Consider the sequence $f_n(x) = (\sin(πnx))^n , n = 1, 2, ...,$ on the interval $[0,1].$ Consider the sequence $f_n(x) = (\sin(πnx))^n , n = 1, 2, ...,$ on the interval $[0,1].$ Prove that for any $δ > 0$ there is a set $E ⊂ [0,1]$ with $m(E) > 1−δ,$ and a subsequence $f_{n_k} (x), k = 1, 2, 3...,$ such that $\lim_{k→∞} f_{n_k} (x) = 0$ for $x ∈ E.$
Not sure what to do.  Want to construct a set $E$ as above with $f_n$ going to zero in $L_1$ norm.  Then it would follow.
 A: Let us estimate
\begin{eqnarray*}
\left\Vert f_{n}\right\Vert _{1} & = & \int_{0}^{1}\left|\sin\left(\pi nx\right)\right|^{n}\, dx\\
 & \overset{y=\frac{n}{2}x}{=} & \frac{2}{n}\cdot\int_{0}^{n/2}\left|\sin\left(2\pi y\right)\right|^{n}\, dy\\
 & \leq & \frac{2}{n}\cdot\int_{0}^{n}\left|\sin\left(2\pi y\right)\right|^{n}\, dy\\
 & = & 2\int_{0}^{1}\left|\sin\left(2\pi y\right)\right|^{n}\, dy,
\end{eqnarray*}
where I used that $y\mapsto\left|\sin\left(2\pi y\right)\right|$
is $1$-periodic in the last step.
Now observe that $\left|\sin\left(2\pi y\right)\right|^{n}\leq1$
and $\left|\sin\left(2\pi y\right)\right|^{n}\xrightarrow[n\rightarrow\infty]{}0$
as long as $\left|\sin\left(2\pi y\right)\right|\neq1$. But $\left|\sin\left(2\pi y\right)\right|=1$
holds only for $y\in\left\{ \frac{1}{4},\frac{3}{4}\right\} $, i.e.
on a finite set, hence on a null-set. Thus, $\left|\sin\left(2\pi y\right)\right|^{n}\xrightarrow[n\rightarrow\infty]{}0$
almost everywhere.
Using dominated convergence, we conclude $\left\Vert f_{n}\right\Vert _{1}\xrightarrow[n\rightarrow\infty]{}0$.
It is well-known (cf. Subsequence convergence in $L^p$) that every $L^{1}$-convergent sequence $\left(f_{n}\right)_{n\in\mathbb{N}}$
has a subsequence $\left(f_{n_{k}}\right)_{k\in\mathbb{N}}$ so that
$f_{n_{k}}\left(x\right)\xrightarrow[n\rightarrow\infty]{}f\left(x\right)$
almost everywhere, where $f$ is the $L^{1}$-limit.
Hence, we can even take $E\subset\left[0,1\right]$ with $m\left(E\right)=1$,
because there is a subsequence $\left(f_{n_{k}}\right)_{k\in\mathbb{N}}$
converging to zero almost everywhere.
