The statement of exercise 1.2.5 is:
Theorem 1.2.2 can be generalized to any bounded increasing function. More generally, let $f$ be the difference of two bounded increasing functions on $(-\infty, \infty)$; such a function is said to be of bounded variation there. Define its purely discontinuous and continuous parts and prove the corresponding decomposition theorem.
Theorem 1.2.2 as follows:
Theorem 1.2.2
Let $F$ be a distribution function. Suppose that there exist a continuous function $G_c$ and a function $G_d$ of the form
$$G_d(x) = \sum_{j}b'_j\delta_{a'_j}(x)$$
[where {$a'_j$} is a countable set of real numbers and $\sum_{j}|b'_j|<\infty$], such that
$$F=G_c+G_d,$$
then
$$G_c=F_c,\; G_d=F_d$$
Here, $F_d$ is a purely jump(discontinuity) part of $F$.
My idea is:
1. Put $f=f_1-f_2$, where $f_1,f_2$ are two bounded increasing function.
2. Let {$a_j$}($j=1,2,3,...$) be a set of discontinuity of $f_1$ and {$a'_j$} that of $f_2$. Let {$b_j$},{$b'_j$} be the jumpsize of $f_1,f_2$, respectively. Pick a point $a \in (-\infty,\infty)$ where $f$ is continuous.(Indeed, such point exists since the set of discontinuity of bounded variation function is at most countable)
3. Define $f_{1d}$:
$$f_{1d}(x)\begin{cases} f_{1}(a)&\text{if }x=a\\
f_1(a)+\sum_{a<a_j<x}b_j +f_{1}(x)-f_{1}(x-)&\text{if }x>a\\
f_1(a)-\sum_{x<a_j<a}b_j -f_{1}(x+)+f_{1}(x)&\text{if }x>a
\end{cases}$$
That is, $f_{1d}$ is purely jump part of $f_1$. Similarly, define $f_{2d}$.
4. Define $f_{1c} = f_1-f_{1d},\;f_{2c} = f_2-f_{2d}$.(then $f_{1c}, f_{2c}$ are continuous.)
5. Define $f_d = f_{1d}-f_{2d}$ and $f_c=f-f_d$.
Then I must show that $f_d$ is purely discontinuous part of $f$, which means that
$$\text{the set of discontinuity of }f = \text{the set of discontinuity of }f_d$$
I have a trouble with this point, because it cannot be ensured that function of the difference of two discontinuous function at some point is discontinuous at the same point. Anyone help to comment what I miss and any help would be appreciated. Thanks in advance.