Does the sequence $f_n(x) = \sin^n(x) + \cos^n(x)$ converge on $\mathbb{R}$?
a) almost everywhere
b) according to Lebesgue measure
My solution:
a) $\lim\limits_{n \rightarrow \infty} (\sin^n(x) + \cos^n(x)) = 0$ (almost everywhere)
Therefore yes
b) $\lim \mu ${$ x: |\sin^n(x) + \cos^n(x)| \geq \varepsilon $}
and I don’t know what to do next, please push me to a solution
Thanks!
The answer is $\{f_n\}$ does not converge in measure. First we show that if $\psi$ is a function on $\mathbb{R}$ that is not [zero almost everywhere], then $f_n\not\to\psi$ in measure. Next we show $f_n\not\to\psi$ in measure neither if $\psi\equiv 0$ a.e. $x\in\mathbb{R}$.
Claim: If $\psi:\mathbb{R}\to \mathbb{R}\cup\{+\infty,-\infty\}$ is not [zero almost everywhere], then $f_n\not\to\psi$ in measure.
In fact, assume $f_n\to\psi$ in measure, and denote the constant zero function by $f$ (i.e. $f\equiv0\ \forall x$). Then Riesz's theorem says exists a subsequence $\{f_{n_k}\}$ s.t. $f_{n_k}\to \psi$ a.e. $x\in\mathbb{R}$. Since $f_n\to f$ a.e. $x\in\mathbb{R}$, a fortiori we have $f_{n_k}\to f$ a.e. $x\in\mathbb{R}$. Since $f_{n_k}$ converges both to $\psi$ and $f$ almost everywhere, we infer $\psi=f$ a.e, that is $\psi\equiv 0$ a.e. $x\in\mathbb{R}$, contradicting the hypothesis that $\psi$ is not [zero almost everywhere]. Therefore $f_n\not\to\psi$ in measure. We have verified our first claim.
Claim: If $\psi\equiv 0$ a.e. $x\in\mathbb{R}$, then neither does $\{f_n\}$ converges to $\psi$ in measure.
In fact, select a fixed $n\in\mathbb{Z}_+$. Since $f_n(0)=1$, therefore by continuity, given any $\epsilon\in (0,1)$, exists $d\in (0,2\pi)$ s.t. $|f_n(x)|>\epsilon$ for all $x\in(0,d)$, i.e. \begin{equation} \{x\in (0,2\pi):\ |f_n(x)|>\epsilon\}\supseteq (0,d) . \end{equation} Hence for all $n\in\mathbb{Z}_+$, for all $\epsilon\in (0,1)$, \begin{equation} m(\{x\in (0,2\pi):\ |f_n(x)|>\epsilon\})>d>0. \end{equation} Therefore \begin{equation} m(\{x\in \mathbb{R}:\ |f_n(x)|>\epsilon\})=\sum_{n=-\infty}^{+\infty}a_n=+\infty, \end{equation} where $a_n\equiv m(\{x\in (0,2\pi):\ |f_n(x)|>\epsilon\})\ \forall n $ is a constant positive sequence. Therefore $\{f_n\}$ does not converge to $\psi\equiv 0\ (\forall x\in\mathbb{R})$ in measure, giving pur second claim.
To sum up, for all measurable function $\psi:\mathbb{R}\to \mathbb{R}\cup\{+\infty,-\infty\}$, $\{f_n\}$ does not converge to $\psi\equiv 0\ (\forall x\in\mathbb{R})$ in measure.