Showing if functions are equal almost everywhere then

Question

Say $f_n$ and $g_n$ are measurable $f_n=g_n$ almost everywhere for $n=1,2,3,...$ then how can we show that $\sup f_n = \sup g_n$ almost everywhere?

I have tried to show that:

$$m(\{\sup f_n \neq \sup g_n \}) = m( \bigcap_{n} \{f_n \neq g_n\} )$$

But I'm not at all sure if this is correct?

Thanks

Answer 1

Let $E_n = \{x : f_n(x) \ne g_n(x)\}$, and let $f, g$ denote the respective suprema. If $$f(x) \ne g(x)$$ then $x \in E_n$ for some $n$ (do you see why?). Thus

$$\{x : f(x) \ne g(x)\} \subseteq \bigcup_{n = 1}^{\infty} E_n$$

which has what measure?