total variation and monotonicy I am stuck with this problem : I have a finite variation function $f$ and I have proved that the total variation on the interval $[0,t]$, denoted with $S_t^f$ for $t\in[0,T]$, is increasing. How can I prove that the function $S_t^f+f(t)$ is increasing too ?

Thank you very much for any help you can provide :)
Well you are not getting much help here and perhaps the homework deadline is looming--so here is a push.

The function $f:[0,T]\to \mathbb{R}$   has bounded variation there. You already have seen, for any $0\leq a&lt;b\leq T$, that $$V(f,[a,b]) + f(b)-f(a)\geq 0$$ is trivial.

What you need to show now is that $$V(f,[0,a]) + V(f,[a,b])= V(f,[0,b]).$$
As is often the case, to show equality you show inequality--two of them.  That is you have two inequalities to establish:
$$V(f,[0,a]) + V(f,[a,b]) \leq V(f,[0,b])$$ and
$$V(f,[0,a]) + V(f,[a,b]) \geq V(f,[0,b]).$$
One is easy and the other will require you to get your hands dirty with partitions and the like.  (Think of a partition of $[0,b]$ and what would happen if you add "$a$" to it.)

[Note that I am delaying the notation $S^f_t=V(f,[0,t])$ for later since it is not very intuitive and can easily mislead you.]



Alternatively, you can just search around for textbooks that give a proof of Jordan's theorem for BV functions, that characterizes them as differences of monotone functions (which is what you are doing here).  Some use exactly this method; Royden's Real Analysis uses the clever idea of defining two half-variations (a positive variation and a negative variation) that ends up in the same place.  My favorite presentation of this topic is in Zygmund and Wheeden, Measure and Integral.  The techniques needed to solve this problem are, while elementary, absolutely essential to an ability to understand and do real analysis.  It is not a waste of time to linger on this problem until it is all completely transparent to you.
For any fixed partition $a=x_0\leq\dots\leq x_n =b$ we have
$$
  S^f(b) -S^f(a) + f(b) - f(a) \geq \sum_{i = 0}^{n-1} |f(x_{i+1}) - f(x_i)| + \sum_{i = 0}^{n-1} (f(x_{i+1}) - f(x_i)) \geq 0
$$
since $y + |y| \geq 0$ for all $y$. Hence, when you take $\sup$ over partitions, the inequality still holds.