Show that there exists a sequence of positive real numbers $C_{n}$ such that $sup_{n}\dfrac{|f_{n}(x)|}{C_{n}}<\epsilon$ for all $x\in X\setminus E $

Question

I am trying the prove following question: Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$, and let $(f_{n})$ be a sequence of almost everywhere finite measurable functions. Show that for every $\epsilon>0$ there exists a measurable set $E$ with $\mu(E)<\epsilon$ and a sequence of positive real numbers $C_{n}$ such that $sup_{n}\dfrac{|f_{n}(x)|}{C_{n}}<\epsilon$ for all $x\in X\setminus E $.
My ideas on the proof: Let $\epsilon>0$ and $n\in \mathbb{N}$. Then, by Lusin's theorem, there exists a closed set $F_{n}$ such that $\mu( X\setminus F_{n} )<\dfrac{1}{2^{n}}$ and $f_{n}$ is continuous on $F_{n}$. Now let $F=\bigcap F_{n}$ and $E=X\setminus F$. Then we have $\mu(E)<\epsilon$ and $f_{n}$ is continuous on $F$ for all $n$. After these steps, I want to say that $f_{n}$ is bounded on $F$. If I can say this, then there exists $C_{n}$ such that $|f_{n}(x)|<C_{n}\epsilon$ for all $n$. So I can finish the proof. But I am not sure that I can say this. $F$ is closed and bounded. But I can not say that $F$ is compact because I do not know anything about $X$ except finiteness. How can I show that $f_{n}$ is bounded on $F$ for all $n$? Or if this proof does not work, how can I prove this question?

Answer 1

Here is a proof that does not use any topological properties of $X$, as Lusin's Theorem does.

Let $n \in \mathbb{Z}_+$. First, note that $\{x: |f_n|(x) < \infty\} = \cup_{m \in \mathbb{Z}_+} \{x: |f_n|(x) < m\}$. By monotone convergence: $$\mu(\{x: |f_n|(x) < \infty\}) = \lim_{m \to \infty} \mu(\{x: |f_n|(x) < m\})$$ By almost everywhere finiteness of $f_n$: $$\mu (X) = \mu(\{x: |f_n|(x) < \infty\})$$ Putting these two statements together we have: $$\mu(X) = \lim_{m \to \infty} \mu(\{x: |f_n|(x) < m\})$$Hence, given $\epsilon > 0$, and using the finiteness of $\mu(X)$, we can find $Z_n$ such that $\mu(X) - \mu(\{x: |f_n|(x) < Z_n\}) < \frac{\epsilon}{2^n}$. Again using finiteness of $\mu(X)$, $$\frac{\epsilon}{2^n} > \mu(X) - \mu(\{x: |f_n|(x) < Z_n\}) =\mu(\{x: |f_n|(x) \geq Z_n\})$$

Let $E_n = \{x: |f_n|(x) \geq Z_n\}$ and let $C_n = \frac{Z_n}{\epsilon}$.

Now, set $E = \cup_n E_n$. Then $\mu(E) \leq \sum_n \mu(E_n) < \sum_n \epsilon / 2^n = \epsilon$.

Consider some arbitrary $n$. For $x \in X \setminus E$, $x \notin E_n$, so $|f_n|(x) < Z_n$. Thus, $\frac{|f_n|(x)}{C_n} < \frac{Z_n}{Z_n / \epsilon} = \epsilon$. Since $x$ and $n$ were arbitrary, it follows that $\sup_n \frac{|f_n|(x)}{C_n} \leq \epsilon$ for all $x \in X \setminus E$.