If $f$ is of bounded variation on $[a+\epsilon, b]$, does it imply $f$ is of bounded variation on $[a,b]$? The problem goes like:
Suppose that $f\in B[a,b]$. If $V^b_{a+\epsilon}f\leq M$ for all $\epsilon >0$, does it follow that f is of bounded variation on $[a,b]$?
I think the answer is yes. Since $V^b_{a+\epsilon}f$ is uniformly bounded and f is also bounded, also since $\epsilon$ is arbitrary, we have f is of bounded variation of $[a,b]$. Is this correct?
 A: Indeed, if $a$ is a point of a given partition, then it is sufficient to take $\epsilon$ sufficiently small to make sure that the next point is already in $[a+\epsilon,b]$. Boundedness of $f$ takes care of the first interval.
A: Since the other answer is more a wave of the hand than an answer one would submit to 
your instructor I thought a proper, economical, write-up might be welcome.   (Not that there is anything wrong with "hand-waving"
we all do it:  the intention is "Here is the idea -- You write it up yourself now.")
Also the other
 poster was less than pleased by the discussion and asked (one presumes) for details.

Problem.  Let $f:[a,b]\to \mathbb R$ have the property that the variation $V_c^b f \leq M$ for all $a<c<b$.  Show that $f$ has bounded
  variation on $[a,b]$ and that $$V_a^b f \leq M + |f(a+)-f(a)|.$$ In
  particular, if $f$ is continuous on the right at $a$, then  $V_a^b f
  \leq M$.

First observe that, for any $a<x<b$ 
$$ |f(b) -f(x)| \leq V_x^b f  \leq M.$$
Hence $|f|$ is bounded on $(a,b]$ by $M+|f(b)|$.
Now take any  subdivision
$a=a_0<a_1  <a_2< \dots < a_n=b$ and 
any $a<t<a_1$.  Observe that
$$ \sum_{i=1}^n |f(a_{i+1})-f(a_i)| \leq 
 |f(t)-f(a)| +  |f(a_1)-f(t)| + \dots +|f(a_n)-f(a_{n-1})|
$$
$$
  \leq  |f(t)-f(a)|  + M  .$$
This is true for all such $t$ and $f$ is bounded, so
$$ \sum_{i=1}^n |f(a_{i+1})-f(a_i)| \leq \liminf_{t\to a+}
|f(t)-f(a)|  + M< \infty.$$
Since this is true for all such subdivisions of $[a,b]$ 
we have 
$$ V_a^b f  \leq \liminf_{t\to a+}
|f(t)-f(a)|  + M$$
and so $f$ has bounded variation on $[a,b]$.
All functions of bounded variation are regulated (i.e. they have finite right and left hand limits at each point) and
so 
$$\liminf_{t\to a+}
|f(t)-f(a)| = \lim_{t\to a+}
|f(t)-f(a)| = |f(a+)-f(a)|$$
completing the proof.
