How to prove $\int_a^b f\,d\alpha =\int_a^b g\,d\alpha$? 
Theorem:
  Let $\alpha\in BV\left([a,b]\right),\, c\in [a,b],$ and $\,f,g:[a,b]\to\mathbb R$ be functions such that $f(x)=g(x)$ on $[a,b]-c $.
  If $f\in R(\alpha) $ on $[a,b]$ and $\alpha$ is continuous, then $g\in R(\alpha)$ on $[a,b]$ and $\int_a^b f\,d\alpha =\int_a^b g\,d\alpha$.

How could I prove the theorem??
I think an idea to prove that $g\in R(\alpha)$ on $[a,b]$  is using the definition (but using $f$). So I have 
$$\left\lvert\sum_{k=1}^nf(t_k)\left(\alpha(x_k)-\alpha(x_{k-1})\right)-\int_a^bf\,d(\alpha(x)) \right\rvert <\varepsilon.$$ 
How can I make a relation between $g$ and $f$ on the definition?
NOTE: $BV$ means bounded variation and $R(\alpha)$ means Riemann-Stieltjes integrable on $\alpha$.
 A: It suffices to show that if $h$ is zero everywhere except at one point then it is $\alpha$ integrable and $\int h=0$. 
Indeed, if we show that then $g-f \in R(a)$ and since $f\in R(a)$ then $ (g-f)+f \in R(a)$ and $\int g = \int ((g-f) +f) = \int g-f + \int f =0+\int f$.
Let $\epsilon >0$, from continuity of $\alpha$ we can find $\delta  $ such that 
$|a(x) -a(c)    |<\dfrac{\epsilon   }{2| g(c) -f(c)|}$ whenever $|x-c|<\delta$.
We will actually show that $|f-g| \in R(a)$ and $\int |f-g|=0$ and at the end we will see that this is indeed enough.
Take any partition $P$, with points $x_1,x_2,\cdots, x_n$ and mesh less than $\delta$ i.e. $|x_{i+1}-x_i|<\delta$
Now, $\text{U}(|g-f|) = \sum_{i=1}^n  M(|g-f|)_{x_i,x_{i+1}}(a(x_{i+1}) - a(x_i) )$. Notice that  $M(|g-f|)_{x_i,x_{i+1}}$ is zero when $c\not \in [x_i,x_{i+1})$ since $|g-f|$ is everywhere zero there. So assume that $c\in [x_j,x_{j+1})$ then 
$M(|g-f|)_{x_j,x_{j+1}} = | g(c) -f(c)|$.
So, 
\begin{align}
\text{U}(|g-f|)&= |g(c) - f(c)| |a(x_j) - a(x_{j+1})|\\
&\leq|g(c) - f(c)| |a(x_j) -a(c)|+|g(c) - f(c)| |a(c) - a(x_{j+1})|\\
&<\frac{\epsilon}{2} + \frac{\epsilon}{2}\\
&=\epsilon
\end{align}
Notice that $|g-f| =g-f$ if $g(c)-f(c)>0$ or $|g-f|= f-g$ otherwise. In the second case you can just use that if $g\in R(a)$ then $-g \in R(a)$ and $\int -g = -\int g$.
