Baire 1 functions are closed under uniform convergence

Question

May I know what does it mean by "functions are closed" under uniform convergence?

Also, how do I then prove that Baire-1 functions are closed under uniform convergence using the epsilon-delta definition of Baire-1 functions?

Answer 1

So uniform convergence, $f_n \twoheadrightarrow f$ , means that for all $\epsilon>0$ there exists $N_\epsilon$, such that if $n>N_\epsilon$, then $\sup_{x\in\mathbb{R}}\vert f_n(x)-f(x)\vert <\epsilon$.

What you want to show is that $f$ is Baire-1. So you need to find a function $\delta'_\epsilon:\mathbb{R}\to (0,\infty)$ such that if $\vert x-y \vert< \min\{ \delta'_\epsilon(x),\delta'_\epsilon(y) \}$, then $\vert f(x)-f(y)\vert<\epsilon$.

Every $f_n$ is Baire-$1$, so has such a $\delta^{(n)}_{ \frac{ \epsilon}{3} }$ function. By the triangle inequality, you obtain

$$ \vert f(x)-f(y) \vert\leq \vert f(x)-f_n(x) \vert+ \vert f_n(x)-f_n(y) \vert +\vert f_n(y) -f(y) \vert $$

for all $n$. The two outer most terms on the RHS are both smaller than $\frac{\epsilon} {3}$ when $n>N_{\frac{\epsilon}{3}}$ by uniform convergence. The middle term of the RHS, when $n=N_{\frac{\epsilon}{3}}+1$, is smaller than $\frac{\epsilon}{3}$ when $\vert x-y\vert<\min\Big\{ \delta^{(N_{\frac{\epsilon}{3}}+1)}_{\frac{\epsilon}{3}}(x),\delta^{(N_{\frac{\epsilon}{3}}+1)}_{\frac{\epsilon}{3}}(y) \Big\}$, since $f_{N_{\frac{\epsilon}{3}}+1}$ is a Baire-$1$ function.

So using the function $\delta'_\epsilon=\delta^{(N_{\frac{\epsilon}{3}}+1)}_{\frac{\epsilon}{3}}$, you get that $f$ is a Baire-$1$ function.

This is considered easy since bounding the difference from above by $k$ terms, and then bounding each of the $k$-terms by $\frac{\epsilon}{k}$ is used alot when showing uniform convergence preserve certain properties.

Edit: I just looked up the paper, and saw that this is literally Theorem 4 in it.