In the article The Preservation of Convergence of Measurable Functions Under Composition, by R.G. Bartle and J.T. Joichi, there is the following theorem and proof:

Theorem 2. The function $\phi$ preserves uniform convergence, almost uniform convergence, or convergence in measure of sequences of measurable functions if and only if $\phi$ is uniformly continuous.

Proof. If $\phi$ is uniformly continuous and if $|\phi(x) - \phi(y)| \geq \epsilon$, then $|x-y| \geq \delta(\epsilon)$. Hence \begin{equation} A = \{s \in S: |\phi \circ f_n(s) - \phi \circ f(s)| \geq \epsilon \} \subseteq \{s \in S: |f_n(s) - f(s)| \geq \delta(\epsilon) \} =B \end{equation} It follows readily that $\phi$ preserves the stated modes of convergence. Q.E.D.

I did't get why "it follows readily", although is probably quite a lot obvious why it is.

From what I understood, the proof uses some kind of contrapositive in the definition of uniformly continuous for the first claim. I understood too why the set A is contained in the set B. But how this relation proves that $\phi$ preserves the convergences?


Suppose $f_n \to f$ uniformly. Let $\epsilon >0$ and choose $\delta >0$ such that $|\phi(x)-\phi(y)| <\epsilon$ whenever $|x-y| <\delta$. Now $|phi(f_n(x))-\phi(f(x))| <\epsilon$ if $|f_n(x)-f(x)| <\delta$. We can choose $N$ such that $|f_n(x)-f(x)| <\delta$ for $n >N$ (for all $x$). This proves that $\phi\circ f \to \phi \circ f$ uniformly.

Suppose $f_n \to f$ in measure. Then $\mu(\phi(f_n(x))-\phi(f(x))| >\epsilon)\leq \mu(|f_n(x)-f(x)| >\delta) \to 0$ as $ n \to \infty$.

I will let you handle almost uniform convergence.


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