On the Lebesgue measure of the set of small values of a real function

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $$\left\{ \begin{gathered} \frac{{{C_1}{{\left| {\sin \left( {\lambda x} \right)} \right|}^a}}} {{{{\left| x \right|}^b}}} \leqslant \left| {f\left( x \right)} \right| \leqslant \frac{{{C_2}{{\left| {\sin \left( {\lambda x} \right)} \right|}^a}}} {{{{\left| x \right|}^b}}},& \forall \left| x \right| > M \\ f\left( x \right) \ne 0,& \forall \left| x \right| \leqslant M \\ \end{gathered} \right.$$ Here $0<C_1<C_2$, $\lambda>0$, $a\geqslant 1$, $b>0$, $M>0$. We can see that $$\left\{ {x \in \mathbb{R}:f\left( x \right) = 0} \right\} = \left\{ {\frac{{k\pi }} {\lambda }:k \in \mathbb{Z}} \right\}$$ and so the Lebesgue measure $m\left( {\left\{ {x \in \mathbb{R}:f\left( x \right) = 0} \right\}} \right) =0$. Now I focus on the question as follows: for each $\varepsilon>0$, put ${B_\varepsilon } = \left\{ {x \in \mathbb{R}:\left| {f\left( x \right)} \right| < \varepsilon } \right\}$, the set of small values of $f$. Is the Lebesgue measure $m\left( {{B_\varepsilon }} \right)$ tends to $0$ as $\varepsilon\to 0$ ?

Answer 1

Fix $\epsilon>0$. Now choose (large) $N>0$ so that $\frac{C_2}{N^b}<\epsilon$. For all $x>N$, we have $$|f(x)|\le \frac{C_2|\sin (\lambda x)|^a}{|x|^b}\le \frac{C_2}{|x|^b}\le \frac{C_2}{N^b}<\epsilon.$$

Hence $(-\infty,-N)\cup (N,\infty)\subseteq B_\epsilon$.