Uniform convergence and $L^p$-convergence Let $f_n,f \in L^p$ and $f_n \to f$ uniformly. Further let $g \in L^q$. Now, clearly $\int (f_n - f) g$ is defined. How do I show that 
$$\int |f_n - f||g|$$
can be made arbitrary small if I choose $n$ large enough? If we just naively apply Hölder we will not be able to make it small but just finite. Further, we cannot just replace $|f_n - f|$ by the bound "$\epsilon$" because that is larger than this. Does someone have a hint?
 A: Outside of a sufficiently large compact set, $\|g\|_q <\epsilon$. Use Hölder naively, and on that portion the integral is bounded above by $(\|f_n\|_p + \|f\|_p)\epsilon$. On the inside of the compact set $\|g\|_1 \leq C \|g\|_q$ where $C$ is given by the volume of the compact set. Then you can replace $|f_n - f|$ by $\epsilon$. 
Note that this requires $\|f_n\|_p$ be uniformly bounded, which would be true if you additionally assume $f_n \to f$ in $L^p$. If you don't assume that:
Consider over $\mathbb{R}$, take $f_n = \frac1n \chi_{[2^n,2^{n+1}]}$, where $\chi$ denotes the characteristic function. $f_n \to 0$ uniformly as a sequence of functions (i.e. $f_n\to f$ in $L^\infty$). For this sequence $\|f_n\|_p = \frac1n \cdot 2^{n/p} \nearrow \infty$ (unless $p = \infty$). (Note that since it is not a bounded sequence, you cannot assert that it has a weakly converging subsequence.) 
Now take $g$ be the function such that $g(x) = 2^{-n/(q-\delta)}$ if $2^n < |x| \leq 2^{n+1}$. By construction it is a function in $L^q$. 
But $|f_n - f| |g| = \frac1n \cdot 2^{-n / (q - \delta)} \chi_{[2^n,2^{n+1}]}$. If you integrate it, you get
$$ \frac1n \cdot 2^{n( 1 - \frac{1}{q - \delta}) }$$
So unless $q = 1$ and $p = \infty$, you can always choose $\delta$ sufficiently small such that the exponent in the integrated expression is positive, and provide a counterexample to what you want proved. (And of course, in the case where $p = \infty$, you can just replace $|f-f_n|$ by $\epsilon$ and be done with it.  
