Convergence in $L_p$, Vitali's theorem, and convergence in measure Working a measure theory question for practice from Bartle. 
Assume that 


*

*$(X,\mathbb{X},\mu)$ is a finite measure space

*$f_{n}\rightarrow f$ in $L_{p}(X,\mathbb{X},\mu)$

*$\varphi$ is a real-valued continuous function on the real line s.t. there exists a positive number K with $\vert\varphi(t)\vert<K\vert t\vert$ if $\vert t\vert>K$ 


Claim: $\varphi\circ f_{n}\rightarrow\varphi\circ f$ in $L_{p}(X,\mathbb{X},\mu)$
My approach thus far has been to use the Vitali convergence theorem. I haven't been able to show that $\varphi\circ f_{n}\rightarrow\varphi\circ f$ in measure without using the continuous mapping theorem. However, Bartle's book doesn't include this theorem so either 


*

*I'm not supposed to use the Vitali theorem since we can't prove the convergence in measure without the continuous mapping theorem.

*There's a way to prove convergence in measure without the continuous mapping theorem.


I'm a little lost with how to proceed.
 A: Sorry, you are absolutely right. We don't need Lipschitz continuity. However, as you suggested Vitali's theorem (or sometimes a generalized Lebesgue's theorem) is applicable.
So the measure space $X$ is finite. Since $f_n\to f$ in $L$ we find that, for a subsequence, $f_n(x)\to f(x)$ for a.e. $x$. By continuity of $\varphi$ we find for that subsequence $$g_n(x):=\varphi(f_n(x))\to \varphi(f(x))=:g(x)$$. 
We shoe that $g(x)\in L^1(X)$. This implies $|g(x)|<\infty$ a.e. in $X$ (if note the integral is necessarily unbounded).
$$|g(x)|=|\varphi(f(x))|\leq \max_{t\in[-K,K]}+ K|f(x)|$$
Integration yields
$$\int_X |g(x)| \leq \int C+K|f(x)|\leq C_1(1+||f||_1)<\infty.$$
Furthermore, the last estimate yields uniform integrability since $||f_n||<C_2$ uniformly in $n$.
To get from the subsequence to the full sequence we use the following
$\mathbf{Lemma}$. Let $(X,d)$ be a metric space and let $x_n,x\in X$ be given. Then there holds: $x_n\to x\Leftrightarrow$ for every subsequence $n'$ of $n$ there is a further subsequence $n''$ of $n'$ such that $x_{n''}\to x$.
$\mathbf{Proof}$ The implication is obvious. Hence, for the other statement, assume that $x_n\not\to x$. But this means, that there is some $\varepsilon >0$ and a subsequence $n'$ such that $d(x_{n'},x)\geq \varepsilon$ for all $n'$. Due to our assumption there is a subsequence $n''$ of $n'$ such that $d(x_{n''},x)<\varepsilon$ for large $n''$. Contradiction.
