Generalized Egoroff's Theorem

Question

Let $(X, \mathcal{A}, \mu)$ be a measure space. If $(f_n)_{n \geq 1}$ is a sequence of functions such that $|f_n| \leq g$, for such $g \in L^1(X, \mu)$, $f_n: X \rightarrow \mathbb{R}$, $f: X \rightarrow \mathbb{R}$, then

$f_n \rightarrow f$ almost-uniformly $\iff f_n \rightarrow f$ almost everywhere.

My attempt:

We know that even without the assumption that $|f_n| \leq g$, for an integrable $g$, $f_n \rightarrow f$ almost-uniformly $\implies f_n \rightarrow f$ almost everywhere.

For the other part, I intend to use the original Egoroff's Theorem, which has the assumption $\mu(X) < +\infty$ instead of the limitation by one integrable function.

Let be $\epsilon > 0$ given.

Consider the sets $B = \{ x \in X : g(x) > 1\}$ and $B_k = \{ x \in X : \frac{1}{2^k} \leq g(x) \leq \frac{1}{2^{k -1}}\}$. Thus, we have that

$X = B \cup (\cup_{k \geq 1} B_k )\cup \{x \in X : g(x) = 0\}$.

Also, as $g$ is integrable, $\mu(B) < +\infty$ and $\mu(B_k) < +\infty$, for every $k \in \mathbb{N}$.

Applying Egoroff's original theorem for those sets, there exists $A \subset B$ and for each $k \in \mathbb{N}$, there is $A_k \subset B_k$ such that $\mu(A) < \epsilon/2, \mu(A_k) < \frac{\epsilon}{2^{k+1}}$ and $f_n \rightarrow f$ almost everywhere on $C = (B \setminus A) \cup (\cup_{k \geq 1} B_k \setminus A_k) \cup \{x \in X : f_n(x) = 0, \forall n\}$.

I'm having trouble showing that $\mu(C^c) < \epsilon$.

Answer 1

We have

$$C^c \subset D := A \cup \bigcup_{k \geqslant 1} A_k \tag{$\ast$}$$

and consequently

$$\mu(C^c) \leqslant \mu(D) \leqslant \mu(A) + \sum_{k \geqslant 1} \mu(A_k) < \frac{\epsilon}{2} + \sum_{k \geqslant 1} \frac{\epsilon}{2^{k+1}} = \epsilon$$

by the assumption on $A$ and the $A_k$.

To see $(\ast)$, let $x \in X$ arbitrary. If $g(x) = 0$ then $f_n(x) = 0$ for all $n$, and hence $x \in C$. If $g(x) > 1$ then $x \in B$ and hence either $x \in A \subset D$ or $x \in (B\setminus A) \subset C$. Otherwise there is a $k \geqslant 1$ with $x \in B_k$, and then either $x \in A_k \subset D$ or $x \in (B_k \setminus A_k) \subset C$. So $X = C \cup D$.