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$\newcommand{\sgn}{\operatorname{sgn}}$

$\sgn(x)$ is the Signum-function, $F$ is an antiderivative of $f$ and $S(x) := F(x) \cdot \sgn(f(x))$

$$ \int \left|f(x)\right| \, dx = S(x) + \left(\sum\limits_{p=1}^{q}\sgn(x-z_p) \lim_{x \to z_p-}S(x)\right) $$

$ z_1, z_2, \ldots, z_q$ are the real roots of $f$

How can I prove that

$ S(x) + \left(\sum\limits_{p=1}^{q}\sgn(x-z_p) \lim_{x \to z_p-}S(x)\right) $ is a continuous function, if $f$ is an continuous function?

Regards

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I suggest you add a more specific tag such as real-analysis, and a summary of what you have done so far. –  Hew Wolff Nov 12 '12 at 4:54
    
I haven't done anything so far, because I have no idea how I could prove it. I have just an example. $$ \int \left|x^2-1\right| \, dx$$ $$ = \left(\frac{x^3}{3}-x\right) \cdot sgn\left(x^2-1\right) + \frac{2}{3} \cdot sgn(x+1) + \frac{2}{3} \cdot sgn(x-1)$$ $$ \left(\frac{x^3}{3}-x\right) \cdot sgn\left(x^2-1\right) + \frac{2}{3} \cdot sgn(x+1) + \frac{2}{3} \cdot sgn(x-1)$$ is continuous, $ \left(\frac{x^3}{3}-x\right) \cdot sgn\left(x^2-1\right)$ is not. –  Gunnar Nov 12 '12 at 5:06

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