Let $f, \alpha:[a,b]\subset\mathbb{R}\longrightarrow\mathbb{R}$ such that both are discontinuous at $c\in[a,b]$. I want to show that there does not exist the Stieltjes integral
\begin{equation*} \int_a^bfd\alpha. \end{equation*}
If they are discontinuous in the same side, O.K., I got a contradiction by the Cauchy Criterion. But if they are discontinuos on opposite sides (one right and one on the left for example) I still did not get!
Somebody have any hints?
Thanks.
Hint: Given $\delta>0$, we can construct two tagget partitions $P_\delta^*$, $Q_\delta^*$ such that $|P_\delta^*|$, $|Q_\delta^*|<\delta$ and $$\left|\sum(Q^*_\delta)-\sum(P^*_\delta)\right|=|f(x_\delta)-f(c)||\alpha(c)-\alpha(y_\delta)|,$$ where $|f(x_\delta)-f(c)|\geq \varepsilon_1$ for some $\varepsilon_1>0$ and $|\alpha(c)-\alpha(y_\delta)|\geq \varepsilon_2$ for some $\varepsilon_2>0$.
Solution: Since $f$ is discontinuous at $c$, there exists $\varepsilon_1>0$ satisfying the following condition: $$\text{For all} \ \delta>0,\text{ there exists}\ x_\delta\in(c-\delta,c+\delta)\text{ such that }|f(x_\delta)-f(c)|\geq \varepsilon_1.$$ Analogously, there exists $\varepsilon_2>0$ such that $$\text{For all} \ \delta>0,\text{ there exists}\ y_\delta\in(c-\delta,c+\delta)\text{ satisfying }|\alpha(y_\delta)-\alpha(c)|\geq \varepsilon_2.$$
Given $\delta>0$, take a tagged partition $P^*=(P,\xi)$ of $[a,b]$, let's say
$$P=\{t_0<\cdots<t_{j-1}<t_{j}<\cdots<t_k\},\quad \xi=(\xi_1,...,\xi_{j-1},\xi_{j},\xi_{j+1},...,\xi_k),$$ such that $|P|<\delta$ and $c\in(t_{j-1},t_{j})$ for some $j\in\{1,...,k\}$. Without loss of generality, we can assume $x_\delta,y_\delta\in[t_{j-1},t_{j}]$.
Define a new tagged partition $Q^*=(Q,\zeta)$ by putting $$Q=\{t_0<\cdots<t_{j-1}<c<t_{j}<\cdots<t_k\},\quad \zeta=(\xi_1,...,\xi_{j-1},c_1,c_2,\xi_{j+1},...,\xi_k).$$ Notice that $$\left|\sum(Q^*)-\sum(P^*)\right|=\left|f(c_1)[\alpha(c)-\alpha(t_{j-1})]+f(c_2)[\alpha(t_{j})-\alpha(c)]-f(\xi_{j})[\alpha(t_{j})-\alpha(t_{j-1})]\right|$$
If $y_\delta<c$ and $x_\delta<c$, take $t_{j-1}=y_\delta$, $\xi_j=c_2=c$ and $c_1=x_\delta$. By the above calculation we see that this choice produces two tagged partitions $P_\delta^*$, $Q_\delta^*$ such that $$\left|\sum(Q^*_\delta)-\sum(P^*_\delta)\right|=|f(x_\delta)-f(c)||\alpha(c)-\alpha(y_\delta)|\geq \varepsilon_1\varepsilon_2$$
If $y_\delta<c$ and $x_\delta>c$, take $t_{j-1}=y_\delta$, $\xi_j=c_1=c$ and $c_2=x_\delta$. Then $$\left|\sum(Q^*_\delta)-\sum(P^*_\delta)\right|=|f(c)-f(x_\delta)||\alpha(c)-\alpha(y_\delta)|\geq \varepsilon_1\varepsilon_2$$
The other possible cases can be dealt analogously.
Thus there exists $\varepsilon=\varepsilon_1\varepsilon_2$ satisfying the following condition: $$\text{For all} \ \delta>0,\text{ there exists}\ P^*_\delta,Q_\delta^*\text{ such that } |P^*_\delta|,|Q^*_\delta|<\delta \text{ but }\left|\sum(Q^*_\delta)-\sum(P^*_\delta)\right|\geq \varepsilon.$$
So, the Cauchy Criterion is not satisfied.
PS: This solution doesn't suppose $f$ or $\alpha$ continuous at any point and thus is valid for any possible type of discontinuity at $c$ ("same side" or not).
