Given two real-valued functions $f_1$ and $f_2$ on a common domain $E.$ Define $f^*=\max(f_1,f_2)$ and $f_*=\min(f_1,f_2).$
Theorem. If $f_1$ and $f_2$ are measurable functions, then the functions $f^*$ and $f_*$ are measurable.
In my book, the proof starts with for any real $a,$ we have:
$E(f^*>a)=E(f_1>a)\cup E(f_2>a),\\\\E(f_*>a)=E(f_1>a)\cap E(f_2>a).\tag1$
From $(1)$, the result follows immediately.
Here, the notation $E(f>a)=\{x\in E:f(x)>a\}.$
But I am not able to understand how the relations in $(1)$ are valid actually. According to me, if at a point $x\in E,$ $f_1>f_2$ then $f^*=f_1$ and $f_*=f_2.$ So the relations in $(1)$ do not hold good exactly. Please help me to see what I am missing here. Please give some insights. Thank you in advance.
Yes they do. We take any $a\in\mathbb R$ and ask: When is $\max(f_1,f_2)>a$? This is exactly the case when at least one of $f_1>a$ or $f_2>a$ holds. So thus you get the first property. Then when is $\min(f_1,f_2)>a$? Exactly if both $f_1>a$ and $f_2>a$. This gives you the second property.
While what you are saying is valid, it does not really affect anything of this. What you are stating there does by no means disprove (1). If $f_1(x)>f_2(x)$ then $f^\ast(x) > a$ if $f_1(x)>a$. But if $f_2(x)>a$ then especially also $f_1(x)>a$, so no invalid points can be added by forming this union.