Can the measure of $\{ x \in \Omega : 0 < f_n(x) < k_n\}$ be bounded by $C/n$ if $f_n$ converges to a non-positive function?

Question

Let $f_n \to f$ in $C^0(\Omega)$ on a bounded domain where $f \leq 0$.

Take a sequence $k_n \to 0$, each $k_n > 0$. Consider $$M_n := \{ x \in \Omega : 0 < f_n(x) < k_n\}.$$

Is it possible to pick the sequence $k_n$ such that $|M_n| \leq \frac Cn$ where $C$ is some constant? Here $|M_n|$ means the Lebesgue measure of $M_n$.

I think the uniform convergence helps but I can't prove it/

Answer 1

All we need to show is that for a given continuous $f$ the measure of the set $\{x \in \Omega: 0 < f(x) < \varepsilon \} =:M (\varepsilon) $ converges to $0$ as $\varepsilon\to 0$. Indeed, once we have this, then for each $n\in \mathbb{N}$ fixed we can choose $\varepsilon = k_n$ small enough so that $|M_n (\varepsilon)|$ becomes smaller than $1/n$. Now to see that $|M (\varepsilon)| \to 0$ as $\varepsilon \to 0 $ observe that, in view of the continuity of measure we have $ \lim\limits_{\varepsilon \to 0} |M (\varepsilon)| = \mu( \bigcap\limits_{\varepsilon > 0} M(\varepsilon) ) = \mu(\emptyset) = 0$.