# Prove that for $E$ of finite measure, $T(f) = \int_E\phi\circ f$ is continuous on $L^p(E)$ if $\phi(x)$ in continuous on $R$ and $\phi(x)<a+b|x|^p$.

In Chapter 8 of Real analysis, 4th edition by Royden, a functional is continuous if $f_n \rightarrow f$ in $L^p$ implies $T(f_n)\rightarrow T(f)$. Royden had given a proof of this proposition in Collary 18 in that Chapter but there seems to be a mistake in it.

Let $\{f_n\}$ be a sequence in $L^p$ that convergences strongly to $f$ in $L^p(E)$. By taking a subsequence if necessary and relabelling, we suppose $\{f_n\}$ is rapidly Cauchy. Therefore, according to Theorem 6 of Chapter 7, $\{f_n\}$ convergences pointwise a.e. on $E$ to $f$. Since $\phi$ is continuous, ${\phi\circ f_n}$ convergences pointwise a.e. on E to $\phi\circ f$. Moreover, by the completeness of $L^p(E)$, since ${f_n}$ is rapid Cauchy in $L^p(E)$, the function $$g = |f_1| + \sum_{n=1}^{\infty} |{f_{n+1}-f_n}|$$ belongs to $L^p(E)$. It is clear that $$|f_n| \le g \text{ a.e. on E for all } n.$$ and hence, by the inequality (33), $$|\phi\circ f_n| \le a + b \cdot |f_n|^p \le a + b\cdot g^p \text{ a.e. on E for all }n.$$ We infer from the Dominated Convergence Theorem that $$\lim_{n\rightarrow \infty} \int_E \phi\circ f_n = \int_E\phi\circ f.$$ Therefore T is continuous on $L^p(E)$.

But, why could one take a subsequence if necessary as in the boldface text above? If this was done, it occur to me that it is only proven that for the subsequence $T(f_{n_k})\rightarrow T(f)$ rather than for the original sequence $T(f_n)\rightarrow T(f)$.

This is enough to prove the claim because we now know that any sequence $f_n\to f$ has a subsequence such that $T(f_{n_k})\to T(f)$.
In particular, assume $T(f_n)\not\to T(f)$. Then there is a subsequence such that $T(f_{n_k})-T(f)$ is uniformly large. However, $f_{n_k}$ has a subsequence $f_{n_{k_\ell}}$ such that $T(f_{n_{k_\ell}})\to T(f)$, which is a contradiction.