Convergence in measure of sequence $f_n(x)=|\sin(a_n x+b_n)|^{p_n}$ with $p_n\rightarrow\infty$.

Question

This is a problem from a set of old Analysis qualifying exams I am using to prepare for my upcoming graduate qualifying test.

Suppose $a_n$, $b_n$ and $p_n$ are numerical sequences such that, $a_n>0$, $\liminf_na_n>0$, $p_n\geq0$ and $\lim_\limits{n \to \infty}p_n=\infty$. Define the sequence of functions $$f_n(x)=|\sin(a_n x+b_n)|^{p_n}, \quad x\in[0,2\pi].$$ Show that $f_n$ converges to $0$ in measure. Here, the measure space is $[0,2\pi]$ with the Borel $\sigma$-algebra and the Lebesgue measure.

If the $a_n$ were integers, the problem simplifies considerably since $f_n$ and $f(x)=\sin x$ have the same distribution (as random variables over the normalized Lebesgue measure on $[0,2\pi]$) which can be check easily: $f$ has period $2\pi$, hence $$\frac{1}{2\pi}\int^{2\pi}_0 \mathbf{1}_E(\sin_n(a_nx+b_n))\,dx=\frac{1}{2\pi a_n}\int^{a_n2\pi+b_n}_{b_n}\mathbf{1}_E(\sin(x))\,dx=\frac{1}{2\pi}\int^{2\pi}_0\mathbf{1}_E(\sin(x))\,dx$$ for any measurable set $E\subset[0,2\pi]$. I am having a but of a struggle to carry on with a similar argument for $A_n$ not integer.

I will appreciate any help, hint or a solution to this. Thank you!

Edit: This is a two part problem. I just wrote down the first part, and perhaps the easiest part, of the problem. The second part asks whether convergence in probability can be improved to convergence point wise almost surely for $f_n$. I may post the second part in a different posting later in case I don't get too far.

Answer 1

Instead of estimating the measure of the sets $E_n(\varepsilon)=\{x\in[0,2\pi]: |f_n(x)|>\varepsilon\}$ directly (as I think the OP is suggesting), it is much easier to show that $\int^{2\pi}_0|f_n|\xrightarrow{n\rightarrow\infty}0$. The condition $\liminf_na_n>0$ implies that there is $a>0$ such that $a_n>a$ for all $n$ large enough. The periodicity of $\sin$ implies that $$m(|f_n|>\varepsilon]\leq\frac{1}{\varepsilon}\int^{2\pi}_0|f_n|\leq\frac{1}{\varepsilon a_n}\int^{2\pi\lceil a_n\rceil + b_n}_{b_n}|\sin x|^{p_n}\, dx=\frac{\lceil a_n\rceil}{\varepsilon a_n}\int^{2\pi}_0|\sin x|^{p_n}\,dx $$ where $x\mapsto\lceil x\rceil$ is the ceiling function. The factor $\frac{\lceil a_n\rceil}{a_n}$ is bounded above since $a_n>a$ for all $n\geq N$. The conclusion follows by dominated convergence.

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Answer 2

Here's a hint: when $a_n$ is not integer, you can just extend $[0,2\pi]$ to contain an integer number of periods of $\sin(a_n x + b_n)$ and look at the measure of the points in the extended set where where $|\sin(a_n x + b_n)|^{p_n} > \epsilon$ and show this goes to zero. By the periodicity of $\sin$ it's the same for all values of $b_n$. The condition that $\liminf_n a_n $ is nonzero is there to ensure the functions don't "spread out" faster than the increasing $p_n$ can "tamp down" the graphs.