Weak Limit of $f_n = \sum_{i=1}^{2^n} (-1)^{i-1} 1_{\big[ \frac{i-1}{2^n}, \frac{i}{2^n} \big) }$ in $L^p$ Consider $\Omega=[0,1]$ equipped with Borel $\sigma$-algebra and Lebesgue measure. Consider the sequence $f_n = \sum_{i=1}^{2^n} (-1)^{i-1} 1_{\big[  \frac{i-1}{2^n}, \frac{i}{2^n} \big) }$ in $L^p$ where $1<p<+\infty.$ I believe that $f_n \rightarrow 0$ weakly in $L^p$ yet I am not able to show it. 
We know by Banach-Alaoglu theorem that the closed unit ball in $L^p$ is weakly compact. In other words every bounded sequence has a weakly convergent sequence and the sequence $(f_n)_n$ is clearly bounded. At least we know that 
$$ f_{n_k} \rightarrow g \text{ (weakly)}$$ 
for some $g \in L^p.$ Yet I could not continue further. Are there more practical ways of varifying weak convergence rather than using the definition? 
Any help is appreciated, thanks in advance!
 A: Indeed, it is usually not an easy task to show that $$\tag{*}\int f_n h\to \int fh$$ for all $h\in L^{p'}$ in general, especially when we don't know a priori the limiting function. However, using the fact that $(f_n)_{n\geqslant 1}$ is bounded in $L^p$ and Hölder's inequality, it suffices to show, by approximation by step functions, that $(*)$ holds when $h$ is the indicator function of a Borel set. Using regularity of the Lebesgue measure, we can see that $f_n\to f$ weakly in $L^p$ if and only if (*) holds for $h=\mathbf 1_{O}$ where $O$ is an open subset of $[0,1]$. Since such sets can be writing as a countable disjoint union of open intervals, we derive that $f_n\to f$ weakly in $L^p$ if and only if $\sup_{n\geqslant 1}\lVert f_n\rVert_p$ is finite and 
$\int_I f_n \to \int_I f$ for all interval $I$. 
Now, using density of $\{k2^{-N},N\geqslant 1, 0\leqslant k\leqslant 2^N\}$, it suffices to consider intervals $I$ whose end points are $k2^{-N}$ and $k'2^{-N}$ respectively. In this case, one can compute explicitely $\int_{(k2^{-N},k'2^{-N})}f$.
