To find a sequence on $L^1$-norm equal to 2, converging a.e. to a function of $L^1$norm equal to 1. I'm having some difficulty with the following two part problem:
(a) Find a sequence of functions $f_n$ in $L^1[0,1]$ converging a.e. to a function $g$ so that $\|f_n\|_1=2, \forall n$ but $\|g\|_1=1$.
(b) Show that for a sequence $f_n$ as above,
$$\lim_{n\to\infty}\int_0^1|f_n(x)-g(x)|dx=1$$
For part (a), I'm not even sure where to begin looking for such a sequence. I understand that I need a sequence of functions whose area under the curve is constant 2, but that of the limit is 1. How does one go about constructing a sequence such as this?
For part (b), I applied Egoroff's theorem but I'm unable to proceed: Given $\epsilon>0$, there exists a set $E\subset[0,1]$ such that $m([0,1]\setminus E])<\epsilon$ and $f_n\to f$ uniformly on  $E$.
$$\left|\int_0^1|f_n(x)-g(x)|dx-1\right|\leq \int_0^1||f_n(x)-g(x)|-1|dx=\int_E||f_n(x)-g(x)|-1|dx+\int_{[0,1]\setminus E}||f_n(x)-g(x)|-1|dx$$
How do I proceed from here? Thanks for your time.
 A: For the first part, you may choose that $$g(x) = \left\{\begin{matrix}2\quad 0\leq x \leq \frac12\\0\quad \frac12 <x\leq 1\end{matrix}\right.$$
And for $n\geq2$
$$f_n(x) = \left\{\begin{matrix}2\quad 0\leq x \leq \frac12\\\quad0\quad \frac12<x \leq 1-\frac1n \\\quad n\quad 1-\frac1n <x\leq 1\end{matrix}\right.$$
It is easy to check that $(f_n)$ converges to g a.e. . However, $\Vert{f_n}\Vert = 2$ and $\Vert g \Vert=1$.
For problem(b), by applying Egroff's theorem, you can get such a subset E that $m(E)=1-\epsilon$,$m([0,1]\backslash E)=\epsilon$ and$(f_n)$ converges to $g$ uniformly on E.Then, you may write the integral as follows:$$\lim_n\int_0^1|f_n(x)-g(x)|=\lim_n(\int_E|f_n(x)-g(x)|+\int_{[0,1]\backslash E}|f_n(x)-g(x)|)$$
Claim:$$\lim_n\int_{[0,1]\backslash E}|f_n(x)-g(x)|=1$$
First observe that, $$\int_{[0,1]\backslash E}|g(x)| <\delta$$by the fact $m(E)=1-\epsilon$ and the absolute continuousness of L-intergral. And$$\int_{[0,1]\backslash E}|f_n(x)|\to 1$$ since $\int_{E}|f_n(x)-g(x)|\to 0 $ and $1\geq \int_{E}|g(x)| \geq 1-\delta$.
And the conclusion follows.
