# Darboux sums elementary question - am I correct

I'm very new to this material and I would like someone more experienced to give input if possible.

$$Q \subset \mathbb R^n$$ is a box and $$f: Q \to \mathbb R$$ is a function. Let $$\Xi_Q$$ be the set of all sub-boxes of $$Q$$, and $$F: \Xi_Q \to \mathbb R$$ be another function.

Suppose the following 2 statements hold:

1) For any $$B \in \Xi_Q$$ we have: $$v(B)\cdot \inf_{x \in B}f(x) \leq F(B) \leq v(B) \cdot \sup_{x \in B}f(B)$$.

2) For any $$B_1,B_2, \dots, B_k \in \Xi_Q$$ which are disjoint in pairs we have: $$F(B_1\cup \dots \cup B_k) = \sum_{i=1}^{k}F(B_i)$$

Show that $$\int_{\overline{B}} f \leq F(B)\leq\int^{\overline{}}_{B}f$$ For all $$B \in \Xi_Q$$.

Clarification: $$\int_{\overline{B}}f = \sup_{\Pi}\underline{S}(f,\Pi)$$ and $$\int_{B}^{\overline{}}f=\inf_{\Pi}\overline{S}(f,\Pi)$$ where $$\Pi$$ is a partition of the box $$B$$ and $$\underline{S},\overline{S}$$ are lower and upper Darboux sums respectively.

What I did:

$$\int_{\overline{B}}f = \sup_{\Pi}\underline{S}(f,\Pi) = \sup_\Pi\{\sum_{R \in \Pi}\inf_{x \in R}f(x)\cdot v(R)\} \leq \sup_\Pi\{\sum_{R \in \Pi}F(R)\} = \sup_\Pi\{F(B)\} = F(B)$$

The inequality comes from statement number $$1$$, as $$R \in \Xi_Q$$ if I understand correctly. The equality after it comes from the fact that $$\Pi$$ is a partition of $$B$$, so all $$R \in \Pi$$ are disjoint, and their union is $$B$$, so from statement number $$2$$ the equality follows.

Now for the other direction, a very similar idea:

$$\int^{\overline{}}_{B}f = \inf_{\Pi}\overline{S}(f,\Pi) =\inf_\Pi\{\sum_{R \in \Pi}\sup_{x \in R}f(x)\cdot v(R)\} \geq \inf_\Pi\{\sum_{R \in \Pi}F(R)\} = \inf_\Pi\{F(B)\} = F(B)$$

Which proves what we wanted to show.

Is this the correct method? I know it's simple but I'm very new to this material and I missed the lecture sadly.