Ex2-5

Question: 
Suppose f is finite on [a,b]and of bounded variation on every interval [a+\epsilon ,b] , \epsilon >0, with V[f; a+\epsilon , b ] \leq M < +\infty
. Show that V[f; a , b ] < +\infty
. Is V[f; a , b ] \leq M
? If not, what additional assumption will make it so? 
Solution part 1: (Show that V[f; a , b ] < +\infty
. Is V[f; a , b ] \leq M
) 
Note f is finite of [a,b], this implies that f(a)
 is finite. 
By the definition of variation, consider any partition of [a,b] \Gamma = \{x_0,x_1,...,x_n\} . 
\epsilon >0 can be chosen arbitrary small, implies we can always choose \epsilon = x_1 - x_0 . which equivalence to have a_1=x_1. 
define a new partition on [a+\epsilon ,b] , \Gamma '= \Gamma - \{x_0 \} = \{x_1,x_2,...,x_n\} . 
\begin{align}
V[f;a,b]&=\sup_\Gamma \sum ^n_{j=1} |f(x_j)-f(x_{j-1})|\\
&= \sup_\Gamma |f(x_1)-f(x_0)| + \sup_\Gamma \sum ^n_{j=2} |f(x_j)-f(x_{j-1})|\\
&= \sup_\Gamma |f(x_1)-f(x_0)| + \sup_{\Gamma'} \sum ^n_{j=2} |f(x_j)-f(x_{j-1})|\\
&= \sup_\Gamma |f(x_1)-f(x_0)| + V[f; a+\epsilon , b ]
\end{align}
Recall that f is finite function on [a,b], with bounded variation on [a-\epsilon ,b] . This shows the sum is finite
We can now conclude that f has bounded variation on [a,b], V[f; a , b ] < +\infty
.
Solution part 2: (Is V[f; a , b ] \leq M
? If not, what additional assumption will make it so?)
Is V[f; a , b ] \leq M
? Generally no. From above we have equation V[f;a , b ] = \sup_\Gamma |f(x_1)-f(x_0)| + V[f; a+\epsilon , b].
It is only possible when \sup_\Gamma |f(x_1)-f(x_0)|=0 , equivalence saying as x_1 get closer to x_0, then f(x_1) get closer to f(x_0).
This happens when f is continuous around x_0=a from positive direction. And this is the additional condition required.