counter example pointwise convergence implies weak convergence I'm looking for a counterexample to pointwise convergence implying weak convergence.
I.e., I'm looking for $f, f_1, f_2, ... \in L^1$ such that $\displaystyle{\lim_{n \to \infty} f_n=f \ }$  a.e., but there exists a bounded measurable $g$ such that $\displaystyle{\lim_{n\to \infty} \int f_n g \neq \int f g}$.
 A: It follows from the uniform boundedness principle that a weakly convergent sequence in a normed space is bounded, so any pointwise convergent and unbounded sequence would do.  However, you do not need this to come up with explicit counterexamples.  A relatively simple way to approach this is to consider the case $f\equiv 0$ and $g\equiv 1$.
It follows from the Lebesgue dominated convergence theorem that if the $f_n$'s are all dominated by a fixed $L^1$ function, then the sequence will converge weakly, so when looking for counterexamples, avoiding this would be a good place to start.  

Now that you have found an example (given in a comment on the question), I thought I would elaborate a bit.  If you restrict to the case where $f\equiv 0$ and $g\equiv 1$, you are just looking for an example of a sequence $f_1,f_2,\ldots$ in $L^1$ converging pointwise a.e. to $0$ such that $\lim_n\int f_n$ doesn't exist or is nonzero.  This can be done as in your example by ensuring that $\int f_n=1$ for all $n$ while the sequence of supports of the functions decreases to a null set.  You could also find examples where the supports are pairwise disjoint.
Basically, any "counterexample" to Lebesgue's dominated convergence theorem would work.  If you have any sequence of integrable functions $f_1,f_2,\ldots$ with $f_n\to f$ a.e., $f\in L^1$, but $\int f_n \not\to\int f$, then taking $g\equiv 1$ shows that you have a counterexample to your problem.  Considering the sequence $f_1-f,f_2-f,\ldots$ puts this in the "simpler" formulation mentioned above.
A: Let $f_n(x) = n^2x\cdot\chi_{[0,\frac{1}{n}]}-n^2x\cdot\chi_{[\frac{1}{n},\frac{2}{n}]} \in L^1(0,1)$.
Clearly, we have $f_n(x)\to 0$ for every x.
But $\int_{[0,1]}f_n dx = -1 \neq 0 $. 
