Proving a left-hand limit exists 
Let $f$ be a bounded and continuous function on $(0, 1)$. Also,
  suppose $f$ has anti-derivative $F$ on $(0, 1)$. Prove that the
  quantity
$$\lim_{x\to 1-} F(x) $$
exists.

I know that when you're showing a limit exists, you need to show that the left-hand limit and right-hand limits are equal; however, I'm not sure about how to approach this problem. If someone can help me out, that would be great.
 A: We have that
$$F(x)=F(0)+\int_0^{x}f(t) dt$$
and $$\lim_{x\to 1^-} F(x)=F(0)+\int_0^{1}f(t) dt$$
which exists since $f$ is countinuos and bounded.
Refer also to the related


*

*Is any continuous bounded function on $(a,b)$ Riemann integrable?
A: To say that $f(x)$ is a bounded and continuous function on $(0,1)$ implies that it is continuous almost everywhere on $[0,1]$, so we can apply the Lebesgue's criterion for Riemann integrability to say that the integral 
$$
\int_0^1 f(t)dt= \lim _{x\to 1^-} \int_0^x f(t)dt
$$ 
exists.
A: Since $f$ is bounded and continuous on $(0,1)$ the Riemann integral $\int_{0}^{1}f(x)\,dx$ exists. By Fundamental Theorem of Calculus the function $G:[0, 1]\to\mathbb {R} $ defined by $$G(x) =\int_{0}^{x}f(t)\,dt$$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with $G'(x) =f(x) $ for $x\in(0,1)$. 
Further it is given that $F'(x) =f(x) $. Thus derivative of $F-G$ vanishes on $(0,1)$ so that $F(x) - G(x) =k$ for some constant $k$ and all $x\in(0,1)$. From the equation $F(x) =G(x) +k$ it is clear that $\lim_{x\to 1^{-}}F(x)=G(1)+k$ exists (as $G$ is continuous at $1$). Similarly $\lim_{x\to 0^{+}}F(x)=G(0)+k$ also exists. 
A: We don't need continuity of $f.$ If $F$ is differentiable on $(0,1)$ and $F'$ is bounded there, then $F$ is uniformly continuous on $(0,1).$ (The MVT in fact shows $F$ is Lipschitz.)  Now it is well known that a uniformly continuous function on $(0,1)$ extends to be continuous on $[0,1].$ Thus the limit in question exists.
