Convergence in $L_p$, Vitali's theorem, and convergence in measure

Question

Working a measure theory question for practice from Bartle.

Assume that

1. $(X,\mathbb{X},\mu)$ is a finite measure space
2. $f_{n}\rightarrow f$ in $L_{p}(X,\mathbb{X},\mu)$
3. $\varphi$ 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

1. I'm not supposed to use the Vitali theorem since we can't prove the convergence in measure without the continuous mapping theorem.
2. There's a way to prove convergence in measure without the continuous mapping theorem.

I'm a little lost with how to proceed.

Answer 1

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.