Function converging in $L^1$ I've been having trouble with rigorously showing that on the interval $[0, 2\pi] $ the function $\sin^n(nx) \rightarrow 0$ in $L^1$ convergence. I've convinced myself that this is true intuitively by looking at graphs of the function but I can't put my finger on a rigorous solution and I was wondering if anybody knows how.
 A: By change of variables $x\to nx$, \begin{align}
 \int_0^{2\pi}|\sin^n(nx)| \ dx &= \frac1n\int_0^{2\pi n}|\sin^n(x)| \ dx \notag \\
   &=\frac{n}{n} \int_0^{2\pi} |\sin^n(x)| \ dx \tag{1} \\
   &= \int_0^{2\pi} |\sin^n(x)| \ dx, \notag
\end{align}
where (1) follows from the $2\pi$-periodicity of $\sin$. The integrand on the right goes to 0 pointwise a.e. and is dominated by $g\equiv 1$ on $[0,2\pi]$. Therefore, by the Dominated Convergence Theorem the integral goes to 0 as $n\to\infty$. 
A: Hint: You have convergence in measure.
A: Given $\epsilon \in (0,1)$, $|\sin(t)| > \epsilon^{1/n} = 1 - \dfrac{|\ln(\epsilon)|}{n} + O\left(\dfrac{1}{n^2}\right)$ on an interval of length $O(1/\sqrt{n})$ around each odd multiple of $\pi/2$. Then
$|\sin(n x)| > \epsilon^{1/n}$ on a set of measure $O(1/\sqrt{n})$ in $[0,2\pi]$
consisting of $O(n)$ intervals, each of length $O(n^{-3/2})$. So
$\int_0^{2\pi} |\sin(nx)|^n \; dx \le \epsilon + O(1/\sqrt{n})$, and 
$\limsup_{n \to \infty} \|\sin(nx)^n\|_1 \le \epsilon$.  Since this is true
for every $\epsilon$, we conclude that the limit is $0$.
