Let $f \in L^{1}(\mathbb R)$ and $|f| \leq 1$ $\mu-$a.e. then $\lim_{ n \to \infty}\int_{\mathbb R} |f(x)|^{n}d\lambda (x) = \lambda(\{|f|=1\})$
My idea:
Define $f_{n}(x):=|f(x)|^{n}$
Note that $f_{n}\leq |f| \leq 1$ $\mu-$a.e.
$\lim_{ n \to \infty}\int_{\mathbb R}f_{n}(x)d\lambda (x)=\lim_{ n \to \infty}\int_{\mathbb R}|f(x)|^{n}d\lambda (x)=\lim_{ n \to \infty} \int_{\mathbb R}|f(x)|^{n}1_{\{|f|=1\}}+|f(x)|^{n}1_{\{|f|<1\}}d\lambda (x)$
It follows from the DCT (given that $f \in L^{1}$):
$\lim_{ n \to \infty}\int_{\mathbb R}|f(x)|^{n}1_{\{|f|=1\}}+|f(x)|^{n}1_{\{|f|<1\}}d\lambda (x)=\int_{\mathbb R}\lim_{ n \to \infty}1_{\{|f|=1\}}+|f(x)|^{n}1_{\{|f|<1\}}d\lambda (x)=\int_{\mathbb R}1_{\{|f|=1\}}d\lambda=\lambda(\{|f|=1\})$
The idea is correct, but perhaps it will be more save if we write it like that : we put $f_n=f^n$ and we consider $A=\{x\in\mathbb{R} \; ; |f(x)|<1\}$, since $f\leq 1$ a.e we get $A^c=\{x\in\mathbb{R} \; ; |f(x)|=1\}$ so : $$ |f_n(x)|\longrightarrow g(x)\left\{\begin{array}{lcr} 0 & \rm{if}& x\in A \\1 & \rm{if}& x\in A^c \end{array} \right. $$ since $|f_n|\leq |f|\in L^1(\mathbb{R})$, by DCT we get $$ \lim_n\int_\mathbb{R} |f_n|d\lambda=\int_\mathbb{R} \lim_n|f_n|d\lambda=\int_\mathbb{R} g d\lambda=\int_{A^c} d\lambda=\lambda(A^c). $$