The same question was posted here - but I am unsure why the difference is a measurable function. To summarize:
Let $f_n: X \rightarrow \mathbb{R}\cup \{+\infty\}$ be increasing functions on measure space $(X, \mathbf{X}, \mu)$ that converges to $f$. Let $\phi:X \rightarrow \mathbb{R}$ be a simple measurable function satisfying $0 \le \phi \le f$. Let $\alpha \in (0,1)$ and define $$ A_n = \{x \in X: f_n(x) \ge \alpha \phi(x) \} .$$ Then $A_n$ is a measurable set for all $n$.
My proof would go something like this:
$$ A_n = \{ x \in X : f_n(x) - \alpha \phi(x) \ge 0 \} $$ $$ = \bigcup _{q \in \mathbb{Q}} \{ x \in X : f_{n}(x) \ge q \} \cap \{ \phi (x) \ge q/\alpha \} $$
As both $f_n(x)$ and $\phi(x)$ are measurable functions, $\{ x \in X : f_{n}(x) \ge q \} \cap \{ \phi (x) \ge q/\alpha \} $ is the intersection of two measurable sets, so is also a measurable set. Lastly, the countable union is also a measurable set.
Is this proof correct? Thanks!
The argument I had in mind was the following.
Sums and products of measurable functions are measurable; see Proposition 2.6 of Real Anaylsis by Folland for example. Therefore, as $f_n$ and $\phi$ are measurable, so is $f_n - \alpha\phi$. Now $A_n = (f_n-\alpha\phi)^{-1}([0, \infty))$ is measurable as $[0, \infty)$ is Borel.
Your proof doesn't work because if $f(x) \geq q$ and $\phi(x) \geq q/\alpha$, then $f(x) - \alpha\phi(x)$ need not be non-negative. If you replace $\phi(x) \geq q/\alpha$ by $\phi(x) \leq q/\alpha$, you certainly obtain a set which is contained in $A_n$, but you would need an argument to justify why it is equal to $A_n$ (it is not clear to me whether it is).