求证：Baire 1函数序列的一致极限仍为Baire 1函数

Alright, let's work through this proof step by step. First, let's restate the key definitions clearly to make sure we're on the same page:

Baire 1 Function: A function $f:[a,b]\to\mathbb{R}$ is Baire 1 if it is the pointwise limit of a sequence of continuous real-valued functions on $[a,b]$.

Uniform Convergence: A sequence $(g_n)$ converges uniformly to $f$ on $[a,b]$ if for every $\varepsilon>0$, there exists an $N\in\mathbb{N}$ such that for all $n\geq N$ and all $x\in[a,b]$, $|g_n(x)-f(x)|<\varepsilon$.

Our goal is to show that if each $g_n$ is Baire 1, and $(g_n)$ converges uniformly to $f$, then $f$ must also be Baire 1. Here's how we do it:

Step 1: Use the Baire 1 property of each $g_n$ #

Since every $g_n$ is a Baire 1 function, by definition, for each $n\in\mathbb{N}$, there exists a sequence of continuous functions $(f_{n,k}){k=1}^\infty$ such that:
$$\lim{k\to\infty} f_{n,k}(x) = g_n(x) \quad \text{for all } x\in[a,b].$$
In plain terms, for any fixed $n$ and $x$, we can make $|f_{n,k}(x)-g_n(x)|$ as small as we want by choosing a sufficiently large $k$.

Step 2: Leverage uniform convergence to bound $|g_n(x)-f(x)|$ #

Because $(g_n)$ converges uniformly to $f$, for every $m\in\mathbb{N}$ (we'll use $1/m$ as our target $\varepsilon$), there exists some $N_m\in\mathbb{N}$ such that for all $n\geq N_m$ and all $x\in[a,b]$:
$$|g_n(x)-f(x)| < \frac{1}{2m}.$$
Let's pick $n_m = N_m$ for each $m$—this gives us a subsequence $(g_{n_m})$ where each term is uniformly close to $f$ across the entire interval:
$$|g_{n_m}(x)-f(x)| < \frac{1}{2m} \quad \text{for all } x\in[a,b].$$

Step 3: Construct the continuous sequence for $f$ #

Now, for each $m$, we use the pointwise convergence of $(f_{n_m,k})$ to $g_{n_m}$. For any fixed $x\in[a,b]$, since $\lim_{k\to\infty} f_{n_m,k}(x) = g_{n_m}(x)$, we can choose a $k_m\in\mathbb{N}$ such that:
$$|f_{n_m,k_m}(x)-g_{n_m}(x)| < \frac{1}{2m}.$$

Define the sequence of continuous functions $(h_m)$ where $h_m = f_{n_m,k_m}$. Each $h_m$ is continuous because every $f_{n,k}$ is continuous by definition.

Step 4: Prove $(h_m)$ converges pointwise to $f$ #

Take any fixed $x\in[a,b]$ and any $\varepsilon>0$. We need to show there exists an $M\in\mathbb{N}$ such that for all $m\geq M$, $|h_m(x)-f(x)| < \varepsilon$.

Choose $M$ large enough so that $1/m < \varepsilon$ for all $m\geq M$. Then for each $m\geq M$:
$$
\begin{align*}
|h_m(x)-f(x)| &= |f_{n_m,k_m}(x)-f(x)| \
&\leq |f_{n_m,k_m}(x)-g_{n_m}(x)| + |g_{n_m}(x)-f(x)| \
&< \frac{1}{2m} + \frac{1}{2m} \
&= \frac{1}{m} < \varepsilon.
\end{align*}
$$
This means $\lim_{m\to\infty} h_m(x) = f(x)$ for every $x\in[a,b]$.

Conclusion #

Since $f$ is the pointwise limit of the continuous sequence $(h_m)$, $f$ is a Baire 1 function. That's exactly what we needed to prove!

内容的提问来源于stack exchange，提问作者Idonknow