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Let $f$ be a function of bounded variation on $[a, b]$ and $T_{a}^{b}(f)$ its total variation. We do not assume that $f$ is continuous. Show that $$\int_{a}^{b}|f'(t)|\, dt \leq T_{a}^{b}(f).$$

I know that if we assume that $f$ is continuous, then the above equation is true because we have the ability to use the Mean Value Theorem. What can I do if we don't assume $f$ is continuous?

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There are some comments here that may help. –  t.b. Mar 31 '12 at 14:02
    
If $f^\prime$ exists, then $f$ is continuous. –  Patrick Mar 31 '12 at 14:13
    
But since $f$ is of bounded variation, $f'$ exists almost everywhere and so $f$ is continuous almost everywhere. But I need $f$ to be continuous everywhere to use the MVT. –  938049 Mar 31 '12 at 14:16

1 Answer 1

Since $f$ is of bounded variation, we can write $f = g -h$ where $g$ and $h$ are monotone increasing and $f' = g' -h'$ a.e. and $$|f'| \leq |g'| + |h'| = g' + h'.$$ So

$$ \int_a^b |f'| \leqslant\int_a^b g' + \int_a^b h' \leqslant \ldots $$

Can you take it from here?

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What I would want to do is: $\int_{a}^{b}g' \leq g(b) - g(a)$ and similarly for $h$. Then $\int_{a}^{b}|f'| \leq g(b) - g(a) + h(b) - h(a)$. But then this would require $g$ and $h$ to be also continuous, which isn't necessarily true. Hmmm...another thought, we can decompose $g = g_{A} + g_{S}$ where $g_{S}' = 0$ a.e. and $g_{A}$ is absolutely continuous and increasing. Then $\int_{a}^{b}g' = \int_{a}^{b}g_{A}' \leq g_{A}(b) - g_{A}(a)$ –  938049 Mar 31 '12 at 14:54
    
@938049: the first part of what you wrote is true for increasing real-valued functions on $[a,b]$ that are differentiable a.e. So it works. –  Nana Mar 31 '12 at 15:01

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