Show the sequence $f_n(x)=\frac{1}{n}\chi_{[0,n]}$ has no weakly convergent subsequence in $L^1$. Show the sequence $f_n(x)=\frac{1}{n}\chi_{[0,n]}$ has no weakly convergent subsequence in $L^1[R]$.
My observations:
Assume $f_{n_k} \to f$ weakly, then:
The sets where $f$ is positive or where it is negative have to have infinite measure. I denote them $
A^+,A^-$. Also $\int f=1$
I am not sure how to proceed.
 A: Suppose that there is ea weakly convergent subsequence $f_{n_k}$ with limit $f\in L_1$. Then $\int_{(a,b]}f=\lim_k\langle f_{n_k},\mathbb{1}_{(a,b]}\rangle =\frac{1}{n_k}(b-a)=0$. This means that $f\equiv0$ almost surely by monotone class arguments, but
$$\int f=\langle f,\mathbb{1}\rangle=\lim_k\langle f_{n_k},\mathbb{1}\rangle =1$$
which is a contradiction.

Here is a monotone class argument based on Dynkin systems.
Let $\mathcal{L}$ be the collection of all measurable sets for which $\inf_Af=0$. $\mathcal{L}$ contains the mutilicative class (a $\pi$-system) $\mathcal{C}$ of all bounded intervals $(a,b]$. Since $\mathbb{R}=\bigcup_n(-n,n]$, by dominated convergence (with $|f|$ as dominating function. $\int_{\mathbb{R}}f=0$. It follows that $\mathbb{R}\in\mathcal{L}$. Furthermore, it is easy to show that $\mathcal{L}$ is a $d$-system. Thus it contains the $\sigma$-algebra generated by $\mathcal{C}$, which happens to be the whole collection of Borel sets.
In particular, $\int_{\{|f|>0\}}f=0$ and so, $f=0$ a.s.
