$K\subseteq\Bbb R^n$ compact, $f:\Bbb R^n\rightarrow\Bbb R$ Lebesgue measurable then $\lim_{R\rightarrow +\infty}|\{x\in K:|f(x)|>R \}|=0$

Question

(Verification)As the title says, if $K \subseteq \Bbb R^n$ is compact and $f:\Bbb R^n \rightarrow \Bbb R$ is a Lebesgue measurable function I would like to prove $$\lim_{R \rightarrow +\infty} |\{ x \in K : |f(x)|>R \}|=0$$ I don't even know if this is true so I would like to know if my proof is correct. Given $\epsilon >0$ by Lusin's Theorem there is $F\subseteq K$ closed (therefore compact) such that $f$ is continuous over $F$ and $|K-F|< \epsilon$. Then we have an $M \in \Bbb R_{>0}$ with $|f(x)|<M$ for every $x \in F$ (because $f$ is continuous over $F$ wich is compact). Let $R>M$ then we have $$\{x \in K : |f(x)|>R \}=\{x \in F : |f(x)|>R \} \, \cup \, \{x \in K-F : |f(x)|>R \}$$ It's clear $\{x \in F : |f(x)|>R \}= \emptyset$ because $|f(x)|<M<R$ for every $x \in F$. Then $$|\{x \in K : |f(x)|>R \}|=|\{x \in K-F : |f(x)|>R \}| \le |K-F| \lt \epsilon$$ So for every $\epsilon >0$ we found $M>0$ with $|\{x \in K : |f(x)|>R \}|<\epsilon$ for every $R \ge M$. We proved $$\lim_{R \rightarrow +\infty} |\{ x \in K : |f(x)|>R \}|=0$$ Is there any mistake in my proof?