Let Q be an n-rectangle, and let $f: Q \rightarrow \mathbb{R}$ be integrable. Suppose $g: Q \rightarrow \mathbb{R}$ is a bounded function such that $U(f,P) \geq U(g,P)$ and $L(f,P) \leq L(g,P)$ for every partition P. Show g is integrable and that $\int_Q g=\int_Q f$.
Theorem: Let Q be a rectangle, and let $f: Q \rightarrow \mathbb{R}$ be a bounded function. Then $\underline{\int_Q} f \leq \overline{\int_Q}f$; equality holds if and only if ggiven $\epsilon>0$, $\exists$ a corresponding partition P of Q for which $U(f,P)-L(f,P)<\epsilon$.
For g to be integrable, $U(g,P)=L(g,P)$. Since f is integrable, that means $U(f,P)=L(f,P)$. Then I feel it is very obvious in this case. Can I argue like that?
Best Answer
You can also argue directly without the equivalent $\epsilon$-criterion; it's just a matter of carefully applying the definitions of $\sup$ and $\inf$. Let $P$ be any partition. Then, we have \begin{align} \overline{\int_Q}g \leq U(g,P) \leq U(f,P) \end{align} (first inequality is obvious while second inequality is by hypothesis). This means $\overline{\int_Q}g$ is a lower bound to the set $\{U(f,P)| \, \, \text{$P$ is a partition of the rectangle $Q$}\}$. Hence, by definition of the upper integral (which is defined as the infimum of this set), we have \begin{align} \overline{\int_Q}g\leq \overline{\int_Q}f\tag{$*$} \end{align}
If you reason similarly with the lower sums, we'll find that \begin{align} \underline{\int_Q}f \leq \underline{\int_Q}g\tag{$**$} \end{align} So, we have \begin{align} \underline{\int_Q}f \leq \underline{\int_Q}g \leq \overline{\int_Q}g\leq \overline{\int_Q}f \end{align} (the middle inequality is always true for any bounded function). Finally, since $f$ is assumed to be integrable, we have $\int_Qf := \underline{\int_Q}f = \overline{\int_Q}f$. Thus, the inequality above shows that the upper and lower integrals of $g$ agree and are equal to $\int_Qf$. Hence, $g$ is also integrable with $\int_Qg = \int_Qf$.