# Proof that a uniform continuous function preserves uniform convergence of sequence of measurable functions

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 $$$$A = \{s \in S: |\phi \circ f_n(s) - \phi \circ f(s)| \geq \epsilon \} \subseteq \{s \in S: |f_n(s) - f(s)| \geq \delta(\epsilon) \} =B$$$$ It follows readily that $$\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 $$n \to \infty$$.