For $f,g$ real-valued functions, $f$ weakly increasing and continuous, $A\subseteq \mathbb{R}$, can we say \begin{align*} \sup_{x\in A}f(g(x))=f(\sup_{x\in A}g(x)) \end{align*}
I ask because I notice in wiki the likelihood ratio test statistic is written as the difference between the log of sups, while in class notes I see it written as the difference in sup of logs.
The equality is true.
Let $s = \sup \limits_{x \in A} g(x)$. Then for $x \in A$ we have $g(x) \leqslant s$ so $f(g(x)) \leqslant f(s)$, which means that $f(s)$ is an upper bound of the set $\{ f(g(x)) : x \in A \}$. It remains to show that it is the least such bound.
Take any $\varepsilon > 0$. By continuity of $f$ there is $\delta > 0$ such that $f(s) - \varepsilon < f(y) < f(s) + \varepsilon$ whenever $|y-s| < \delta$. Since $s-\delta < s$, there is $x \in A$ such that $s-\delta < g(x)$ and of course $g(x) \leqslant s$, hence $|g(x) - s| < \delta$. Then we get $f(s) - \varepsilon < f(g(x))$, so $f(s) - \varepsilon$ is not an upper bound of $\{ f(g(x)) : x \in A \}$, QED.
