Let $\psi:\mathbb{R}\to\mathbb{R}$ be a continuous function. Let's consider the next "function" $$\Phi_X:\underset{f}{L^p(X,\mathcal{A},\mu)}\underset{\mapsto}{\to}\underset{\psi\circ f}{L^p(X,\mathcal{A},\mu)} $$ What condition on $\psi$ will be sufficient and necessary for $\Phi_X$ to be well defined? In this case, $\Phi_X$, will it be continuous?
My idea: For the function to be well defined we must prove that $\psi\circ f$ belongs to $L^p(X,\mathcal{A},\mu)$, that is, $\psi\circ f:X\to\mathbb{R}$ must be $\mathcal{A}-$measurable and $\int_X|\psi\circ f|^pd\mu=\underset{\varphi\in I_{|\psi\circ f|^p}}{\sup}\int_X\varphi d\mu<+\infty$ (be finite), where $I_{|\psi\circ f|^p}$$=\{\varphi\in\mathscr{S}_+:0\leq\varphi\leq f\}$.
- $\mathcal{A}-$measurable: Let $\alpha\in\mathbb{R}$, as $\psi$ is continuous then $\psi^{-1}(\langle -\infty,\alpha\rangle)\in B(\mathbb{R})$, also $f$ is measurable, so $f^{-1}(\psi^{-1}(\langle -\infty,\alpha\rangle))\in \mathcal{A}$, i.e. $(\psi\circ f)^{-1}(\langle -\infty,\alpha\rangle))\in \mathcal{A}$.
- $\int_X|\psi\circ f|^pd\mu<+\infty:$ I have tried to find some necessary and sufficient condition on $\psi$ for this to be possible, but I have not succeeded. If you could give me any idea or suggestion I will be very grateful.
I think your best bet is to ensure that your $\varphi$ satisfies something like $$\|\varphi\circ f\|_p \leq C\| f\|_p$$ for all $f\in L^p(X,\mathcal{A},\mu)$ to ensure that the function is well defined. You will immediately get continuity for $\Phi_X$. A sufficient condition would be that $\varphi$ is Lipschitz continuous to ensure that $\Phi_X$ is continuous. Lipschitz continuity would give us that for all $x\in X$ we would have $|\varphi(f(x))-\varphi(g(x))|\leq D| f(x)-g(x)|$. Hence $\|\Phi_X(f)-\Phi_X(g)\|_p\leq D\|f-g \|_p$.