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 $\dots < m_1 < m_0 < m_1 <m_2 <\dots$ for which $m_{i+1} - m_i < 1/2k$. Consider the set
$E_i = \{x \in (\alpha, \beta) | m_i < f(x) < m_{i+2}\}$
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.
