Find a sequence $f_n$ so that $\int_0 ^1 |f_n(x)| = 2$ and $\lim_{n \to \infty} f_n(x) = 1$. The entire problem reads:
(a) Find a sequence $f_n: [0,1] \rightarrow \mathbb{R}$ such that $\int_0 ^1 |f_n(x)| = 2$ for all $n \in \mathbb{N}$ and $\lim_{n \to \infty} f_n(x) = 1$ for all $x \in [0,1]$. (b) If the $f_n$ are as in part (a), then prove $\lim_{n \to \infty} \int_0 ^1 |f_n(x) -1| dx = 1$. 
I'm having trouble finding the proper $f_n$ to fit this situation. Any help is appreciated. Once I find $f_n$, I may need tips for part (b) as well, but I think I could figure that part out given (a). Thanks in advance. 
 A: Define $f_n$ as a triangle on $[0,1/n]$ with $f(x) = 1$ for $x = 0$ and $x \in [1/n,1]$ (the triangle top is choosen such that the total integral is $2$). 
A: For part b). By Egorov Theorem, for every $\epsilon>0$, there exists some measurable set $M_{\epsilon}\subseteq[0,1]$ such that $M_{\epsilon}^{c}:=[0,1]-M_{\epsilon}$ satisfies $|M_{\epsilon}^{c}|<\epsilon$ and $f_{n}\rightarrow 1$ uniformly on $M_{\epsilon}$.
We see that
\begin{align*}
\int_{0}^{1}|f_{n}-1|&=\int_{M_{\epsilon}}|f_{n}-1|+\int_{M_{\epsilon}^{c}}|f_{n}-1|\\
&\leq\int_{M_{\epsilon}}|f_{n}-1|+\int_{M_{\epsilon}^{c}}|f_{n}|+|M_{\epsilon}^{c}|\\
&=\int_{M_{\epsilon}}|f_{n}-1|+\int_{0}^{1}|f_{n}|-\int_{M_{\epsilon}}|f_{n}|+|M_{\epsilon}^{c}|\\
&=\int_{M_{\epsilon}}|f_{n}-1|-\int_{M_{\epsilon}}|f_{n}|+2+|M_{\epsilon}^{c}|\\
&<\int_{M_{\epsilon}}|f_{n}-1|-\int_{M_{\epsilon}}|f_{n}|+2+\epsilon,
\end{align*}
taking $n\rightarrow\infty$, we get 
\begin{align*}
\limsup_{n}\int_{0}^{1}|f_{n}-1|\leq 0-1+2+\epsilon=1+\epsilon,
\end{align*}
so 
\begin{align*}
\limsup_{n}\int_{0}^{1}|f_{n}-1|\leq 1.
\end{align*}
On the other hand,
\begin{align*}
\int_{0}^{1}|f_{n}-1|\geq\int_{0}^{1}|f_{n}|-1=2-1=1.
\end{align*}
A: Hint for (b): Apply Fatou's lemma to $|f_n| + 1- |f_n-1|.$
A: Choose $$f(x):=4n^2x+1 \text{ for }x\in \left[0,\frac{1}{2n}\right]\\
f(x):=-4n^2x+1+4n \text{ for }x\in \left[\frac{1}{2n},\frac{1}{n}\right]\text{and }f(x)=1\text{ otherwise}.$$
