Let $Q$ be a rectangle in $\mathbb{R}^n$; let $f:Q\to \mathbb{R}$; assume $f$ is integrable over $Q$.

Show that if $f(x) \geq 0$ for $x \in Q$ then $\int_Q f \geq 0$.

This seems easy to me because obviously we have that $\underline{\int}_Q f = \sup_P \left\{L(f,P)\right\} = \overline{\int_Q} f \geq 0$ since $f$ is integrable on $Q$ and always greater than $0$. However I think the proof should be more involved.


As you said, it is quite straighforward.

Suppose $f(x) \neq 0$ for all $x \in Q$. If $P$ is any partition of $Q$ and $R$ is a subrectangle determined by $P$, then $m_R(f) \geq 0$, so that $\sum_Rm_r(f)V(R) =L(f,P)\geq 0$. Taking the supremum we get $\int_{\_}f \geq 0$. Since $\int f$ exists, $\int_{\_}f=\int f$ and hence $\int f \geq 0$


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