rightarrow 0$ for all $S\subset{A}$ (measurable) where $A$ is a compact subset of $\mathbb{R}$ and want to show $\int_A f_jg\rightarrow 0$.
Since $A$ is bounded, $g\in L^q\implies g\in L^\infty$ so I want to write down something like $|\int _A f_j g|\le ||g||_{L^\infty} |\int f_j|$. But this is false. I can only use the tools that I know if I control $\int |f|$, but I've already come up with counterexamples to show we have no control over that so considering $\int |f_jg|$ is useless.  Another idea was to use Cauchy schwarz but we need not have $\int f_j^2 \rightarrow 0$
 A: The conclusion is not true. Let $p>1,$ $A=[0,1]$ and $f_j(x) =j\,(\log \,j)^{-1}$ for $0\le x\le j^{-1}$ and $f(x)=0$ otherwise. Then $\|f_j\|_1\to 0. $ On the other hand $$ \|f_j\|_p^p=j^{p-1} (\log\,j)^{-p}\to \infty.$$
As $f_j$ are not bounded in $L^p,$ in view of the Banach-Steinhaus theorem, there is $g\in L^q$ such that the sequence $$\int\limits_0^1 f_j(x)g(x)\,dx$$ is unbounded.
Remark The conclusion holds under additional assumption that  $\|f_j\|_p$ are uniformly bounded, say by $M.$ Then we can approximate $g\in L^q$ by linear combinations of indicator functions. Indeed, for $g\in L^q$ and   $N>0$ let
$$A_N=\left \{x\in A\,:\,|g(x)|\le N\right \}$$
Then $$\lim_N\int\limits_{A\setminus A_N}|g(x)|^q\,dx =0.$$
For a given $\varepsilon>0$ there is $N$ such that
$$\int\limits_{A\setminus A_N}|g(x)|^q\,dx<{1\over 2}\varepsilon^q$$
For $n\in \mathbb{N}$ and $1\le k\le n$ let
$$B_{k,n}=\left \{x\in A\,:\,{(k-1)N\over n}<|g(x)|\le {kN\over n}\right \}$$
Define
$$g_n(x)=\sum_{k=1}^n {kN\over n} 1{\hskip -2.5 pt}\hbox{I}_{B_{k,n}}$$
Next $$\|g_n-g\|_q^q=\int\limits_A|g_n(x)-g(x)|^q\,dx=\int\limits_{A_N}|g_n(x)-g(x)|^q\,dx+\int\limits_{A\setminus A_N}|g(x)|^q\,dx\\ =\sum_{k=1}^n\int\limits_{B_{k,n}}|g_n(x)-g(x)|^q\,dx+{1\over 2}\varepsilon^q
\le {N^q\over n^q}\sum_{k=1}^nm(B_{k,n}) +{1\over 2}\varepsilon^q\le {N^q\over n^q}m(A)+{1\over 2}\varepsilon^q$$
Fix $n$ such that $\displaystyle{N^q\over n^q}m(A)\le {1\over 2}\varepsilon^q.$ Then
$\|g_n-g\|_q\le \varepsilon.$
Next
$$\int\limits_Af_jg\,dx =\int\limits_Af_j[g-g_n]\,dx+\int\limits_Af_jg_N\,dx$$
Hence $$\left |\int\limits_Af_jg\,dx\right |\le \int\limits_A|f_j|\,|g-g_n|\,dx+\left |\int\limits_Af_jg_n\,dx\right | \\ \le 
\|f_j\|_p\|g-g_n\|_q+ \left |\int\limits_Af_jg_n\,dx\right | \le M\varepsilon +\left |\int\limits_Af_jg_n\,dx\right | $$
By assumptions we have $$\lim_j \int\limits_A f_jg_n\,dx =0$$ Thus the conclusion follows.
