real analysis - Pointwise limit of continuous functions is 1) measurable and 2) pointwise discontinuous

I'm looking for the full proof of some classic theorem stated in 1 of my textbooks with only a sketch proof.

For Baire class 1 functions - pointwise limit of continuous functions: How to show that 1) they are measurable and 2) they can be discontinuous, but the set of discontinuous points is 'small'. In particular, a function f is the limit of continuous functions if and only if it is pointwise discontinuous (the set of discontinuous points is of first category / countable union of nowhere dense sets).

Sketch of Lebesgue's proof (1904) - from "A Radical Approach to Lebesgue's Theory of Integration" by David Bressoud:

Let f be the limit of continuous functions, $f_n \rightarrow f$, on $[a, b]$. Let $P_k$ denote the set of points at which the oscillation of $f$ is greater than or equal to $1/k$. If we can show that each $P_k$ is nowhere dense, then $f$ is pointwise discontinuous. We take any open interval $(\alpha, \beta) \subset [a, b]$ and partition the entire $y$-axis from $-\infty$ to $\infty$ using points $\dots < m_1 < m_0 <
m_1

$E_i = \{x \in (\alpha, \beta) | m_i < f(x) < m_{i+2}\}$

We have that

$(\alpha, \beta)= \bigcup_{i=-\infty}^{\infty}E_i$ and $x_1, x_2 \in E_ i \Rightarrow |f(x_1) - f(x_2)| < m_{i+2} - m_i < \frac{1}{k}$

The oscillation of $f$ on $E_i$ is less than $1/k$.

Lebesgue begins by using the fact that $f$ is the limit of continuous functions to prove that each $E_i$ is a countable union of closed sets (this will lead to $f$ being measurable). In fact, he does more than this. He proves that $f$ is a limit of continuous functions if and only if, for each $k \in \mathbb{N}$, the domain can be represented as a countable union of closed sets so that the oscillation of $f$ on each set is strictly less than $1/k$.

He next proves that given any set $E$ that is a countable union of closed sets, we can construct a function for which the points of discontinuity are precisely the points of $E$. Let $\phi_i$ be a function on $(\alpha, \beta)$ for which the points of discontinuity are precisely the points in $E_i$. Could all of the functions $\phi_i$, $-\infty < i < \infty$, be pointwise discontinuous? If they were, then there would be a point in $(\alpha, \beta)$, call it $c$, where all of them are continuous. But $f(c) \in E_j$ for some $j$, and that means that $j$ is not continuous at $c$, a contradiction. At least one of the $\phi_i$ must be totally discontinuous.

If $\phi_j$ is totally discontinuous on $(\alpha, \beta)$, then there is an open subinterval of $(\alpha, \beta)$ for which $j$ is discontinuous at every point of this subinterval. By the way we defined $j$, the set $E_j$ contains an open subinterval of $(\alpha, \beta)$. From the definition of $E_j$, the oscillation is less than $1/k$ at every point in this subinterval. We have shown that $P_k$ is nowhere dense, and, therefore, $f$ is pointwise discontinuous.

In the other direction, if $f$ is not the limit of continuous functions, then there is some $k$ for which the domain cannot be expressed as a countable union of closed sets with oscillation strictly less than $1/k$ on each set. Lebesgue uses this to find an open interval contained in $P_k$. The function $f$ must be totally discontinuous.

Answer

Since continuous functions are measurable and pointwise limits of measurable functions are measurable (most measure theory textbooks prove this, see Theorem 4.9 on page 166 of Real analysis by Bruckner, Bruckner & Thomson), Baire class 1 functions are measurable.

On page 20 of the aforementioned book it is proven that every Baire 1 function is continuous except at the points of a set of the first category.

However the converse does not hold: there is a function that is continuous except at the points of a set of the first category but is not in the Baire 1 class. One such function is the characteristic function of the set of the non-endpoints of the Cantor set.

The correct characterization of the Baire 1 class is: A function is Baire 1 if and only if every restriction of the function to any nonempty perfect set has a point of continuity.