Let $f_t(x)=\left|\sin tx\right|^t$, $x\in[0,1]$, $t\in(0,+\infty)$. Does the parametric family converge on $[0,1]$ in measure with respect to the Lebesgue measure as $t$ approaches $+\infty$? Does it converge almost everywhere?
My intuition about the problems tells me that the family is converging in measure on the interval $[0,1]$ and that the convergence isn't almost everywhere, but I can't seem to formulate any concrete ideas or proofs. Any thoughts?
1) If $x=0$ it does converges. Otherwise, consider two sequences $$ t_n'=\frac{\pi n}{x}\qquad t_n''=\frac{\pi+2\pi n}{2x} $$ tending to infinity. Clearly, $$ \lim\limits_{n\to\infty}f_{t_n'}(x)=0\qquad\lim\limits_{n\to\infty}f_{t_n''}(x)=1 $$ so $\lim\limits_{t\to\infty} f_t(x)$ does not exist by Heine's definition of continuity. Therefore $(f_t(x))_{t>0}$ converges only for $x=0$.
2) Fix $n>3$ and $t>2n$. Let $k=\lceil \frac{t}{2\pi}\rceil$, then $\frac{k}{t}\leq\frac{1}{2\pi}+\frac{1}{t}\leq\frac{1}{\pi}$. Now, note that $$ \begin{align} \int_0^1|\sin tx|^tdx &\leq\int_0^1|\sin tx|^{2n}dx\qquad\\ &=\frac{1}{t}\int_0^t\sin^{2n} sds \\ &\leq \frac{1}{t}\int_0^{2\pi k}\sin^{2n} sds \\ &= \frac{k}{t}\int_0^{2\pi}\sin^{2n} sds \\ &\leq \frac{1}{\pi}\frac{{2n \choose n}}{2^{2n}} \\ \end{align} $$ For evaluation of the integral see this post. Since $t>2n$ we get that $$ \lim\limits_{t\to\infty}\int_0^1|\sin tx|^tdx\leq \frac{1}{\pi 2^{2n}}{2n \choose n} $$ Since $n>3$ is arbitrary, we conclude $$ \lim\limits_{t\to\infty}\int_0^1|\sin tx|^tdx\leq\lim\limits_{n\to\infty}\frac{1}{\pi 2^{2n}}{2n \choose n}=0 $$ In evaluation of the last limit we used this asymptotics.
Thus we proved that $(f_t)_{t>0}$ $L_1$-converges to $0$. Hence it converges to $0$ in measure.
