Simple calculation of integration The problem says:
Let $f:[a,b]\rightarrow\mathbb{R}$
  be Riemann integrable and $|f(x)|\leq M$
 . Let $F(x)=\int_{a}^{x}f(t)dt$
 . Prove that $|F(y)-F(x)|\leq M|y-x|$
 . Deduce that $F$
  is continuous.
The latter part is trivial since if F satisfies that condition, then F satisfies Lipschitz property. 
What I wonder is the first part of problem.
I solved it as follows:
$f$
  be continuous on [a,b]
  but for n-many finite point $a_{1},\cdots,a_{n}$
  in [a,b]
  with $a<a_{1}$
 . This is becase since f
  is Riemann integrable on [a,b]
 , f
  must be continuous on all the point in [a,b]
  but at most finite many points Thus, for any point $x,y\in[a,b]$
 , there are finitely many discontinuous points $b_{1},\cdots b_{m}$
  of f
  between x
  and y
 . Then $|F(y)-F(x)|=|\int_{x}^{b_{1}}f(t)dt+\int_{b_{2}}^{b_{3}}f(t)dt+\cdots+\int_{b_{m}}^{y}f(t)dt|$
  by FTC. Then $|F(y)-F(x)|=|\int_{x}^{y}f(t)dt|$
  Thus, it suffices to show that $|\int_{x}^{y}f(t)dt|\leq M|y-x|$
 . Then either $x<y$
  or $x<y$
  or $x=y$
 . 
• If $x=y$
 , then since F
  is also a function, $F(y)=F(x)$
 . Thus, $0=|F(y)-F(x)|\leq M|y-x|=M\cdot0=0$
 . 
• If $x<y$
 , then $|\int_{x}^{y}f(t)dt|=|\int_{x}^{y}f(t)dt|$
 . Then since $|f(x)|\leq M$
 , $|\int_{x}^{y}f(t)dt|\leq\int_{x}^{y}|f(t)|dt\leq\int_{x}^{y}Mdt=M|y-x|$
 .
• If $x>y$
 , then $|\int_{x}^{y}f(t)dt|=|-\int_{y}^{x}f(t)dt|$
 . Then since $|f(x)|\leq M$
 , $|\int_{x}^{y}f(t)dt|\leq\int_{y}^{x}|f(t)dt\leq\int_{y}^{x}Mdt=M|y-x|$
But I found that the fact that $f$  is Riemann integrable doesn't imply it is continuous all but finitely many points and bounded. Actually the opposit direction holds .  So, I want to know how to I use FTC to show that $|F(y)-F(x)|=|\int ^x _y$ f(t)dt. 
That is, How to use FTC if f is not continuous?

You need only the following: (assume $x\le y$)
$$\begin{split}
\left| F(y) - F(x)\right| &= \left|\int_a^y f(t)dt - \int_a^x F(t) dt\right|\\
&=\left| \int_x^y f(t)dt\right|\\
&\le  \int_x^y |f(t)| dt \\
&= M|x-y|
\end{split}$$
Note that it is not advisable to use FTC to show that. After all, before this question you do not even know if $F$ is continuous. Indeed, $F$ might not even be differentiable (If $f$ is not continuous). 
