Integral of $f_n$ from $0$ to $1$ is zero Let $f_n(t)$ be defined as $$f_n(t)=\begin{cases}1, \text{if}\,\,t\in[\frac{p}{2^k},\frac{p+1}{2^k})\\ 0, \text{otherwise}\end{cases}$$ where $n=2^k+p,\,\,0\le p<2^k$. Then, 
a)what can be the value of $\lim\sup f_n(t)$ and $\lim\inf f_n(t)$? 
b)Is $\int_0^1|f_n(t)|\to0$ when $n\to\infty$ true?
For part b), I think that as $n\to\infty$, the function approaches zero because of the constriction of the interval on which it is $1$.Any ideas. Thanks beforehand.

Hints:
a) For a fixed $t \in [0,1]$ and some time $N$, is there an $n > N$ so that $f_n(t) > 0$?
b) Write out an explicit bound to formalize your idea. What is the value of integral at the stage when the interval has length $2^{-n}$?
Bonus: Think about $b$ for the function $2^k f_n$ instead ($n = 2^k + p$). What's the difference?
