Pointwise convergence of $f_n(x) = \sum_{i=1}^n \left( 1_{ [ \frac{2i-2}{2n},\frac{2i-1}{2n})}(x)-1_{ [ \frac{2i-1}{2n},\frac{2i}{2n}]}(x) \right)$ Consider the following sequence of functions:
\begin{align}
f_n(x) = \sum_{i=1}^n  \left( 1_{ [ \frac{2i-2}{2n},\frac{2i-1}{2n})}(x)-1_{ [ \frac{2i-1}{2n},\frac{2i}{2n}]}(x) \right).
\end{align}
We are interested in understanding how does this function convergence pointwise as $n \to \infty$ for all $x\in [0,1]$.
That is, what is
\begin{align}
f(x)=\lim_{n \to \infty} f_n(x)
\end{align}
Some thoughts:
This is a periodic function that oscillates between $1$ and $-1$.  As $n$ increases the period shrinks.
I have a feeling that the limit might not exist here for all $x$.
 A: This in response to a comment by the OP.
Consider
$$f_{2^{m-1}}=\sum^{2^{m-1}}_{j=1}\Big(\mathbb{1}_{\big(\tfrac{2j-2}{2^m},\tfrac{2j-1}{2^m}\big]}-\mathbb{1}_{\big(\tfrac{2j-1}{2^m},\tfrac{2j}{2^m}\big]}\Big)$$
Notice that
$$\begin{align}
\big(\frac{2j-2}{2^m},\frac{2j-1}{2^m}\big]&=\big(\frac{2j-2}{2^m},\frac{4j-3}{2^{m+1}}\big]\cup\big(\frac{4j-3}{2^{m+1}},\frac{2j-1}{2^m}\big]\\
&=\big(\frac{4j-4}{2^{m+1}},\frac{4j-3}{2^{m+1}}\big]\cup\big(\frac{4j-3}{2^{m+1}},\frac{4j-4}{2^{m+1}}\big]
\end{align}$$
Similarly
$$\begin{align}
\big(\frac{2j-1}{2^m},\frac{2j}{2^m}\big]&=\big(\frac{2j-1}{2^m},\frac{4j-1}{2^{m+1}}\big]\cup\big(\frac{4j-1}{2^{m+1}},\frac{2j}{2^m}\big]\\
&=\big(\frac{4j-2}{2^{m+1}},\frac{4j-1}{2^{m+1}}\big]\cup\big(\frac{4j-1}{2^{m+1}},\frac{4j}{2^{m+1}}\big]
\end{align}$$
From this
$$
|f_{2^m}-f_{2^{m-1}}|=2\sum^{2^{m-1}}_{j=1}\mathbb{1}_{\big(\tfrac{4j-3}{2^{m+1}},\tfrac{4j-2}{2^{m+1}}\big]}+\mathbb{1}_{\big(\tfrac{4j-j}{2^{m+1}},\tfrac{4j}{2^{m+1}}\big]}$$
If I did not mess up with the signs, we obtain that
$$\|f_{2^m}-f_{2^{m-1}}\|_1=2\cdot 2^{m-1}\cdot 2\cdot 2^{-(m+1)}=1$$
This shows that $\{f_n\}$ (not $\{f_{2^m}\}$) is not a Cauchy sequence in $L_1$. This also shows that $f_n$ does not converges point wise a.s. (dominated convergence would give you a contradiction).
