Problem 7.V Bartle Elements of Integration $\textbf{Question:}$ Let $(X,\mathcal{F},\mu)$ be an arbitrary measure space. Let $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ be continuous and satisfy for some $K>0$:
$$ \vert \varphi(t) \vert \leq K \vert t \vert, \forall t\in \mathbb{R} (*)$$
If $f \in L^p$, then $\varphi \circ f$ belongs to $L^p$. Conversely, if $\varphi$ does not satisfy (*), there exists a measure space $(X,\mathcal{F},\mu)$ and a function $f \in L^p$ such that $\varphi \circ f$ does not belong to $L^p$.
$\textbf{My attempt:}$ If $\varphi$ satisfy $(*)$ we have for each $(X,\mathcal{F},\mu)$ and $x\in X$
$$ \vert (\varphi \circ f)(x) \vert = \vert \varphi(f(x)) \vert \leq K \vert f(x) \vert $$
$$ \implies \vert \varphi \circ f \vert^p \leq K^p \vert f \vert^p $$
So $\varphi \circ f \in L^p$. I can't solve the other statement, help please.
 A: The result is not true as stated. In the inequality for $\phi$ we should add 'for $|t|$ sufficiently large' to make it correct.
First a counterexample: suppose the measure space is finite and $\phi (t)=\sqrt {|t|}$. Then the inequality fails but $\phi(f) \in L^{p}$ whenever $f \in L^{p}$.
Here is the proof when the question is modified as stated:
If (*) is false after the modification then there exist $t_n$ increasing t0 $\infty$ such that $\phi (t_n) >nt_n$. Consider the real line with Lebesgue measure and define $f(x)$ to be $t_n$ on $(t_n-r_n, t_n+r_n)$ where $r_n$'s are so small that the intervals $(t_n-r_n,t_n+r_n)$ are disjoint. Take $f$ to be $0$ outside these intervals.  Then $\int|f|^{p}=2\sum t_n^{p} r_n$ and $\int|\phi(f)|^{p}>2\sum n^{p}t_n^{p} r_n$. We only have to choose $r_n=\frac 1{n^{p}t_n^{p+1}}$.[ You can always choose $t_n$ to tend to $\infty$ as fast as you want so the disjointness condition can be met easily].
A: Observe that he Converse part is equivalent to the statement
Theorem: If $\phi\in C(\mathbb{R})$ and for any measure space $(X,\mathscr{F},\mu)$
$$\phi\circ f\in L_p(\mu)$$
whenever $f\in L_p(\mu)$, then $\phi$ satisfies (*).
We now prove this Theorem. We argue by contradiction. Suppose $\phi$ satisfies $\phi\circ f\in L_p(\mu)$ whenever $f\in L_p(\mu)$ for any measure space $(X,\mathscr{F},\mu)$, but that that for any $n\in\mathbb{N}$ there is $t_n$ such that $|\phi(t_n)|> n|t_n|$.
In particular, consider the Lebesgue space $(\mathbb{R},\mathscr{B},\lambda_1)$. Since $f=\mathbb{1}_{[-1/2,1/2]}\in L_p$ and $\phi(f)=\phi(0)\mathbb{1}){[-1/2,1/2]^c}+\phi(1)\mathbb{1}_{[-1/2,1/2]}$, it follows that
$$\|\phi\circ f\|^p_p=|\phi(1)|^p+|\phi(0)|^p\cdot\infty<\infty$$
and so, $\phi(0)=0$. This shows that each $|t_n|>0$.
Fix $0<\varepsilon<p$ and define $a_n=\frac{1}{|t_n|^p  n^{1+\varepsilon}}$, and  let $\{A_n:n\in\mathbb{N}\}$ be a sequence  of pairwise disjoint measurable sets in the real line such that $\lambda_1(A_n)=a_n$. Define
$$ f=\sum_nt_n\mathbb{1}_{A_n}$$
Clearly $\|f\|^p_p=\sum_n|t_n|^p\frac{1}{|t_n|^p n^{1+\varepsilon}}<\infty$; however
$$
\|\phi(f)\|_p=\sum_n|\phi(t_n)|^p\lambda_1(A_n)>\sum_nn^p|t_n|^p\frac{1}{|t_n|^pn^{1+\varepsilon}}=\infty$$
contraditing the assumption that $\phi\circ f\in L_p$ whenever $f\in L_p$. Therefore, there exist $k>0$ such that $|\phi(t)|\leq k|t|$ for all $t$.

Some remarks:

*

*The result hold for $0<p<\infty$.

*If $\phi\in C(\mathbb{R})$ and $(\Omega,\mathscr{F},\mu)$ is an arbitrary but fixed nonatomic finite measure space, then the following slightly weaker result holds:
$\phi\circ f\in L_p(\Omega,\mathscr{F},\mu)$ whenever $f\in L_p(\Omega,\mathscr{F},\mu)$ iff there is $k>0$ such that
$$|\phi(t)|\leq k|t|\quad\text{for all}\quad|t|>k$$
https://math.stackexchange.com/a/3755045/121671
