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Here is the exercise

Let $f:[a,b]\rightarrow \mathbb{R}$ be Riemann-integrable. Prove that $f^+$, $f^-$ and $|f|$ are also Riemann-integrable, when

$$f^+=\begin{cases} f(x) & f(x)\geq 0 \\ 0 & otherwise \end{cases}$$

$$f^-=\begin{cases} -f(x) & f(x)\leq 0 \\ 0 & otherwise \end{cases}$$

This problem seems so obvious. Why wouldn't $f^+$ be integrable? Anyway, I need to prove this using this hint:

"$f$ is Riemann-integrable if with every $\epsilon>0$ there exists step functions $h\leq f\leq g $ such that $\int h -\int g <\epsilon$."

I have no idea where to start...

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  • $\begingroup$ I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+\leq f^+ \leq g^+ $ and $\int h^+ -\int g^+ <\epsilon$. $\endgroup$ Jan 25, 2019 at 15:51
  • $\begingroup$ Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $h\leq f\leq g$. Now it's obvious that $h^+\leq f^+\leq g^+$ and as $f$ is Riemann-integrable, then $\int h^+ -\int g^+<\epsilon$ thus making $f^+$ also Riemann-integrable. $\endgroup$
    – jte
    Jan 25, 2019 at 16:22
  • $\begingroup$ @Matematleta Thanks I'll try to understand that. $\endgroup$
    – jte
    Jan 25, 2019 at 16:22

1 Answer 1

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We prove the Riemann integrability of $f^+$. A similar proof can be done for $f^-$.

As $f$ is Riemann-integrable, for all $\epsilon>0$ there exists step functions $h \leq f\leq g $ such that $\int h -\int g <\epsilon$.

Now define $h^+ = \max(h,0)$ and $g^+ = \max(g,0)$. You can verify that:

  1. $h^+, g^+$ are step functions.
  2. You have $h^+ \le f^+ \le g^+$ for all $x \in [a,b]$.
  3. And also $g^+ - h^+ \le g-h$ for all $x \in [a,b]$. This implies $0 \le \int (g^+-h^+) \le \int (g-h) \le \epsilon$

And concludes the proof.

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  • $\begingroup$ I guess this is as simple as it gets as I immediately understood it. Thanks. $\endgroup$
    – jte
    Jan 25, 2019 at 16:26

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