Q:

Given a Riemann Integrable function $F(x) = \int_0^x f $ . Prove F is continuous

Let $f :\mathbb{R} → \mathbb{R}$ be Riemann integrable on every interval $[a,b]$ and define $F :\mathbb{R} → \mathbb{R}$ by
$F(x) = \int_0^x f$ . Prove that F is continuous.
My thought process behind this proof is that to be Riemann integrable we are assuming f to be bounded, which implies continuity. 
Then the fundamental theorem of calculus states: Let $f : [a, b] → \mathbb{R} $ be continuous and define $ F : [a, b] → \mathbb{R}$ by $F(x)= \int_a^x f $. Then $F$ is differentiable. 
So as we have f a continuous function we can apply the fundamental theorem of calculus. This gives F is differentiable and differentiable functions are continuous so F is continuous. 
Is this correct or have I made wrong assumptions at all? 

A:

Yes, a Riemann integrable function is bounded over closed intervalls. Now consider
$F(x+h)-F(x)=\int_{x}^{x+h}f(y)dy $
We have $|F(x+h)-F(x)|\leq |h|M$ for some $M$ since on $[x,x+h]$, $f$ is bounded. Clearly $F$ is continuous.

