One inequality involving Total variation function

Question

If $F$ is of bounded variation in $[a,b]$, then I need to prove that $$ \int_{a}^{b}|F'(x)| dx \leq T_F(a,b)$$

If $F'$ were Riemann integrable then it was easy to prove (in fact we can prove equality). But, how to proceed here ?

Answer 1

If $F'$ fails to be Riemann integrable you can use the Lebesgue integral instead. Start with the special assumption that $F$ is nondecreasing and bounded. Then, in particular, $T_F = F(b) - F(a)$. Extend the function $F$ to the half-line by defining $F(x) = F(b)$ if $x > b$. Since $F$ is differentiable almost everywhere (with nonnegative derivative) in $(a,b)$, \begin{align*}\int_a^b |F'(x)| \, dx = \int_a^b F'(x) \, dx &= \int_a^b \lim_{n \to \infty} n \left(F (x + 1/n) - F(x) \right) \, dx \\ &\le \liminf_{n \to \infty} \int_a^b n \left(F (x + 1/n) - F(x) \right) \, dx\end{align*} by an application of Fatou's lemma. Then \begin{align*} \int_a^b n \left(F (x + 1/n) - F(x) \right) \, dx &= n \int_a^b F(x+1/n) \, dx - n \int_a^b F(x) \, dx \\ &= n \int_{a+1/n}^{b+1/n} F(x) \, dx - n \int_a^b F(x) \, dx \\ &= n \int_b^{b+1/n} F(x) \, dx - n \int_a^{a+1/n} F(x) \, dx.\end{align*} Now, if $x \in [b,b+1/n]$ then $F(x) = F(b)$, and if $x \in [a,1+a/n]$ then $F(x) \ge F(a)$. Thus $$n \int_b^{b+1/n} F(x) \, dx - n \int_a^{a+1/n} F(x) \, dx \le F(b) - F(a).$$ (Note: if you are a bit more careful you can do a little better and obtain the upper bound $F(b^-) - F(a^+)$). Anyway, this is independent of $n$, so we end up with $$\int_a^b |F'(x)| \, dx \le T_F.$$

What if $F$ fails to be nondecreasing? Define $V(x) = T_F[a,x]$ to be the total variation of $F$ on $[a,x]$, $a \le x \le b$. Then $V$ is nondecreasing, so $$ \int_a^b |V'(x)| \, dx \le T_V.$$ Note that for any two points $x,y \in [a,b]$ you have $ |f(x) - f(y)| \le |V(x) - V(y)|$ so that $|F'(x)| \le |V'(x)|$ whenever both derivatives exist. Moreover, since $T_V = V(b) - V(a) = V(b) = T_F$ you havefinally $$\int_a^b |F'(x)| \, dx \le \int_a^b |V'(x)| \, dx = T_V = T_F.$$