Show that $\lim_{n\to\infty}\int_0^1|f_n(x)-1|\,dx = 1$ if $\lVert f_n\rVert_{1} = 2$ and $\lim_{n\to\infty}f_n(x) = 1$

Question

Suppose that $\{f_n\}_{n=1}^\infty$, $f_{n}:[0,1]\to\mathbb{R}$, is a sequence of measurable functions such that for every $x\in [0,1]$ $$\lim_{n\to\infty}f_n(x)=1$$ and for every $n$: $$\int_{0}^{1}|f_{n}(x)|\,dx = 2.$$ Prove that $$\lim_{n\to\infty}\int_{0}^{1}|f_{n}(x) - 1|\,dx = 1$$

What I tried/have: Such a sequence is clearly possible since we can take $f_n = 1+n\chi_{(0,\frac{1} {n})}$. The proposition is also trivial if $f_n\geq 1$. I also managed to get, using the reverse triangle inequality that $$\liminf_n \int_0^1|f_n(x) - 1|\,dx\geq \liminf_n\int_0^1|f_n(x)|\,dx - 1 = 1$$ but I'm quite stuck on proving that the $\limsup$ is less than or equal to 1. Since it is not possible to find a dominating function (eg. consider the sequence I stated before) and there is no clear way to construct a monotone sequence, I feel like I need to use Fatou's Lemma somehow. Could someone please give me some tips on how I might show that the limit superior is at most $1$? Or maybe, if there is an easier approach, how I would directly compute the limit?

Answer 1

It is sufficient to show $\limsup_n \int |f_n-1| \le 1$.

Let $\epsilon >0$ then Egorov gives some measurable $B \subset [0,1]$ such that $m B < \epsilon$ and $f_n \to 1$ uniformly on $B^c$.

Then $\int |f_n-1| = \int_B|f_n-1|+ \int_{B^c}|f_n-1| \le \int_B|f_n| + mB + \int_{B^c}|f_n-1|$ so $\limsup_n \int |f_n-1| \le \limsup_n \int_B|f_n| + \epsilon$.

Since $\int |f_n| = \int_B |f_n|+ \int_{B^c}|f_n| = 2$, we see that $\lim_n \int_B |f_n| = 2-m B^c$, that is $\limsup_n \int |f_n-1| \le 2-m B^c + \epsilon \le 1+2 \epsilon$. Since $\epsilon>0$ was arbitrary we have the desired result.