# Riemann-integration problem

## 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...

• 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 $\int h^+ -\int g^+ <\epsilon$. – mathcounterexamples.net Jan 25 at 15:51
• 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. – jte Jan 25 at 16:22
• @Matematleta Thanks I'll try to understand that. – jte Jan 25 at 16:22

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.

• I guess this is as simple as it gets as I immediately understood it. Thanks. – jte Jan 25 at 16:26